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Balbharti Maharashtra State Board 11th Biology Textbook Solutions Chapter 7 Cell Division Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Biology Solutions Chapter 7 Cell Division

1. Choose correct option

Question (A)
The connecting link between Meiosis – I and Meiosis – II is ……….. .
(a) interphase – I
(b) interphase – II
(c) interkinesis – III
(d) anaphase – IV
Answer:
(c) interkinesis – III

Question (B)
Synapsis is pairing of ………………. .
(a) any two chromosomes
(b) non – homologous chromosomes
(c) sister chromatids
(d) homologous chromosomes
Answer:
(d) homologous chromosomes

Question (C)
Spindle apparatus is formed during which stage of mitosis?
(a) Prophase
(b) Metaphase
(c) Anaphase
(d) Telophase
Answer:
(b) S-phase

Question (D)
Chromosome number of a cell is almost doubled up during _______ .
(a) G₁ – phase
(b) S – phase
(c) G₂-phase
(d) G₀-phase
[Note: Due to DNA replication the DNA content of cell doubles during S-phase. But the number of chromosomes remain the same.]
Answer:
(b) S – phase

Question (E)
How many meiotic divisions are necessary for formation of 80 sperms?
(a) 80
(b) 40
(c) 20
(d) 10
Answer:
(c) 20

Question (F)
How many chromatids are present in anaphase – I of meiosis – I of a diploid cell having 20 chromosomes?
(a) 4
(b) 6
(c) 20
(d) 40
Answer:
(d) 40

Question (G)
In which of the following phase of mitosis chromosomes are arranged at equatorial plane?
(a) Prophase
(b) Metaphase
(c) Anaphase
(d) Telophase
Answer:
(b) Metaphase

Question (H)
Find incorrect statement.
(a) Condensation of chromatin material occurs in prophase.
(b) Daughter chromatids are formed in anaphase.
(c) Daughter nuclei are formed at metaphase.
(d) Nuclear membrane reappears in telophase.
Answer:
(c) Daughter nuclei are formed at metaphase.

Question (I)
Histone proteins are synthesized during
(a) G₁ phase
(b) S – phase
(c) G₂ – phase
(d) Interphase
Answer:
(b) S – phase

2. Answer the following questions

Question (A)
While observing a slide, student observed many cells with nuclei. But some of the nuclei were bigger as compared to others but their nuclear membrane was not so clear. Teacher inferred it as one of the phase in the cell division. Which phase may be inferred by teacher?
Answer:
Prophase.

Question (B)
Students prepared a slide of onion root tip. There were many cells seen under microscope. There was a cell seen under microscope. There was a cell with two groups of chromosomes at opposite ends of the cell. This cell is in which phase of mitosis?
Answer:
Anaphase.

Question (C)
Students were shown some slides of cancerous cells. Teacher made a comment as if there would have been a control at one of its cell cycle phase, there wouldn’t have been a condition like this. Which phase the teacher was referring to?
Answer:
The phase teacher was referring would be G_i phase.

Question (D)
Some Mendelian crossing experimental results were shown to the students. Teacher informed that there are two genes located on the same chromosome. He enquired if they will be ever separated from each other?
Answer:

1. Genes are located on chromosomes at specific distance and position.
2. The greater this distance, the greater the chance that a crossover can occur between the genes and the greater the chances of recombination.
3. The chances of recombination are less between the genes that are placed closed to each other on the chromosome.
4. Therefore, due to recombination the two genes located on the same chromosome have possibility of separating from each other.

Question (E)
Students were observing a film on Paramoecium. It underwent a process of reproduction. Teacher said it is due to cell division. But students objected and said that there was no disappearance of nuclear membrane and no spindle formation, how can it be cell division? Can you clarify?
Answer:

1. Paramoecium is a unicellular organism. The division in Paramoecium occurs by amitosis.
2. It is the simplest mode of cell division.
3. In amitosis, nucleus elongates and a constriction appears. This constriction deepens and divides the nucleus in two daughter nuclei followed by the division of cytoplasm.

Question (F)
Is the meiosis responsible for evolution? Justify your answer.
Answer:

1. Meiosis ensures that organisms produced by sexual reproduction contain correct number of chromosomes.
2. Meiosis exhibits genetic variation by the process of recombination.
3. Variations increase further after union of gametes during fertilization creating offspring with unique characteristics. Thus, it creates diversity of life and is responsible for evolution.

Question (G)
Why mitosis and meiosis – II are called as homotypic division?
Answer:
1. In mitosis, the chromosome number and genetic material of daughter cells remain same as that of the parent cell.
2. In meiosis – II, two haploid cells formed during first meiotic division divide further into four haploid cells. This division is identical to mitosis. The daughter cells formed in second meiotic division are similar to their parent cells with respect to the chromosome number formed in meiosis -1. Hence mitosis and meiosis – II are called homotypic division.

Question (H)
Write the significance of mitosis.
Answer:

1. As mitosis is equational division, the chromosome number is maintained constant.
2. It ensures equal distribution of the nuclear and the cytoplasmic content between the daughter cells, both quantitatively and qualitatively. Therefore, the process of mitosis also maintains the nucleo-cytoplasmic ratio.
3. The DNA is also equally distributed.
4. It helps in growth and development of organisms.
5. Old and worn-out cells are replaced through mitosis.
6. It helps in the asexual reproduction of organisms and vegetative propagation in plants.

Question (I)
Enlist the different stages of prophase – I.
Answer:
1. Prophase -I:
It is the most complicated and longest phase of meiotic division.
It is further divided into five sub-phases viz. leptotene, zygotene, pachytene, diplotene and diakinesis.

a. Leptotene:
The volume of the nucleus increases.
The chromosomes become long distinct and coiled.
They orient themselves in a specific fonn known as bouquet stage. This is characterized with the ends of chromosomes converged towards the side of nucleus where the centrosome lies. j Lep
The centriole duplicates into two and migrates to opposite poles. [Note: Centrioles divide during Gj phase of interphase.]

b. Zygotene:
Pairing of non-sister chromatids of homologous chromosomes takes place by formation of synaptonemal complex. This pairing is called synapsis.
Each pair consists of a maternal chromosome and a paternal chromosome. Chromosomal pairs are called bivalents or tetrads.

c. Pachytene:
Each individual chromosome begins to split longitudinally into two similar chromatids. Therefore, each bivalent now appears as a tetrad consisting of four chromatids.
The homologous chromosomes begin to separate but they do not separate completely and remain attached to one or more points. These points are called chiasmata (Appear like a cross-X).
Chromatids break at these points and broken segments are exchanged between non-sister chromatids of homologous chromosomes resulting in recombination.

d. Diplotene:
The chiasma becomes clearly visible in diplotene due to beginning of repulsion between synapsed homologous chromosomes. This is known as desynapsis. Synaptonemal complex also starts to disappear.

e. Diakinesis:
The chiasmata begin to move along the length of chromosomes from the centromere towards the ends of chromosomes. The displacement of chiasmata is termed as terminalization.
The terminal chiasmata exist till the metaphase.
The nucleolus and nuclear membrane completely disappear and spindle fibres begin to appear.

3. Draw labelled diagrams and write explanation

Question (A)
With the help of suitable diagram, describe the cell cycle.
Answer:
1. Series of events occurring in the life of a cell is called cell cycle. Interphase and M – phase are the two phases of cell cycle.
2. Interphase: It is the stage between two successive cell divisions. It is the longest phase of a cell cycle during which the cell is highly active and prepares itself for cell division.
The interphase is subdivided into three sub-phases as G₁ – phase, S-phase and G₂-phase.
a. G₁ – phase (First gap period/First Gap Phase):
It begins immediately after cell division.
RNA (mRNA, rRNA and tRNA) synthesis, protein synthesis and synthesis of membranes take place during this phase.
b. S – phase (Synthesis phase):
In this phase DNA is synthesized (replicated), so that amount of DNA per cell doubles.
Synthesis of histone proteins takes place in this phase.
c. G₂ – phase (Second growth phase/Second Gap Phase):
Metabolic activities essential for cell division occur during this phase.
Various proteins which are necessary for the cell division are also synthesized in this phase.
Apart from this, RNA synthesis also occurs during this phase.
In animal cells, a daughter pair of centrioles appears near the pre-existing pair.

Question (B)
Distinguish between mitosis and meiosis.
Answer:

Question (C)
Draw labelled diagrams and write explanation Draw the diagram of metaphase.
Answer:
Metaphase:
a. Chromosomes are completely condensed and appear short.
b. Centromere and sister chromatids become very prominent.
c. All the chromosomes are arranged at equatorial plane of cell. This is called metaphase plate.
d. Mitotic spindle is fully formed in this phase.
e. Centromere of each chromosome divides horizontally into two, each being associated with a chromatid. [Note: The centromeres divide at the beginning of anaphase so that the two chromatids of each chromosome become separated from each other.
Source: Cell Division, Donald B. McMillan, Richard J. Harris, in An Atlas of Comparative Vertebrate Histology, 2018.]

Question 4.
Match the following column – A with column – B

Answer:

Question 5.
Is the given figure correct? Why?

Answer:
1. The given figure is incorrect as the spindle fibres are not attached to centromere of the chromosomes.
2. During metaphase, chromosomes are attached to spindle fibres with the help of centromeres.

Question 6.
If an onion has 16 chromosomes in its leaf cell, how many chromosomes will be there in its root cell and pollen grain.
Answer:
1. The chromosomes in root cell will be 16 as root cell is a diploid cell.
2. The chromosomes in pollen grain will be 8 as pollen grain is a haploid cell.

7. Identify the following phases of mitosis and label the ‘A’ and ‘B’ given in diagrams.

Question (i)

Answer:
The diagram shown is of Metaphase.
A: Chromosomes arranged on metaphase plate

Question (ii)

Answer:
The diagram shown is of Anaphase.
B: Chromatids moving to opposite poles.

Practical / Project:

Question 1.
Fix the onion root tips at different durations of the day starting from 6am up to 9am at the intervals of half an hour. Prepare the slide of each fixed root tip and analyse the relation between time and phase of mitosis.
Answer:
Mitotic division is an equational division in which one parent cell give rise to two daughter cells with equal number of chromosomes in daughter cells and mother cell. It has four sub phases: prophase, metaphase, anaphase, telophase.

Mitosis is affected by temperature and time. Mitotic index is high in morning so the mitosis is observed clearly in the morning. (Mitotic index is defined as the ratio between the number of cells in a population undergoing mitosis to the total number of cells in a population. )
[Note: Students catt use above information for reference and perform this activity on their own.]

11th Biology Digest Chapter 7 Cell Division Intext Questions and Answers

Can you recall? (Textbook Page No. 76)

How do your wounds heal?
Answer:
a. A wound is an injury to living tissue.
b. Healing of wound take place by mitosis.
c. Repetitive mitotic divisions near the site of injury results in healing of wound.

Can you tell? (Textbook Page No. 79)

What is cell cycle?
Answer:

1. Sequential events occurring in the life of a cell is called cell cycle.
2. Interphase and M – phase are the two phases of cell cycle.
3. Cell undergoes growth or rest during interphase and divides during M – phase.

Discuss with teacher (Textbook Page No. 76)

Some cells do not have gap phase in their cell cycle whereas some cells spend maximum part of their life in gap phase. Search for such cells. Some cells are said to be in G₀ phase. What is this G₀ phase?
Answer:

1. G₀ is the phase of the cell cycle in eukaryotes in which many cell types stop dividing. It is also called a quiescent stage.
2. If cells are deprived of appropriate growth factors, they stop at the Gi checkpoint of the cell cycle. Their growth and division are arrested and they remain in G₀ phase.
3. Mature neurons and muscle cells remain in G₀ phase.

Question 5.
Can you tell? (Textbook Page No. 79)
Answer:
1. Series of events occurring in the life of a cell is called cell cycle. Interphase and M – phase are the two phases of cell cycle.
2. Interphase: It is the stage between two successive cell divisions. It is the longest phase of a cell cycle during which the cell is highly active and prepares itself for cell division.
The interphase is subdivided into three sub-phases as G₁ – phase, S-phase and G₂-phase.
a. G₁ – phase (First gap period/First Gap Phase):
It begins immediately after cell division.
RNA (mRNA, rRNA and tRNA) synthesis, protein synthesis and synthesis of membranes take place during this phase.
b. S – phase (Synthesis phase):
In this phase DNA is synthesized (replicated), so that amount of DNA per cell doubles.
Synthesis of histone proteins takes place in this phase.
c. G₂ – phase (Second growth phase/Second Gap Phase):

1. Metabolic activities essential for cell division occur during this phase.
2. Various proteins which are necessary for the cell division are also synthesized in this phase.
3. Apart from this, RNA synthesis also occurs during this phase.
4. In animal cells, a daughter pair of centrioles appears near the pre-existing pair.

Internet my friend (Textbook Page No. 77)

What is Karyogram or Karyotype?
Answer:
1. A karyotype is a representation of condensed chromosomes arranged in pairs.
2. Analysis of the karyotype of a particular individual indicates whether the individual has a normal set of chromosomes or whether there are abnormalities in number or appearance of individual chromosomes.

Can you tell? (Textbook Page No. 79)

Which are the steps of mitosis?
Answer:
Steps in mitosis are Karyokinesis and Cytokinesis. Karyokinesis includes four stages – Prophase, Metaphase, Anaphase and Telophase.

Internet my friend (Textbook Page No. 79)

How the life span of a cell is decided?
Answer:

1. Life span of different cells vary greatly.
2. Life span of a cell is decided by its growth rate, metabolic activities and cell size.
3. The life span of a cell can be analysed in laboratory by applying carbon-14 technique to DNA.
4. This method is commonly used in archaeology and paleontology to find the age of fossils. Same can be applied to determine the life span of a cell.

Do yourself (Textbook Page No. 80)

Write down the explanation of prophase I in your own words.
Answer:
1. Prophase -I:
It is the most complicated and longest phas0e of meiotic division.
It is further divided into five sub-phases viz. leptotene, zygotene, pachytene, diplotene and diakinesis.

a. Leptotene:

1. The volume of the nucleus increases.
2. The chromosomes become long distinct and coiled.
3. They orient themselves in a specific fonn known as bouquet stage. This is characterized with the ends of chromosomes converged towards the side of nucleus where the centrosome lies.
4. The centriole duplicates into two and migrates to opposite poles. [Note: Centrioles divide during G_j phase of interphase.]

b. Zygotene:

1. Pairing of non-sister chromatids of homologous chromosomes takes place by formation of synaptonemal complex. This pairing is called synapsis.
2. Each pair consists of a maternal chromosome and a paternal chromosome. Chromosomal pairs are called bivalents or tetrads.

c. Pachytene:

1. Each individual chromosome begins to split longitudinally into two similar chromatids. Therefore, each bivalent now appears as a tetrad consisting of four chromatids.
2. The homologous chromosomes begin to separate but they do not separate completely and remain attached to one or more points.
3. These points are called chiasmata (Appear like a cross-X).
4. Chromatids break at these points and broken segments are exchanged between non-sister chromatids of homologous chromosomes resulting in recombination.

d. Diplotene:
The chiasma becomes clearly visible in diplotene due to beginning of repulsion between synapsed homologous chromosomes. This is known as desynapsis. Synaptonemal complex also starts to disappear.

e. Diakinesis:

1. The chiasmata begin to move along the length of chromosomes from the centromere towards the ends of chromosomes. The displacement of chiasmata is termed as terminalization.
2. The terminal chiasmata exist till the metaphase.
3. The nucleolus and nuclear membrane completely disappear and spindle fibres begin to appear.

Curiosity Box: (Textbook Page No. 81)

(i) What is exact structure of synaptonemal complex?
Answer:
Synaptonemal complexes are zipper like structures assembled between homologous chromosomes during the prophase of first meiotic division.
[Source: ncbi.nlm. nih.gov/pubmed/8743892]

(ii) What is structure of chiasma?
Answer:
Chiasma is a X-shaped point of attachment between two non-sister chromatids of a homologous chromosomes.

(iii) Which type of proteins are involved in formation of spindle fibres?
Answer:
Spindle fibres are formed from microtubules with many accessory proteins.

(iv) Why and how spindle fibres elongate and some contract?
Answer:
a. Spindle fibres elongate for assembly of chromosomes at equatorial plane of the cell during metaphase and spindle fibres contract for pulling chromosomes towards opposite poles during anaphase.
b. The spindle fibres elongate (polymerize) by incorporating subunits of the protein tubulin and contract

(v) What is the role of centrioles in formation of spindle apparatus?
Answer:
Centriole plays an important role in cell division. Centrioles help organize microtubule assembly and forms spindle apparatus that separate the chromosomes during cell division.

Curiosity box (Textbook Page No. 81)

What would have happened in absence of meiosis?
Answer:

1.  Gametes are produced by the process of meiosis which are essential for sexual reproduction.
2. Diploid organisms have two set of chromosomes (one paternal and one maternal).
3. For a diploid organism to undergo sexual reproduction it needs to create gametes that contain only one set of chromosomes so the number of chromosomes remains same in the next generation.
4. In absence of meiosis, the chromosome number of parents and their offsprings will differ in every generation; hence no species will hold its characters.
5. Also, there will be no crossing over of homologous chromosomes. Thus, there will be no variations with respect to the changing environment in progeny to maintain their existence, which may lead to extinction of species.

Can you tell? (Textbook Page No. 82)

(i) What is the difference between mitosis and meiosis?
Answer:

(ii) What is difference between meiosis – I and meiosis – II?
Answer:

(iii) Elaborate the process of recombination.
Answer:
a. Recombination is exchange of genetic material between paternal and maternal chromosomes during gamete formation.
b. The points where crossing over takes place is known as chiasmata.
c. Chromatids acquire new combinations of alleles by physically exchanging segments in crossing-over.
d. The exchange of genetic material between homologous chromosomes involves accurate breakage and joining of DNA molecules through a complex mechanism.
e. It is catalyzed by enzymes.

Do Yourself (Textbook Page No. 82)

Prepare a concept map on cell division in following box.
Answer:
Refer Quick Review

Internet My Friend (Textbook Page No. 82)

Different types of proteins like cyclins, maturation promoting factor (MPF), cyclosomes, enzymes like cyclin dependent kinases (CDK) play important role in control of cell cycle. Collect more information about these proteins and enzymes from internet, prepare a power-point presentation and present it in the class.
Answer:

1. The regulation of the cell cycle involves an internal control system consisting of proteins called cyclins and enzymes called cyclin-dependent kinases.
2. A Cdk is a protein kinase. When the kinase of the Cdk is activated upon binding to a cyclin, it phosphorylates target proteins in the cell, regulating their activities.
3. Those proteins play important roles in initiating or regulating significant events of the cell cycle, such as DNA replication, mitosis, and cytokinesis.
4. Maturation Promoting Factor (MPF) triggers the cell’s passage into the mitotic phase.
   [Note: Students are expected to perform the above activity by their own with the help of information provided in the answer.]

Balbharti Maharashtra State Board Bookkeeping and Accountancy 11th Solutions Chapter 4 Ledger Textbook Exercise Questions and Answers.

Maharashtra State Board Bookkeeping and Accountancy 11th Solutions Chapter 4 Ledger

1. Answer in one sentence only.

Question 1.
What is Ledger?
Answer:
Ledger is an important book of account in which individual records business transactions with respect to persons, properties, expenses, or losses are maintained.

Question 2.
What is ledger posting?
Answer:
Transferring the entry passed in the journal to the ledger for the individual record is called ledger posting.

Question 3.
When does an account show a nil balance?
Answer:
When the total credit side of an account equals the total of the debit side, such an account shows a nil balance.

Question 4.
What is Folio?
Answer:
Page number of the account opened in the ledger is called Ledger folio.

Question 5.
Where is the statement form of ledger A/c is used in actual practice?
Answer:
The statement form of ledger A/c is used in the banks and financial institutions to prepare the client’s account showing balances of accounts after each transaction is complete.

Question 6.
Why Proprietor’s Capital account is a liability for the business?
Answer:
Capital invested in the business by the proprietor is an asset for the proprietor and liability for the business.

Question 7.
Why does a cash account never shows a credit balance?
Answer:
Available cash with the business is an asset of the business and the account of every asset shows debit, cash account always shows debit balance.

Question 8.
What is ‘Trial Balance’?
Answer:
Trial Balance is an abstract or list of all the ledger accounts as on a specified date showing debit total and the credit total of all the accounts or their net balance.

2. Write the word, term, phrase, which can substitute each of the statements.

Question 1.
Principal Book of accounts.
Answer:
Ledger

Question 2.
Transferring a journal entry to the appropriate accounts in the Ledger.
Answer:
Posting

Question 3.
Page number of Ledger to which an entry is posted.
Answer:
Ledger folio

Question 4.
The process of extracting the balance and inserting it on the lesser side of an account.
Answer:
Balancing

Question 5.
A debit balance to Personal Accounts.
Answer:
Debtor

Question 6.
A credit balance to Bank Account.
Answer:
Bank overdraft

Question 7.
An account to be debited for goods damaged by fire.
Answer:
Loss by fire

Question 8.
A Trial Balance in which only net balances of all ledger accounts are transferred.
Answer:
Net Trial Balance

3. Select appropriate alternatives from those given below and rewrite the sentences.

Question 1.
In case of a credit transaction one of the account must be a ______________ account.
(a) Cash
(b) Credit
(c) Personal
(d) Debit
Answer:
(c) Personal

Question 2.
‘c/d’ indicates ______________ balance.
(a) Opening
(b) Closing
(c) Positive
(d) Negative
Answer:
(b) Closing

Question 3.
______________ Column of ledger is used for writing page number of Journal.
(a) J.F.
(b) L.F.
(c) Date
(d) Particulars
Answer:
(a) J.F.

Question 4.
Debtors Account shows ______________ balance.
(a) Real
(b) Negative
(c) Credit
(d) Debit
Answer:
(d) Debit

Question 5.
______________ is the process of deriving the difference between totals of the debit and credit side of each ledger a/c.
(a) Totalling
(b) Journalizing
(c) Balancing
(d) Posting
Answer:
(c) Balancing

Question 6.
Total of Purchase book is ______________ to Purchase Account.
(a) posted
(b) moved
(c) given
(d) entered
Answer:
(a) posted

Question 7.
Real account always shows ______________ balance.
(a) minimum
(b) maximum
(c) debit
(d) credit
Answer:
(c) debit

Question 8.
______________ is prepared to test arithmetical accuracy of Books of Accounts.
(a) Trial Balance
(b) Ledger
(c) Journal
(d) List
Answer:
(a) Trial Balance

4. State whether the following statements are ‘True or False’ with reasons.

Question 1.
Ledger is a book of Original Entry.
Answer:
This statement is False.
Ledger is a book of Secondary Entry.
Journal is a book of Original Entry: First, all transactions are recorded to journal or subsidiary books, and then they are pasted to the respective ledger accounts.

Question 2.
The process of recording a transaction in the Journal is called Posting.
Answer:
This statement is False.
The process of recording a transaction in the Journal is called Journalising. Posting means transferring journal entries to respective ledger accounts.

Question 3.
A cash withdrawal from business by the trader should be credited to Drawings A/c.
Answer:
This statement is False.
Cash withdrawn from the business by the trader should be debited to Drawing A/c. It is a personal account and as per the golden rules of a personal account Debit the receiver.

Question 4.
Balances of Nominal Accounts are carried forward to the next year.
Answer:
This statement is False.
Balances of Nominal Accounts are transferral to Trading and Profit and loss accounts of the year to find Gross Profit and Net Profit.

Question 5.
When the debit side of an account is greater than the credit side, the account shows a debit balance.
Answer:
This statement is True.
While balancing the ledger account the side which is greater is the balance of that account so when the debit side of an account is greater the account shows debit balances.

Question 6.
The name of an account written on top of each account is called ‘Head of Account’.
Answer:
This statement is True.
There are many ledger accounts in the ledger book. To identity, the name of the account, every account on the Top Head of Account is written.

Question 7.
Agreement of Trial Balance always proves accounting accuracy.
Answer:
This statement is False.
Even though the Trial balance is tally there may be some mistake like the complete omission of transaction or compensatory error so Agreement of Trial Balance does not always prove accounting accuracy.

Question 8.
The trial balance is based on the double-entry principle that for every debit there is an equal amount of corresponding credit.
Answer:
This statement is True.
Trial balance is an extract of ledger balances. Ledger is prepared of journal book which follows the Double Entry System of book-keeping. When both the effects of debit and credit with equal amount is given. The trial balance will be tally.

5. Fill in the blanks.

Question 1.
______________ Balance on Nominal Account shows expenses or loss.
Answer:
debit

Question 2.
Cash account always shows ______________ balance.
Answer:
debit

Question 3.
The right hand side of an account is called ______________ side.
Answer:
credit

Question 4.
Creditors shows ______________ balance.
Answer:
credit

Question 5.
______________ accounts are closed by transferring its balances to Profit and Loss Account.
Answer:
Nominal

Question 6.
‘b/d’ means ______________
Answer:
brought down

Question 7.
Rent paid for the residential quarter will be debited to ______________ account.
Answer:
Drawings

Question 8.
Sold goods of ₹ 24,000 at 20% Profit on cost, the purchase price of the goods is ______________
Answer:
₹ 20,000

6. Complete the following table.

Question 1.

Answer:
Ledger

Question 2.

Answer:
Return outward

Question 3.

Answer:
J.F.

Question 4.

Answer:
Credit balance

Question 5.

Answer:
Nominal A/c

7. Put ‘4’ mark for the nature of balance for the following.

Question 1.


Answer:

Practical Problems

Question 1.
Give Journal entries of the following posting from the ledger account.
In the books of Sopan



Solution:
Journal of Sopan

Question 2.
Prepare necessary Ledger Accounts from the following Subsidiary Books.
Purchase Book

Purchase Return Book

Solution:
In the Ledger of ______________

Question 3.
From the following transactions prepare necessary Ledger Accounts in the Books of Vinay and balance the same.
2019 Jan.
1 Started business with Cash ₹ 10,000
6 Bought goods from Vikas ₹ 3,000
9 Sold goods to Bhushan ₹ 2,400
12 Paid to Vikas on account ₹ 1,600
19 Received on account from Bhushan ₹ 1,000
25 Cash Purchases ₹ 3,600
30 Cash Sales ₹ 5,000
31 Paid Wages ₹ 400
Solution:
In the Ledger of Vinay

Question 4.
Journalise the following transactions and prepare Cash A/c only.
2019 July
1 Hardik started the business with Cash ₹ 15,000 and Machinery ₹ 20,000.
4 Purchased goods for ₹ 9,000 less 10% Cash Discount.
9 Sold goods to Amar ₹ 3,000.
12 Distributed goods worth ₹ 700 as free samples.
14 Bought Stationery for ₹ 550 for office use.
18 Received ₹ 950 from Dhanashree, a customer, whose account was earlier written off as a bad debt.
21 Abhiram invoiced us goods worth ₹ 3,000.
24 Settled Abhiram’s account, he allowed 5% cash discount.
27 Exchanged goods worth ₹ 2,500 against Furniture of the same amount.
29 Withdrawn cash from ATM ₹ 5,000 for office use and ₹ 3,000 for personal use.
Solution:
Journal of Hardik ______________


Ledger of Hardik

Question 5.
Prepare Aparna’s account in the books of Suparna.
2019 Jan.
1 Balance due from Aparna ₹ 60,000
4 Sold goods to Aparna ₹ 15,000 at 10% Trade Discount.
7 Goods returned by Aparna ₹ 1,500 (Gross)
11 Received crossed cheque from Aparna ₹ 50,000
17 Invoiced goods to Aparna ₹ 12,000
25 Sold goods to Aparna in cash ₹ 6,000
30 Received cash from Aparna ₹ 33,000 in full settlement of her account.
Solution:
Ledger of Suparna

Working Notes:
Jan. 4:
Trade discount = 10% on ₹ 15,000
= \(\frac {10}{100}\) × ₹ 15,000
= ₹ 1,500
Net Selling Price = Catalogue price – Trade discount
= 15,000 – 1,500
= ₹ 13,500

Question 6.
Prepare Cash A/c, Bank A/c, Purchases A/c, Sales A/c, and Capital A/c and balance the same in the books of Madanlal.
2019 Aug.
1 Started business with a bank balance of ₹ 40,000.
2 Purchased goods from Aseem worth ₹ 15,000 less 10% Trade Discount.
3 Sold goods to Arun for ₹ 8,000 in cash.
4 Paid Rent ₹ 3,000 and Electricity bill ₹ 500.
5 Purchased 100 Shares of Perfect Technologies for ₹ 55 per share and paid Brokerage ₹ 250 by transfer through net banking.
6 Withdrawal of goods for personal use ₹ 500.
7 Sold goods for cash ₹ 5,000 less 10% Cash Discount.
8 Deposited cash into Bank ₹ 2,000.
9 Paid ₹ 3,000 for daughter’s tuition fees by Debit Card.
10 Purchased a Table for ₹ 2,000.
19 Received ₹ 1,500 by selling the scrap.
27 Paid cash into a bank in excess of ₹ 2,000
Solution:
In the Ledger of Madanlal

Question 7.
Journalise the following transactions; post them into Ledger for February 2019
1 Sunil Started business with a stock of goods ₹ 20,000 and Cash ₹ 1,70,000 out of which ₹ 50,000 borrowed from his friend Kedar @ 10 p.a.
5 Placed an order for goods worth ₹ 7,000 with Mohan for which an advance of ₹ 5,500 was paid.
9 Purchased Stationery for office use ₹ 4,500
12 Goods distributed as free samples ₹ 2,000
17 Paid Freight ₹ 400 on behalf of Mr. Dev.
24 Received Goods from Mohan as per our order dated 5th Feb and settled his account.
27 Bought goods from Shekhar on two months credit for ₹ 7,000 at 20% Trade Discount with instructions to send them to Sagar.
28 Sent to Sagar Outward Invoice for goods supplied by Shekhar, at list price less 10% Trade Discount.
Solution:
Journal of Sunil

Ledger of Sunil

Question 8.
Journalise the following transactions and Prepare ledger accounts in the books of Sanjeev.
2019 June
1 Cash Received from Raju ₹ 10,000 for commission.
3 Intra-state sale to Rakesh ₹ 3,000 and SGST @ 2.5% and CGST @ 2.5% applicable.
5 Received full amount from Rakesh.
8 Intra-state purchases from Mangesh ₹ 2,000 and SGST @ 2.5% and CGST @ 2.5% applicable.
11 Paid the necessary amount to Mangesh.
18 Paid Rent ₹ 2,500
24 Paid mobile bill ₹ 1,000 out of which ₹ 700 for office use and for ₹ 300 for personal use.
Solution:
Journal of Sanjeev


In the Ledger of Sanjeev

Question 9.
The following ledger balances were extracted from the books of Pawan Pawar, Pune as of 1st July 2019.

The following transactions took place during July 2019. Post them into Ledger and prepare Trial Balance as of 31st July 2019.
1 Introduced additional Capital ₹ 40,000
4 Bought goods from Rakesh ₹ 80,000 @ 10% Trade Discount
7 Sold goods to Rashmi ₹ 30,000
9 Returned goods to Rakesh ₹ 20,000 (Gross)
11 Rashmi returned goods to us ₹ 400
14 Paid to Rakesh ₹ 40,000 @ 2% Cash Discount
22 Made purchases ₹ 17,000 and amount paid by cheque
24 Cash Sales ₹ 8,000
27 Bought Stationery ₹ 3,000
28 Received from Rashmi ₹ 39,000 by RTGS and discount allowed ₹ 1000
29 Paid Salary ₹ 10,000
29 Sold goods to Rashmi ₹ 20,000
31 Bought goods from Rakesh ₹ 36,000 and paid by cheque.
Solution:
Ledger of Pawan Pawar, Pune

Trial Balance as of 31st July 2019

Balbharti Maharashtra State Board 11th Biology Textbook Solutions Chapter 6 Biomolecules Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Biology Solutions Chapter 6 Biomolecules

1. Choose correct option

Question (A)
Sugar, amino acids and nucleotides unite to their respective subunits to form ________
(a) bioelements
(b) micromolecules
(c) macromolecules
(d) all of these
Answer:
(c) macromolecules

Question (B)
Glycosidic bond is found in __________ .
(a) Disaccharide
(b) Nucleosides
(c) Polysaccharide
(d) all of these
Answer:
(d) all of these

Question (C)
Amino acids in a polypeptide are joined by _______ bond.
(a) Disulphide
(b) glycosidic
(c) hydrogen bond
(d) none of these
Answer:
(d) none of these

Question (D)
Lipids associated with cell membrane are _________ .
(a) Sphingomyelin
(b) Isoprenoids
(c) Phospholipids
(d) Cholesterol
Answer:
(c) Phospholipids

Question (E)
Linoleic, Linolenic and ________ acids are referred as essential fatty acids since they cannot be synthesized by the body and hence must be included in daily diet.
(a) Arachidonic
(b) Oleic
(c) Steric
(d) Palmitic
Answer:
(a) Arachidonic

Question (F)
Hemoglobin is a type of ________ protein, which plays indispensable part in respiration.
(a) simple
(b) derived
(c) conjugated
(d) complex
Answer:
(c) conjugated

Question (G)
When inorganic ions or metallo-organic molecules bind to apoenzyme, they together form
(a) isoenzyme
(b) holoenzyme
(c) denatured enzyme
(d) none of these
Answer:
(b) holoenzyme

Question (H)
In enzyme kinetics, Km = Vmax/2. If Km value is lower, it indicates _______
(a) Enzyme has less affinity for substrate
(b) Enzyme has higher affinity towards substrate
(c) There will be no product formation
(d) All active sites of enzyme are saturated
Answer:
(b) Enzyme has higher affinity towards substrate

2. Solve the following questions

Question (A)
Observe the following figures and write the differences between them.

Answer:

3. Answer the following questions

Question (A)
What are building blocks of life?
Answer:
Life is composed of four main building blocks: Carbohydrates, proteins, lipids and nucleic acids.

Question (B)
Explain the peptide bond.
Answer:
1. The covalent bond that links the two amino acids is called a peptide bond.
2. Peptide bond is formed by condensation reaction.

Question (C)
How many types of polysaccharides you know?
Answer:
There are two types of polysaccharides:
1. Homopolysaccharides: It contains same type of monosaccharides. E.g. Starch, glycogen, cellulose.
2. Heteropolysaccharides: It contains two or more different monosaccharides. E.g. Hyaluronic acid, heparin, hemicellulose.

Question (D)
Enlist the significance of carbohydrates.
Answer:
Significances of carbohydrates are as follows:

1. Carbohydrates provide energy for metabolism.
2. Glucose is the main substrate for ATP synthesis.
3. Lactose, a disaccharide present in the milk provides energy to babies.
4. Polysaccharide serves as a structural component of cell membrane, cell wall and reserved food as starch and glycogen.

Question (E)
What is reducing sugar?
Answer:
1. A sugar that serves as a reducing agent due to presence of free aldehyde or ketone group is called a reducing sugar.
2. These sugars reduce the Benedict’s reagent (Cu²+ to Cu⁺) since they are capable of transferring hydrogens (electrons) to other compounds, a process called reduction.
3. All monosaccharides are reducing sugars.

Question (F)
Enlist the examples of simple proteins and their significance.
Answer:
Examples of simple proteins are: E.g.: Albumins and histones.
Significance:
1. Albumin:
a. % It is the main protein in the blood.
b. It maintains the pressure in the blood vessels.
c. It helps in transportation of substances like hormone and drugs in the body.
2. Histones:
a. It is the chief protein of chromatin.
b. They are involved in packaging of DNA into structural units called nucleosomes.

Question (G)
Describe the secondary structure of protein with examples.
Answer:

1. There are two types of secondary structure of protein: a-helix and P-pleated sheets.
2. The polypeptide chain is arranged in a spiral helix. These spiral helices are of two types: a-helix (right handed) and P-helix (left handed).
3. This spiral configuration is held together by hydrogen bonds.
4. The sequence of amino acids in the polypeptide chain determines the location of its bend or fold and the position of formation of hydrogen bonds between different portions of the chain or between different chains. Thus, peptide chains form an a-helix structure.
5. Example of a-helix structure is keratin.
6. In some proteins two or more peptide chains are linked together by intermolecular hydrogen bonds. Such structures are called P-pleated sheets.
7. Example of P-pleated sheet is silk fibres.
8. Due to formation of hydrogen bonds peptide chains assume a secondary structure.

Question (H)
Explain the induced fit model for mode of enzyme action.
Answer:
1. The induced fit model shows that enzymes are flexible structures in which the active site continually reshapes by its interactions with the substrate until the time the substrate is completely bound to it. It is also the point at which the final form and shape of the enzyme is determined.
2. Three-Dimensional conformation:
a. All enzymes have specific 3-dimensional conformation.
b. They have one or more active sites to which substrate (reactant) combines.
c. The points of active site where the substrate joins with the enzyme is called substrate binding site.

Question (I)
What is RNA? Enlist types of RNA.
Answer:
1. RNA stands for Ribonucleic Acid. It is a long single stranded polynucleotide chain which helps in protein synthesis, functions as a messenger and translates messages coded in DNA into protein.
2. There are three types of RNA:
mRNA (messenger RNA), rRNA (ribosomal RNA) and tRNA (transfer RNA)

Question (J)
Describe the concept of metabolic pool.
Answer:
1. Metabolic pool is the reservoir of biomolecules in the cell on which enzymes can act to produce useful products as per the need of the cell.
2. The concept of metabolic pool is significant in cell biology because it allows one type of molecule to change into another type E.g. Carbohydrates can be converted to fats and vice-versa.

Question (K)
How do secondary metabolites useful for mankind?
Answer:
1. Drugs developed from secondary metabolites have been used to treat infectious diseases, cancer, hypertension and inflammation.
2. Morphine, the first alkaloid isolated from Papaver somniferum is used as pain reliver and cough suppressant.
3. Secondary metabolites like alkaloids, nicotine, cocaine and the terpenes, cannabinol are widely used for recreation and stimulation.
4. Flavours of secondary metabolites improve our food preferences.
5. Tannins are added to wines and chocolate for improving astringency.
6. Since most secondary metabolites have antibiotic property, they are also used as food preservatives.
7. Glucosinolates is a secondary metabolite which is naturally present in cabbage imparts a characteristic flavour and aroma because of nitrogen and sulphur-containing chemicals. It also offers protection to these plants from many pests.

4. Solve the following questions

Question (A)
Complete the following chart.

Answer:

Question (B)
Answer the following with reference

i. Name the type of bond formed between two polypeptides.
ii. Which amino acid is involved in the formation of such bond?
iii. Amongst I, II, III and IV structural level of protein, which level of structure includes such bond? Answer:
i. Disulfide bond.
ii. Cysteine
iii. Tertiary structure.
[Note: Quaternary structure of protein also have disulfide bond, for stabilization of protein structure.!

Question (C)
Match the following items given in column I and II.

Answer:

5. Long answer questions

Question (A)
What are biomolecules? Explain the building blocks of life.
Answer:
Biomolecules are essential substances produced by our body which are necessary for life.
The building blocks of life are carbohydrates, lipids, proteins and nucleic acids.
1. Carbohydrates:
a. Carbohydrates are biomolecules made from carbon, hydrogen and oxygen.
b. The general formula of carbohydrates is (CH20) n.
c. They contain hydrogen and oxygen in the same ratio as in water (2:1).
d. Carbohydrates can be broken down to release energy.
e. Based on sugar units, carbohydrates are classified into three types: Monosaccharides, disaccharides and polysaccharides.

2. Lipids:
a. These are group of substances with greasy consistency with long hydrocarbon chain containing carbon, hydrogen and oxygen.
b. In lipids hydrogen to oxygen ration is greater than 2:1.
c. Lipid is a broader term used for fatty acids and their derivatives.
d. They are soluble in organic solvents (non-polar solvents).
e. Fatty acids are organic acids which are composed of hydrocarbon chain ending in carboxyl group (COOH) ….
f. These are divided into: Saturated fatty acids and unsaturated fatty acids.
g. Fatty acids are basic molecules which form different kinds of lipids.
h. Lipids are classified into three types:
Simple lipids, Compound lipids, Derived lipids.

3. Proteins:
a. Proteins are large molecules containing amino acid units ranging from 100 to 3000.
b. They have higher molecular weight.
c. In proteins, amino acids are linked together by peptide bonds which join the carboxyl group of one amino acid residue to the amino group of another residue.
d. A protein molecule consists of one or more polypeptide chains.
e. Proteins contain any or all twenty naturally occurring amino acid types.
f. Proteins have different structures like primary structure, secondary structure, tertiary structure and quaternary structure.
g. Proteins are classified into three types:
Simple proteins: Simple proteins on hydrolysis yield only amino acids. E.g. Histones and albumins. Conjugated proteins: It consists of a simple protein united with some non-protein substance. E.g. Haemoglobin.
Derived proteins: These proteins are not found in nature as such but are derived from native protein molecules on hydrolysis. E.g. Metaproteins, peptones.

4. Nucleic Acids:
a. Nucleic acids are macromolecules composed of many small units or monomers called nucleotides.
b. Each nucleotide is formed of three components i.e. pentose sugar, a nitrogen base and a phosphate (phosphoric acid).
c. When sugar combine with nitrogenous base it forms nucleoside. Nucleotides can be called as nucleoside phosphate.
d. There are two types of nucleic acids, i.e. DNA and RNA.
DNA (Deoxyribonucleic acid) is a genetic material of a cell. It is double stranded helix. Each strand of helix is made up of deoxyribose nucleotides.
RNA (Ribonucleic Acid) is a single stranded structure having fewer nucleotides as compared to DNA. The strands may be straight or variously folded upon itself. It is made up of nucleotides.

Question (B)
Explain the classes of carbohydrates with examples.
Answer:
Based on number of sugar units, carbohydrates are classified into three types namely, monosaccharides, disaccharides and polysaccharides.
1. Monosaccharides:
a. Monosaccharides are the simplest sugars having crystalline structure, sweet taste and soluble in water.
b. They cannot be further hydrolyzed into smaller molecules.
c. They are the building blocks or monomers of complex carbohydrates.
d. They have the general molecular formula (CH20)n, where n can be 3, 4, 5, 6 and 7.
e. They can be classified as triose, tetrose, pentose, etc.
f. Monosaccharides containing the aldehyde (-CHO) group are classified as aldoses e.g. glucose, xylose, and those with a ketone(-C=0) group are classified as ketoses. E.g. ribulose, fructose.

2. Disaccharides:
a. Disaccharide is formed when two monosaccharide react by condensation reaction releasing a water molecule. This process requires energy.
b. A glycosidic bond forms and holds the two monosaccharide units together.
c. Sucrose, lactose and maltose are examples of disaccharides.
d. Sucrose is a nonreducing sugar since it lacks free aldehyde or ketone group.
e. Lactose and maltose are reducing sugars.
f. Lactose also exists in beta form, which is made from P-galactose and p-glucose.
g. Disaccharides are soluble in water, but they are too big to pass through the cell membrane by diffusion.

3. Polysaccharides:
a. Monosaccharides can undergo a series of condensation reactions, adding one unit after the other to the chain till a very large molecule (polysaccharide) is formed. This is called polymerization.
b. Polysaccharides are broken down by hydrolysis into monosaccharides.
c. The properties of a polysaccharide molecule depends on its length, branching, folding and coiling.
d. Examples: Starch, glycogen, cellulose.

Question (C)
Describe the types of lipids and mention their biological significance.
Answer:
Lipids are classified into three main types:
1. Simple lipids:
a. These are esters of fatty acids with various alcohols. Fats and waxes are simple lipids.
b. Fats are esters of fatty acids with glycerol (CH₂OH-CHOH-CH₂OH).
c. Triglycerides are three molecules of fatty acids and one molecule of glycerol.
d. Unsaturated fats are liquid at room temperature and are called oils. Unsaturated fatty acids are hydrogenated to produce fats e.g. Vanaspati ghee.

Biological significance:
a. Fats are a nutritional source with high calorific value and they act as reserved food materials.
b. In plants, fat is stored in seeds to nourish embryo during germination.
c. In animals, fat is stored in the adipocytes of the adipose tissue.
d. Fats deposited in subcutaneous tissue act as an insulator and minimize loss of body heat.
e. Fats deposited around the internal organs act as cushions to absorb mechanical shocks.
f. Wax is another example of simple lipid. They are esters of long chain fatty acids with long chain alcohols.
g. They are found in the blood, gonads and sebaceous glands of the skin.
h. Waxes are not as readily hydrolyzed as fats.
i. They are solid at ordinary temperature.
j. Waxes form water insoluble coating on hair and skin in animals, waxes form an outer coating on stems, leaves and fruits.

2. Compound lipids:
a. These are ester of fatty acids containing other groups like phosphate (Phospholipids), sugar (glycolipids), etc.
b. They contain a molecule of glycerol, two molecules of fatty acids and a phosphate group or simple sugar.
c. Some phospholipids such as lecithin also have a nitrogenous compound attached to the phosphate group.
d. Phospholipids have both hydrophilic polar groups (phosphate and nitrogenous group) and hydrophobic non-polar groups (hydrocarbon chains of fatty acids).
e. Glycolipids contain glycerol, fatty acids, simple sugars such as galactose. They are also called cerebrosides.
Biological significance:
a. Phospholipids contribute in the formation of cell membrane.
b. Large amounts of glycolipids are found in the brain white matter and myelin sheath.

3. Derived Lipids:
a. They are composed of fused hydrocarbon rings (steroid nucleus) and a long hydrocarbon side chain.
b. One of the most common sterols is cholesterol.
Biological significance:
a. It is widely distributed in all cells of the animal body, but particularly in nervous tissue.
b. Cholesterol exists either free or as cholesterol ester.
c. Adrenocorticoids, sex hormones (progesterone, testosterone) and vitamin D are synthesized from cholesterol.
d. Cholesterol is not found in plants.
e. Sterols exist as phytosterols in plants.
f. Yam Plant (Dioscorea) produces a steroid compound called diosgenin. It is used in the manufacture of antifertility pills, i.e. birth control pills.

Question (D)
Explain the chemical nature, structure and role of phospholipids in biological membrane.
Answer:
Chemical nature: Phospholipids are amphiphilic in nature. As they have hydrophilic head and hydrophobic tail.
Structure: It contains an alcohol, two fatty acid chains and a phosphate group.
Role: Phospholipids forms the membranes around the cells and cellular organelles. They form a lipid bilayer membrane. The phospholipids are arranged tail to tail. It serves as a barrier against movement of any ions or polar compounds into and out of the cell.

Question (E)
Describe classes of proteins with their importance.
Answer:
On the basis of structure, proteins are classified into three categories:
1. Simple proteins:
a. Simple proteins on hydrolysis yield only amino acids.
b. These are soluble in one or more solvents.
c. Simple proteins may be soluble in water.
d. Histones of nucleoproteins are soluble in water.
e. Globular molecules of histones are not coagulated by heat.
f. Albumins are also soluble in water but they get coagulated on heating.
g. Albumins are widely distributed e.g. egg albumin, serum albumin and legumelin of pulses are albumins.
Importance: They are involved in structural components; they also act as a storage kind of protein.
Some are associated with nucleic acids in nucleoproteins of cell.

2. Conjugated proteins:
a. Conjugated proteins consist of a simple protein united with some non-protein substance.
b. The non-protein group is called prosthetic group e.g. haemoglobin.
c. Globin is the protein and the iron containing pigment haem is the prosthetic group.
d. Similarly, nucleoproteins have nucleic acids.
e. Proteins are classified as glycoproteins and mucoproteins.
f. Mucoproteins are carbohydrate-protein complexes e.g. mucin of saliva and heparin of blood.
g. Lipoproteins are lipid-protein complexes e.g. conjugate protein found in brain, plasma membrane, milk etc. Importance: They are involved in structural components of cell membranes and organelles.
They also act as a transporter.
Some conjugated proteins are important in electron transport chain in respiration.

3. Derived proteins:
a. These proteins are not found in nature as such.
b. These proteins are derived from native protein molecules on hydrolysis.
c. Metaproteins, peptones are derived proteins.
Importance: They act as a precursor for many molecules which are essential for life.

Question (F)
What are enzymes? How are they classified? Mention example of each class.
Answer:
1. Enzymes are biological macromolecules which act as a catalyst and accelerates the reaction in the body.
2. Enzymes are classified into six classes:
a. Oxidoreductases: These enzymes catalyze oxidation and reduction reactions by the transfer of hydrogen and/or oxygen, e.g. alcohol dehydrogenase

b. Transferases: These enzymes catalyse the transfer of certain groups between two molecules, e.g. glucokinase

c. Hydrolases: These enzymes catalyse hydrolytic reactions. This class includes amylases, proteases, lipases etc. e.g. Sucrase

d. Lyases: These enzymes are involved in elimination reactions resulting in the removal of a group of atoms from substrate molecule to leave a double bond. It includes aldolases, decarboxylases, and dehydratases, e.g. fumarate hydratase.

e. Isomerases: These enzymes catalyze structural rearrangements within a molecule. Their nomenclature is based on the type of isomerism. Thus, these enzymes are identified as racemases, epimerases, isomerases, mutases, e.g. xylose isomerase.

f. Ligases or Synthetases: These are the enzymes which catalyze the covalent linkage of the molecules utilizing the energy obtained from hydrolysis of an energy-rich compound like ATP, GTP e.g. glutathione synthetase, Pyruvate carboxylase.

Question (G)
Explain the properties of enzyme? Describe the models for enzyme actions.
Answer:
1. Proteinaceous Nature:
All enzymes are basically made up of protein.

2. Three-Dimensional conformation:
a. All enzymes have specific 3-dimensional conformation.
b. They have one or more active sites to which substrate (reactant) combines.
c. The points of active site where the substrate joins with the enzyme is called substrate binding site.

3. Catalytic property:
a. Enzymes are like inorganic catalysts and influence the speed of biochemical reactions but themselves remain unchanged.
b. After completion of the reaction and release of the product they remain active to catalyze again.
c. A small quantity of enzymes can catalyze the transformation of a very large quantity of the substrate
into an end product.
d. For example, sucrase can hydrolyze 100000 times of sucrose as compared with its own weight.

4. Specificity of action:
a. The ability of an enzyme to catalyze one specific reaction and essentially no other is perhaps its most significant property. Each enzyme acts upon a specific substrate or a specific group of substrates.
b. Enzymes are very sensitive to temperature and pH.
c. Each enzyme exhibits its highest activity at a specific pH i.e. optimum pH.
d. Any increase or decrease in pH causes decline in enzyme activity e.g. enzyme pepsin (secreted in stomach)shows highest activity at an optimum pH of 2 (acidic)

5. Temperature:
a. Enzymes are destroyed at higher temperature of 60-70°C or below, they are not destroyed but become inactive.
b. This inactive state is temporary and the enzyme can become active at suitable temperature.
c. Most of the enzymes work at an optimum temperature between 20°C and 35°C.

There are two types of models:
1. Lock and Key model:
a. Lock and Key model was first postulated in 1894 by Emil Fischer.
b. This model explains the specific action of an enzyme with a single substrate.
c. In this model, lock is the enzyme and key is the substrate.
d. The correctly sized key (substrate) fits into the key hole (active site) of the lock (enzyme).

2. Induced Fit model (Flexible Model):
a. Induced Fit model was first proposed in 1959 by Koshland.
b. This model states that approach of a substrate induces a conformational change in the enzyme.
c. It is the more accepted model to understand mode of action of enzyme.
d. The induced fit model shows that enzymes are rather flexible structures in which the active site continually reshapes by its interactions with the substrate until the time the substrate is completely bound to it.
e. It is also the point at which the final form and shape of the enzyme is determined.
[Note: Temperature is a factor affecting enzyme activity and not a property of enzyme.]

Question (H)
Describe the factors affecting enzyme action.
Answer:
The factors affecting the enzyme activity are as follows:
1. Concentration of substrate:
a. Increase in the substrate concentration gradually increases the velocity of enzyme activity within the limited range of substrate levels.
b. A rectangular hyperbola is obtained when velocity is plotted against the substrate concentration.
c. Three distinct phases (A, B and C) of the reaction are observed in the graph.
Where V = Measured velocity, V_max = Maximum velocity, S = Substrate concentration,
K_m = Michaelis-Menten constant.
d. K_m or the Michaelis-Menten constant is defined as the substrate concentration (expressed in moles/lit) to produce half of maximum velocity in an enzyme catalyzed reaction.
e. It indicates that half of the enzyme molecules (i.e. 50%) are bound with the substrate molecules when the substrate concentration equals the Km value.
f. K_m value is a constant and a characteristic feature of a given enzyme.
g. It is a representative for measuring the strength of ES complex.
h. A low K_m value indicates a strong affinity between enzyme and substrate, whereas a high Km value reflects a weak affinity between them.
1. For majority of enzymes, the K_m values are in the range of 10_-5 to 10_-2 moles.

2. Enzyme Concentration:

a. The rate of an enzymatic reaction is directly proportional to the concentration of the substrate.
b. The rate of reaction is also directly proportional to the square root of the concentration of enzymes.
c. It means that the rate of reaction also increases with the increasing concentration of enzyme and the rate of reaction can also decrease by decreasing the concentration of enzyme.

3. Temperature:

a. The temperature at which the enzymes show maximum activity is called Optimum temperature.
b. The rate of chemical reaction is increased by a rise in temperature but this is true only over a limited range of temperature.
c. Enzymes rapidly denature at temperature above 40°C.
d. The activity of enzymes is reduced at low temperature.
e. The enzymatic reaction occurs best at or around 37°C which is the average normal body temperature in homeotherms.

4. Effect of pH:

a. The pH at which an enzyme catalyzes the reaction at the maximum rate is known as optimum pH.
b. The enzyme cannot perform its function beyond the range of its pH value.

5. Other substances:
a. The enzyme action is also increased or decreased in the presence of some other substances such as co-enzymes, activators and inhibitors.
b. Most of the enzymes are combination of a co-enzyme and an apo-enzyme.
c. Activators are the inorganic substances which increase the enzyme activity.
d. Inhibitor is the substance which reduces the enzyme activity.

Question (I)
What are the types of RNA? Mention the role of each class of RNA.
Answer:
There are three types of cellular RNAs:
1. messenger RNA (mRNA),
2. ribosomal RNA (rRNA),
3. transfer RNA (tRNA). ‘

1. Messenger RNA (mRNA):
a. It is a linear polynucleotide.
b. It accounts 3% of cellular RNA.
c. Its molecular weight is several million. , d. mRNA molecule carrying information to form a complete polypeptide chain is called cistron.
e. Size of mRNA is related to the size of message it contains.
f. Synthesis of mRNA begins at 5’ end of DNA strand and terminates at 3’ end.

Role of messenger RNA:
It carries genetic information from DNA to ribosomes, which are the sites of protein synthesis.

2. Ribosomal RNA (rRNA):
a. rRNA was discovered by Kurland in 1960.
b. It forms 50-60% part of ribosomes.
c. It accounts 80-90% of the cellular RNA.
d. It is synthesized in nucleus.
e. It gets coiled at various places due to intrachain complementary base pairing.
Role of ribosomal RNA: It provides proper binding site for m-RNA during protein synthesis.

3. Transfer RNA (tRNA):
a. These molecules are much smaller consisting of 70-80 nucleotides.
b. Due to presence of complementary base pairing at various places, it is shaped like clover-leaf.
c. Each tRNA can pick up particular amino acid.
d. Following four parts can be recognized on tRNA
1. DHU arm (Dihydroxyuracil loop/ amino acid recognition site
2. Amino acid binding site
3. Anticodon loop / codon recognition site
4. Ribosome recognition site.
e. In the anticodon loop of tRNA, three unpaired nucleotides are present called as anticodon which pair with codon present on mRNA.
f. The specific amino acids are attached at the 3′ end in acceptor stem of clover leaf of tRNA.
Role of transfer RNA: It helps in elongation of polypeptide chain during the process called translation.

Question (J)
What is metabolism? How metabolic pool is formed in the cell.
Answer:

1. Metabolism is the sum of the chemical reactions that take place within each cell of a living organism and provide energy for vital processes and for synthesizing new’ organic material.
2. Metabolic pool in the cell is formed due to glycolysis and Krebs cycle.
3. The catabolic chemical reaction of glycolysis and Krebs cycle provides ATP and biomolecules. These biomolecules form the metabolic pool of the cell.
4. These biomolecules can be utilized for synthesis of many important cellular components.
5. The metabolites can be added or withdrawn from the pool according to the need of the cell.

Question 6.
If double stranded DNA has 14% C (cytosine) what percent A (adenine), T (thymine) and G (guanine) would you expect?
Answer:
A purine always pairs with pyrimidine.
Adenine pairs with thymine and cytosine pairs with guanine.
Therefore, as per the given data If cytosine = 14% then guanine = 14%.
According to Chargaff s rule,
(C+G) = 14+ 14 = 28%
Therefore, (A+T) = 72%
So, A= 36%, T= 36%, G = 14%.

Question 7.
Name
1. The reagent used for testing for reducing sugar.
2. The form in which carbohydrate is transported in a plant.
3. The term that describes all the chemical reactions taking place in an organism.
Answer:
1. Benedict’s reagent
2. Sucrose
3. Metabolism

Practical / Project:

Question 1.
Perform an experiment to study starch granules isolated from potato.
Answer:
Isolation of starch granules from potato:

1. Peal the potato with a clean knife.
2. Grind the potato till the homogenous mixture is formed.
3. Then strain the mixture through a cheese cloth into a beaker.
4. Keep it standing for some time.
5. Throw the supernatant and fill the beaker containing starch with water.
6. Stir it well and again allow the starch to settle.
7. After sometime, again through the supernatant.
8. Repeat this for 2-3 times.
9. Collect the white starch in the watch glass and keep it in the oven for drying.

To study the isolated starch granules:
1. Examination under microscope:
Examine starch granules under microscope by using a mixture of equal volumes of glycerol and distilled water.
Result: The potato starch granules appears transparent granules. They are irregularly shaped.
2. Using Iodine solution:
Boil a little amount of starch with water. Cool it. Add iodine solution to it.
Result: The solution changes colour to blue. This indicates the presence of starch.

Question 2.
Study the action of enzyme urease on urea.
Answer:
Urease is an enzyme which exists in a dimer form. It has two active sites which are highly specific and only bind to urea or hydroxy urea. The active sites of urease contain nickel atoms. Urease catalyzes the hydrolysis of substrate urea into carbon dioxide and ammonia. It attacks the nitrogen and carbon bond in amide compounds and forms alkaline product like ammonia.

11th Biology Digest Chapter 6 Biomolecules Intext Questions and Answers

Can you recall? (Textbook Page No. 59)

(i) Which are different cell components?
Answer:
a. The three main components of any cell are: Cell membrane, Cytoplasm, Nucleus.
b. The components present in both plant and animal cells are: Endoplasmic reticulum, ribosomes, golgi apparatus, lysosomes, mitochondria, vacuoles.
c. The components present in plant cell and not in animal cell: Cell wall and plastids.
d. The components present in animal cell and not in plant cell: Cilia and flagella.

(ii) What is the role of each component of cell?
Answer:
The role of each component of a cell is as follows:
a. Cell membrane: Cell membrane separates the cytoplasmic contents from external environment.
b. Cytoplasm: Site for metabolic activities and organelles.
c. Nucleus: It is the control center of the cell. Genetic material is present in the nucleus.
d. Endoplasmic reticulum: It produces, processes and transports proteins and lipids.
e. Ribosomes: Ribosome is the site for protein synthesis.
f. Golgi apparatus: It is involved in modifying, sorting and packing of proteins for secretion. It also transports lipids around the cell.
g. Lysosomes: It is involved in digestion of worn out organelles and waste removal.
h. Mitochondria: It is responsible for production of energy.
i. Vacuoles: It has various functions like storage, waste disposal, protection and growth.
j. Cell wall: It provides strength and support to the cell.
k. Plastids: They are responsible for production and storage of food. It also contains photosynthetic pigments (Chloroplasts).
l. Cilia and flagella: Help in motility.

Can you tell? (Textbook Page No. 62)

What are carbohydrates?
Answer:

1. The word carbohydrates mean ‘hydrates of carbon’.
2. They are also called saccharides.
3. They are biomolecules made from just three elements: carbon, hydrogen and oxygen with the general formula C_x(H₂0)y.
4. They contain hydrogen and oxygen in the same ratio as in water (2:1).
5. Carbohydrates can be broken down (oxidized) to release energy.

Can you tell? (Textbook Page No. 62)

(i) Enlist the natural sources, structural units and functions of the following polysaccharides.
a. Starch
b. Cellulose
c. Glycogen
Answer:
a. Starch:
1. Natural Sources: Cereals (wheat, maize, rice), root vegetables (potato, cassava etc.)
2. Structural units: Starch consist of two types of molecules – Amylose and amylopectin.
3. Functions: It acts as a reserve food and supply energy.

b. Cellulose:
1. Natural sources: Plant fibers (cotton, flax, hemp, jute, etc.), wood.
2. Structural units: It is made from p glucose molecules.
3. Functions: It in a major component of cell wall. It provides structural support.

c. Glycogen:
1. Natural sources: Fruits, starchy vegetables, whole grain foods.
2. Structural units: It consists of linear chains of glucose residues. The glucose is linked linearly by a (1 → 4) glycosidic bonds and branches are linked to the linear chain by a (1 → 6) glycosidic bonds.
3. Functions: It is stored in liver and muscles and it readily provides energy when the blood glucose level decreases.

(ii) The exoskeleton of insects is made up of chitin. This is a ________.
(A) mucoprotein
(B) lipid
(C) lipoprotein
(D) polysaccharide
Answer:
polysaccharide

(iii) List names of structural polysaccharides.
Answer:
Arabinoxylans, cellulose, chitin, pectin.

(iv) Write a note on oligosaccharide and glycosidic bond.
Answer:
Oligosaccharides:
a. A carbohydrate polymer comprising of two to six monosaccharide molecules is called oligosaccharide.
b. They are linked together by glycosidic bond.
c. They are classified on the basis of monosaccharide units:
Disaccharides: These are the sugars containing two monosaccharide units and can be further hydrolysed into smaller components. E.g.: Sucrose, maltose, lactose, etc.
Trisaccharides: These contain three monomers. E.g. Raffmose.
Tetrasaccharides: These contain four monomers. E.g.: Stachyose.

Glycosidic bond:
a. Glycosidic bond is a covalent bond that forms a linkage between two monosaccharides by a dehydration reaction.
b. It is formed when a hydroxyl group of one sugar reacts with the anomeric carbon of the other.
c. Glycosidic bonds are readily hydrolyzed by acid but resist cleavage by base.
d. There are two types of glycosidic bonds: a-glycosidic bond and P-glycosidic bond.

Can you tell? (Textbook Page No. 63)

What are lipids? Classify them and give at least one example of each.
Answer:
Lipids:
Lipids are a group of heterogeneous compounds like fats, oils, steroids, waxes, etc.
They are macro-biomolecules.
These are group of substances with greasy consistency with long hydrocarbon chain containing carbon, hydrogen and oxygen.

Lipids are classified into:
1. Saturated fatty acids: They contain single chain of carbon atoms with single bonds.
E.g. Palmitic acid, stearic acid
2. Unsaturated fatty acids: They contain one or more double bonds between the carbon atoms of the hydrocarbon chain.
a. Simple lipids: These are esters of fatty acids with various alcohols.
E.g. Fats, wax.
b. Compound lipids: These are ester of fatty acids containing other groups like phosphate (Phospholipids), sugar (glycolipids), etc.
E.g. Lecithin
c. Sterols: They are derived lipids. They are composed of fused hydrocarbon rings (steroid nucleus) and a long hydrocarbon side chain.
E.g. Cholesterol, phytosterols.

Find out (Textbook Page No. 63)

(i) Why do high cholesterol level in the blood cause heart diseases?
Answer:
a. When there is high level of cholesterol in the blood, the cholesterol builds up on the walls of arteries causing a condition called atherosclerosis (a form of heart disease).
b. Because of this the arteries are narrowed and the blood flow to the heart is slowed down.
c. The blood carries oxygen to the heart, but because of this condition enough blood and oxygen does not reach to the heart and causes heart diseases.
d. If the condition increases, the supply of oxygen and blood is completely cut off to the heart and this can lead to heart attack.

(ii) Polyunsaturated fatty acids are believed to decrease blood cholesterol level. How?
Answer:
a. The liver converts polyunsaturated fatty acids into ketones instead of cholesterol.
b. Therefore, polyunsaturated fatty acids are transported directly to tissues for oxidation without leaving behind any lipoprotein in the form of cholesterol as it is seen in the case of saturated fatty acids.
c. Thus, polyunsaturated fatty acids are believed to decrease blood cholesterol level.

Can you tell? (Textbook Page No. 64)

Which of the following is a simple protein?
(A) nucleoprotein
(B) mucoprotein
(C) chromoprotein
(D) globulin
Answer:
Globulin

Can you tell? Textbook Page No. 64)

What are conjugated proteins? How do they differ from simple ones? Give one example of each.
Answer:
1. Conjugated proteins consist of a simple protein attached with some non-protein substance. The non-protein group is called prosthetic group.
2. The conjugated protein functions in interaction with other chemical group whereas simple proteins contain only amino acids and no other chemical group attached to it.
3. Example of conjugated protein is haemoglobin. Globin is the protein and iron containing pigment and haem is the prosthetic group.

Can you tell? (Textbook Page No. 64)

All Proteins are made up of the same amino acids; then how proteins found in human beings and animals may be different from those of other?
Answer:
The proteins found in human beings and animals may be different from those of others because the ratio of amino acids present in the protein differs.

Can you tell? (Textbook Page No. 67)

What is a nucleotide? How is it formed? Mention the names of all nucleotides.
Answer:
1. Nucleotide is a unit which consists of a sugar, phosphate and a base. Nucleotides are basic units of nucleic acids.
2. The nitrogen base and a sugar form a nucleoside. In a nucleoside, nitrogenous base is attached to the first carbon atom (C-1) of the sugar and when a phosphate group gets attached with that of the carbon (C-5) atom of the sugar molecule a nucleotide molecule is formed.
3. The names of all nucleotides are:

Can you tell? (Textbook Page No. 67)

Describe the structure of DNA molecule as proposed by Watson and Crick.
Answer:

1. According to Watson and Crick, DNA molecule consists of two strands twisted around each other in the form of a double helix.
2. The two strands i.e. polynucleotide chains are supposed to be in opposite direction so end of one chain having 3′ lies beside the 5′ end of the other.
3. One turn of the double helix of the DNA measures about 34A.
4. It consists paired nucleotides and the distance between two neighboring pair nucleotides is 3.4A.
5. The diameter of the DNA molecule has been found be 20A.

Can you tell (Textbook Page No. 70)

Name the chemical found in the living cell which has necessary message for the production of all enzymes required by it.
Answer:
DNA found in the nucleus of a living cell has necessary message for the production of all enzymes required by it. DNA forms mRNA through the process of transcription. This mRNA through the process of translation forms proteins.

Can you tell? (Textbook Page No. 67)

Difference between DNA and RNA is because of
(A) sugar and base
(B) sugar and phosphate
(C) phosphate and base
(D) sugar only
Answer:
Sugar and base

Can you tell? (Textbook Page No. 67)

Differentiate between DNA and RNA.
Answer:

Can you tell? (Textbook Page No. 70)

Co-enzyme is ________
(A) often a metal
(B) often a vitamin
(C) always as organic molecule
(D) always an inorganic molecule
Answer:
Always as organic molecule

Can you tell? (Textbook Page No. 70)

(i) Which enzyme is needed to digest food reserve in castor seed?
(A) Amylase
(B) Diastase
(C) Lipase
(D) Protease
Answer:
Lipase

(ii) List the important properties of enzymes.
Answer:
a. Proteinaceous Nature
b. Three-Dimensional conformation
c. Catalytic property
d. Specificity of action
e. Temperature

Try this: (Textbook Page No. 70)

To demonstrate the effect of heat on the activities of inorganic catalysts and enzymes.
Answer:
1. Using MnO₂ and Enzymes without any heat treatment:
Mn0₂ and cellular enzymes (catalase/peroxidase) causes breakdown of H₂0₂ and evolution of oxygen.
2. Using Mn0₂ and Enzymes after heat treatment:
Oxygen evolves in the H₂0₂ solution containing boiled and cooled Mn0₂. But oxygen does not evolve in the tube containing the enzyme.
3. This confirms that heat affects the enzyme and inactivates it whereas heat does not have any effect on inorganic catalyst.

Balbharti Maharashtra State Board Bookkeeping and Accountancy 11th Solutions Chapter 3 Journal Textbook Exercise Questions and Answers.

Maharashtra State Board Bookkeeping and Accountancy 11th Solutions Chapter 3 Journal

1A. Answer in One Sentence:

Question 1.
What is Journal?
Answer:
Journal is a book of account in which all types of day-to-day business transactions are recorded in chronological
order.

Question 2.
What is Narration?
Answer:
Explanation of transaction which is written just below the accounting entry in the particular column is called narration.

Question 3.
What is GST?
Answer:
GST is an abbreviated form of Goods and Service Tax that is levied by the Government on specific goods and services in the place of different taxes levied earlier.

Question 4.
In which year GST was imposed by the Central Government of India?
Answer:
In the year 2017 GST was imposed by the Central Government of India.

Question 5.
What is meant by simple entry?
Answer:
An entry in which only two accounts are affected viz. one account is debited and the other account is credited is called simple entry.

Question 6.
What is the meaning of combined entry?
Answer:
A journal entry that combines more than one debit or more than one credit or both is called a combined/compound entry.

Question 7.
Which account is debited, when rent is paid by Debit card?
Answer:
The rent account is debited when rent is paid by debit card.

Question 8.
Which discount is not recorded in the books of account?
Answer:
Trade discount is not recorded in the books of the account.

Question 9.
In which order monthly transactions are recorded in a Journal?
Answer:
In chronological (date wise) order monthly transactions are recorded in the journal.

Question 10.
Which account is credited, when goods are sold on credit?
Answer:
Sales account is credited, when goods are sold on credit.

2. Give one word/term or phrase for each of the following statements:

Question 1.
A book of prime entry.
Answer:
Journal

Question 2.
The tax imposed by Central Government on Goods and Services.
Answer:
GST

Question 3.
A brief explanation of an entry.
Answer:
Narration

Question 4.
The process of recording transactions in the Journal.
Answer:
Journalising

Question 5.
The French word from which the word Journal is derived.
Answer:
Jour

Question 6.
Concession is given for immediate payment.
Answer:
Cash discount

Question 7.
Entry in which more than one accounts are be debited or credited.
Answer:
Combined Entry

Question 8.
Anything took by the proprietor from the business for his private use.
Answer:
Drawings

Question 9.
Tax payable to the Government on purchase of goods.
Answer:
Input Tax

Question 10.
Page number of the ledger.
Answer:
Ledger Folio

3. Select the most appropriate alternative from the alternatives given below and rewrite the statements.

Question 1.
___________ means explanation of the transactions recorded in the Journal.
(a) Narration
(b) Journalising
(c) Posting
(d) Casting
Answer:
(a) Narration

Question 2.
___________ discount is not recorded in the books of accounts.
(a) Trade
(b) Cash
(c) GST
(d) VAT
Answer:
(a) Trade

Question 3.
Recording of transaction in Journal is called ___________
(a) posting
(b) journalising
(c) narration
(d) prime entry
Answer:
(b) journalising

Question 4.
Every Journal entry require ___________
(a) casting
(b) posting
(c) narration
(d) journalising
Answer:
(c) narration

Question 5.
The ___________ column of the Journal is not recorded at the time of journalising.
(a) date
(b) particulars
(c) ledger folio
(d) amount
Answer:
(c) ledger folio

Question 6.
Goods sold on credit should be debited to ___________
(a) Purchase A/c
(b) Customer A/c
(c) Sales A/c
(d) Cash A/c
Answer:
(b) Customer A/c

Question 7.
Wages paid for installation of Machinery should be debited to ___________
(a) Wages A/c
(b) Machinery A/c
(c) Cash A/c
(d) Installation A/c
Answer:
(b) Machinery A/c

Question 8.
The commission paid to the agent should be debited to ___________
(a) Drawing A/c
(b) Cash A/c
(c) Commission A/c
(d) Agent A/c
Answer:
(c) Commission A/c

Question 9.
Loan taken from Dena Bank should be credited to ___________
(a) Capital A/c
(b) Dena Bank A/c
(c) Cash A/c
(d) Dena Bank Loan A/c
Answer:
(d) Dena Bank Loan A/c

Question 10.
Purchase of animals for cash should be debited to ___________
(a) Livestock A/c
(b) Goods A/c
(c) Cash A/c
(d) Bank A/c
Answer:
(a) Livestock A/c

4. State whether the following statements are True or False with reasons.

Question 1.
Narration is not required for each and every entry.
Answer:
This statement is False.
Correct statement: Narration is required for each and every entry.
Reasons: Narration is a brief explanation of the Journal Entry. It is written in the bracket just below the accounting entry. By reading the narration, the reader understands the meaning and significance of accounting entry and the nature and type of business transactions. Narration should be as short as possible and it should be simple and easy to understand.

Question 2.
A journal voucher is a must for all transactions recorded in the Journal.
Answer:
This statement is True.
Reasons: A voucher is a document that supports a payment made by the businessmen. It is legal evidence that a certain sum of money has been paid to a specific person or party. A Journal voucher is an original or basic voucher on the basis of which business transactions are journalized in the journal. Journal voucher provides legal proof for the business transactions. Therefore a journal voucher is necessary for all transactions recorded in the journal.

Question 3.
Cash discount allowed should be debited to discount A/c.
Answer:
This statement is True.
Reasons: Any allowance or reduction in payment allowed by the seller to the buyer or creditor to the debtor on payment of cash is called cash discount. It is the concession given to encourage prompt payment. The gash discount allowed is an expense or a loss to the receiver. Expenses or Losses are always to be debited. Cash discount allowed is an expense or loss and therefore it is debited to Discount A/c.

Question 4.
Journal is a book of prime entry.
Answer:
This statement is True.
Reasons: Journal is the most important book of accounts. It is a book of daily records. It is the main book of accounts in which transactions are recorded for the first time from source documents. Therefore this book is known as the book of original entry or first entry or prime entry. Business transactions are first entered in the journal and then they are recorded in other accounts book. For these reasons, the journal is called a book of prime entry.

Question 5.
Trade discount is recorded in the books of accounts.
Answer:
This statement is False.
Correct statement: Trade discount is never recorded in the books of accounts.
Reasons: The discount which is allowed or given by the manufacturer to wholesalers and by wholesalers to retailers and retailers to customers on the bulk purchases is called trade discount. By custom or by law trade discount is calculated on the catalog or printed price of the goods. Trade discount is directly deducted from the printed price and net prices of the goods or services are recorded in the books of accounts. A trade discount is given to encourage the buyers to increase their purchases. It is given to traders to enable them to earn a sizeable profit on the printed prices.

Question 6.
Goods lost by theft are debited to the goods account.
Answer:
This statement is False.
Correct statement: Goods lost by theft is credited to the goods account.
Reasons: Goods account is a real account because unsold goods are the property of the business. If goods are purchased or acquired, the Goods account is debited and if goods are sold or lost from the business, they are credited. As per the traditional approach, goods lost means go away from the business, and whatever goes out an account of it is credited. As per the modern approach if loss of business increases account of such loss is credited in the hooks of account.

Question 7.
If rent is paid to the landlord, the landlord’s A/c should be debited.
Answer:
This statement is False.
Correct Statement: If rent is paid to the landlord, the Rent account should be debited.
Reasons: Rent paid is an expense and hence it is a nominal account. When rent is paid to the landlord, the rent account is affected and not the landlord’s account. Rent is an expense to a tenant who pays it and it is an income for the landlord. Such payment is not personally due to the landlord as there is no lending and borrowing of money between landlord and rent payer. As per the rule of nominal account, the rent account is debited because it is an expense. As cash goes out cash account is credited.

Question 8.
Book Keeping records monetary and non-monetary transactions.
Answer:
This statement is False.
Correct Statement: Book Keeping records only monetary transactions.
Reasons: According to the money measurement concept, in the books of accounts accountant records only those business transactions which are monetary or financial in nature and capable to be expressed in monetary terms. It means the qualitative and quantitative aspects which cannot be measured in terms of money are not recorded in the books of account personal or non-monetary transactions are not recorded, in the books of accounts e.g. giving lift from car to neighbour, drinking tea along with friends in the restaurant, etc. are not recorded in the books of account as these transactions are not monetary in nature.

Question 9.
Drawings made by the proprietor increase his capital.
Answer:
This statement is False.
Correct statement: Drawings made by the proprietor decreases his capital.
Reasons: Total amount of goods and services withdrawn by the proprietor from the business from time to time for personal use or family use is called drawings. Withdrawals made by a businessman for business purpose is not treated as drawings. Drawings are always adjusted or deducted from capital. Heavy withdrawals made by a businessman for self-use reduces capital in the business. If the businessman controls the drawings more funds are made available for the development of the business. Drawing made by the proprietor reduces his capital investment.

Question 10.
GST paid on the purchase of goods Input tax A/c should be debited.
Answer:
This statement is True.
Reasons: GST is abbreviated from Goods and Service Tax. GST is levied by the government on the purchases of Goods and Services at a specified rate. Since it is imposed on purchases of goods and services, it increases its cost. GST is added to the purchase price. Purchases are always debited and hence GST i.e. Input tax account is also debited along with purchases. In the case of sales of goods and services, the output tax account is credited.

5. Fill in the blanks.

Question 1.
The first book of original entry is the ___________
Answer:
Journal

Question 2.
The process of recording transaction into journal is called ___________
Answer:
Journalising

Question 3.
An explanation of the transaction recorded in the journal ___________
Answer:
Narration

Question 4.
___________ discount is not recorded in the books of accounts.
Answer:
Trade

Question 5.
___________ is concession allowed for bulk purchase of goods or for immediate payment.
Answer:
Discount

Question 6.
Every Journal Entry requires ___________
Answer:
Narration

Question 7.
___________ discount is always recorded in the books of accounts.
Answer:
Cash

Question 8.
___________ is the document on the basis of which the entry is recorded in the journal.
Answer:
Voucher

Question 9.
There are ___________ parties to a cheque.
Answer:
Three

Question 10.
The ___________ cheque is safer than other cheques as it cannot be encashed on the counter of the bank.
Answer:
Crossed.

6. Specimen and Proforma.

Question 1.
Prepare specimen of Tax Invoice.
Answer:

Question 2.
Prepare specimen of Receipt.
Answer:

Question 3.
Prepare specimen of the Crossed cheque.
Answer:

Question 4.
Prepare specimen of Cash voucher.
Answer:

7. Correct the following statements and rewrite the statements.

Question 1.
All business transactions are recorded in the Journal.
Answer:
Only monetary transactions are recorded in the Journal.

Question 2.
A cash discount is not recorded in the books of accounts.
Answer:
A cash discount is recorded in the books of accounts.

Question 3.
Journal is a book of Secondary entry.
Answer:
Journal is a book of Prime entry.

Question 4.
GST is imposed by the Government of India from 1st July 2018.
Answer:
GST is imposed by the Government of India from 1st July 2017.

Question 5.
The machinery purchased by the Proprietor decreases his capital.
Answer:
Machinery purchased by the proprietor increases his Capital.

8. Do you agree or disagree with the following statements.

Question 1.
Narration is required for every entry.
Answer:
Agree

Question 2.
GST stands for Goods and Sales Tax.
Answer:
Disagree

Question 3.
Trade discount is not recorded in the books of accounts.
Answer:
Agree

Question 4.
Wages paid for the installation of Machinery is debited to Wages Account.
Answer:
Disagree

Question 5.
The process of entering or recording the transactions in a Journal is called Journalising.
Answer:
Agree

9. Calculate the following:

Question 1.
Purchased Motor Car from Tata & Company worth ₹ 2,00,000 at 18% GST. Find out GST amount.
Solution:
Cost of the Motor Car = ₹ 2,00,000
GST @ 18% = 2,00,000 × \(\frac{18}{100}\) = ₹ 36,000
Net value of the Motor Car = ₹ 2,00,000 + ₹ 36,000 = ₹ 2,36,000

Question 2.
Paid Transport charges ₹ 10,000 @ 5% GST. Calculate CGST & SGST.
Solution:
Transport charges = ₹ 10,000 @ 5% GST.
CGST = Transport charges × 2.5%.
= 10,000 × \(\frac{2.5}{100}\)
= 10,000 × \(\frac{25}{1000}\)
= ₹ 250
SGST = Transport charges × 2.5%.
= 10,000 × \(\frac{2.5}{100}\)
= 10,000 × \(\frac{25}{1000}\)
= ₹ 250
Net value = 10,000 + 250 + 250 = ₹ 10,500.

Question 3.
Bought goods from Ranjan ₹ 10,000 @ 5% GST and 10% cash discount. Calculate cash discount.
Solution:
Cost of the goods bought = ₹ 10,000 @ 5% GST and 10% cash discount.
GST on Goods Purchased = Cost of goods × 5%.
= 10,000 × \(\frac{5}{100}\)
= ₹ 500.
Net value of Goods Purchased = 10,000 + 500 = ₹ 10,500
Cash discount = Net value × 10%.
= 10,500 × \(\frac{10}{100}\)
= ₹ 1,050.

Question 4.
Received cheque of ₹ 90,000 from Kiran in full settlement of his account ₹ 1,00,000/-. Calculate discount rate.
Solution:
Discount allowed to Kiran = Amount due – Amount received
= 1,00,000 – 90,000
= ₹ 10,000.
Rate of discount allowed to Kiran = \(\frac{100 \times \text { Total Discount allowed }}{\text { Amount on which discount allowed }}\)
= \(\frac{100 \times 10,000}{1,00,000}\)
= 10%

Question 5.
Sold goods of ₹ 1,00,000 at 10% Trade Discount and 10% cash discount to Ram and received 50% amount by cheque. Calculate the amount of cheque received.
Solution:
Trade Discount = Catalogue price × Rate of trade discount
= 1,00,000 × \(\frac{10}{100}\)
= ₹ 10,000
Net amount receivable = Catalogue price – Trade Discount
= 1,00,000 – 10,000
= ₹ 90,000
50% of net amount received.
∴ Amount receivable = 50% of 90,000
= \(\frac{50}{100}\) × 90,000
= ₹ 45,000
Cash discount allowed = 10% on ₹ 45,000
= \(\frac{10}{100}\) × 45,000
= ₹ 4,500
Amount of cheque received = 50% of total amount – Cash discount
= 45,000 – 4,500
= ₹ 40, 500
Amount received by cheque = ₹ 40,500.

10. Complete the following table.

Question 1.

Answer:
1. Drawings A/c
2. Sonali’s A/c
3. Bad debts A/c
4. Livestock A/c
5. Capital A/c

Practical Problems

Question 1.
Journalise the following transactions in the books of Anand General Merchants.
2019 April
1 Mr. Anand started the business with cash of ₹ 60,000.
5 Purchased goods for cash ₹ 30,000.
7 Sold goods of ₹ 10,000 to Suresh.
10 Purchased Furniture from Mr. Govind on credit ₹ 30,000.
15 Paid for Rent ₹ 3000 and paid by debit card.
21 Purchased goods from Urmila on credit ₹ 70,000.
27 Paid for Transport ₹ 1,000 to United Transport.
30 Paid to Urmila ₹ 20,000 on behalf of Sharmila.
Solution:
In the Journal of Anand General Merchants

Question 2.
Journalise the following transactions in the books of Gajanan.
2019 May
3 Purchased goods for ₹ 90,000 and amounts paid by Bank directly.
7 Sold goods to Satish on credit ₹ 30,000.
9 Paid for Postage ₹ 10,000.
12 Paid for Wages ₹ 15,000.
15 Received cheque of ₹ 30,000 from Satish.
21 Received Dividend ₹ 5000.
25 Purchased Laptop of ₹ 40,000 and paid by cheque.
28 Deposited cash ₹ 10,000 into State Bank of India.
31 Purchased goods for ₹ 40,000 and paid by RTGS.
Solution:
Journal of Gajanan

Question 3.
Journalise the following transactions in the books of Ashok General Stores.
2019 May
1 Received ₹ 5,000 from Ram on behalf of Bharat.
4 Purchased Goods for cash ₹ 55,000.
8 Paid for Salary ₹ 8,000.
12 Purchased goods from Ganesh ₹ 30,000 on credit.
17 Sold goods to Mrs. Neha ₹ 60,000 on credit.
20 Purchased Machinery of ₹ 80,000 @ 12% GST and amount paid by cheque.
25 Paid to SG & Sons by cheque ₹ 30,000.
28 Received Commission ₹ 10,000 from Ganesh.
30 Paid Rent ₹ 5000.
31 Purchased Shares of Atul Company Ltd. for ₹ 10,000 through Demat account.
Solution:
Journal of Ashok General Stores


Working Note:
Dated 20th May 2019:
Calculation of CGST and SGST on Machinery
Catalogue price of ₹ 80,000 @ 12%.
CGST i.e. Central Goods and Services Tax = 12% × \(\frac{1}{2}\) = 6%.
CGST = Price of Machinery × 6%.
= 80,000 × \(\frac{6}{100}\)
= ₹ 4,800.
SGST i.e. State Goods and Service Tax = 12% × \(\frac{1}{2}\) = 6%.
SGST = 80,000 × \(\frac{6}{100}\) = ₹ 4,800.

Question 4.
Journalise the following transactions in the books of Sanjay General Stores.
2019 June
1 Started business with cash ₹ 50,000, Bank ₹ 1,00,000, Goods worth ₹ 50,000.
5 Purchased goods from Mohan on credit ₹ 80,000 at 10% Trade Discount.
9 Sold goods to Urmila ₹ 30,000 at 5% Trade Discount.
12 Paid into Dena Bank ₹ 40,000.
15 Goods worth ₹ 5000 were distributed as a free samples.
22 Paid for Commission ₹ 5,000 to Anand.
24 Received ₹ 28,000 from Urmila in full settlement of her account by Debit Card.
29 Paid for Advertisement ₹ 9,000.
30 Purchased Laptop for ₹ 20,000 @ 28% GST and amount paid by NEFT.
Solution:
Journal of Sanjay General Stores

Working Note:
5th June 2019:
Calculation of Trade discount
Purchased goods for ₹ 80,000 @ 10% Trade discount
Trade discount = 80,000 × \(\frac{10}{100}\) = ₹ 8,000
Net Purchase Price = 80,000 – 8,000 = ₹ 72,000

24th June 2019:
Discount allowed to Urmila = Amount due – Amount received
= 28,500 – 28,000
= ₹ 500

30th June 2019:
Calculation of CGST and SGST
Price of Laptop = ₹ 20,000 @ 28% GST.
CGST = (Price of Laptop) × Rate of CGST
= 20,000 × \(\frac{14}{100}\)
= ₹ 2,800
SGST = (Price of Laptop) × Rate of SGST
= 20,000 × \(\frac{14}{100}\)
= ₹ 2,800

Question 5.
Journalise the following transactions in the books of Kunal Stores.
2018 August
1 Purchased goods of ₹ 90,000 at 10% Trade Discount and 10% Cash Discount from Rakesh and 1/3rd amount paid by cheque.
5 Opened current account in State Bank of India by depositing ₹ 60,000.
8 Cash purchases ₹ 85,000.
10 Goods sold on credit to Tushar ₹ 20,000 @ 10% Trade Discount.
12 Paid Salary ₹ 4,000.
16 Tushar returned goods of ₹ 250.
17 Goods taken by Kunal for his private use ₹ 2,000.
20 Purchased Laptop of ₹ 40,000 from Joshi Electronics @ 18% GST and paid by cheque.
22 Rent paid by cheque ₹ 15,000.
25 Purchased Motor car worth ₹ 2,00,000 for cash @ 18% GST and paid by Bank.
26 Goods distributed as free sample ₹ 4,000.
28 Purchased goods from Amit of ₹ 60,000 on credit.
30 Paid by ECS cash to Amit ₹ 58,500, who allowed us a discount of ₹ 1,500.
30 Sold goods ₹ 5,000 at a loss of ₹ 1,000
31 Sold goods for ₹ 20,000.
Solution:
Journal of Kunal Stores


Working Notes:
1. 2018 Aug. 1st:
Trade discount = 10% on Purchase catalogue price
= \(\frac{10}{100}\) × 90,000
= ₹ 9,000
Net Purchase price = 90,000 – 90,00 = ₹ 81,000
Amount paid = \(\frac{1}{3}\) × 81,000 = ₹ 27,000
Cash discount = 10% on 27,000
= \(\frac{10}{100}\) × 27,000
= ₹ 2700
Amount paid by cheque = 27,000 – 2700 = ₹ 24,300

2. Aug. 10th:
Net price of Goods sold to Tushar = 20,000 – 10% Trade discount
= 20,000 – \(\frac{10}{100}\) × 20,000
= 20,000 – 2,000
= ₹ 18,000

3. 20th Aug. 2018:
Calculation of GST
CGST = 9% on ₹ 40,000
= \(\frac{9}{100}\) × 40,000
= ₹ 3,600
SGST = 9% on ₹ 40,000
= \(\frac{9}{100}\) × 40,000
= ₹ 3,600

4. 25th Aug. 2018:
Calculation of GST
CGST = 9% on ₹ 2,00,000
= \(\frac{9}{100}\) × 2,00,000
= ₹ 18,000
SGST = 9% on ₹ 2,00,000
= \(\frac{9}{100}\) × 2,00,000
= ₹ 18,000

Question 6.
Journalise the following transactions in the books of Nina General Stores.
2018 Sept
1 Sold goods of ₹ 50,000 at 10% Trade Discount and 10% Cash Discount to Raj and received 50% by cheque and 20% by cash.
3 Bought goods worth ₹ 60,000 from Prashant at 7.5% Trade Discount and half amount paid by cash.
5 Returned goods worth ₹ 550 to Prashant.
7 Sold goods worth ₹ 90,000 to Ranvir on credit at 10% Trade Discount.
12 Received Commission ₹ 4,500.
15 Received cheque of ₹ 80,000 from Ranvir in full settlement of his account.
18 Purchased Computer worth ₹ 80,000 from Reliance Company by cheque at 28% GST.
22 Wages paid ₹ 13,000.
23 Paid for Life Insurance premium ₹ 17,000.
27 Sold goods worth ₹ 28,000 to Tushar who paid us ₹ 18,000 immediately
Solution:
Journal of Nina General Stores


Working Notes:
1. 1st Sept. 2018:
Selling (invoice) price = ₹ 50,000
Trade Discount = 10% on ₹ 50,000
= \(\frac{10}{100}\) × 50,000
= ₹ 5,000
Net selling price = 50,000 – 5,000 = ₹ 45,000
50% of the Net selling price received by cheque
Amount of cheque received = 50% of Net selling price – Cash discount
= \(\frac{50}{100}\) × 45,000 – 10% on ₹ 22,500
= 22,500 – 2,250
= ₹ 20,250
20% of Net selling price received by cash.
Net amount of cash received = 20% of Net selling price – Cash discount
= \(\frac{20}{100}\) × 45,000 – Cash discount
= 9,000 – \(\frac{10}{100}\) × 9000
= 9,000 – 900
= ₹ 8,100
30% of Net Selling price is not received
∴ Amount not received = \(\frac{30}{100}\) × 45,000 = ₹ 13,500
Total of cash discount = 2,250 + 900 = ₹ 3,150

2. 3rd Sept. 18:
Trade discount = 7.5% on 60,000
= \(\frac{7.5}{100}\) × 60,000
= ₹ 4,500
Net Purchase price = 60,000 – 4,500 = ₹ 55,500

3. 18th Sept. 18:
Calculation of GST
CGST = (Purchase Price of Computer) × 14%
= 80,000 × \(\frac{14}{100}\)
= ₹ 11,200
SGST = 80,000 × \(\frac{14}{100}\) = ₹ 11,200
Net Purchase price of Computer = 80,000 + 11,200 + 11,200 = ₹ 1,02,400

Question 7.
Journalise the following transactions in the books of Varun
2018 Oct.
1 Purchased Machinery of ₹ 95,000 and paid ₹ 5,000 for freight.
3 Purchased goods for ₹ 1,50,000 and amount paid by Bank.
6 Purchased Laptop from Nagesh & Co. worth ₹ 1,80,000 @ 18% GST.
10 Paid into Bank of Baroda ₹ 70,000.
12 Paid for Rent ₹ 4,000 and Commission ₹ 3,000.
15 Bought goods from Tushar Company Ltd. ₹ 1,20,000 at 12% GST and paid 1/2 amount by RTGS.
16 Cash purchases ₹ 50,000 amount paid by cheque.
20 Invoiced goods to Satish ₹ 80,000 at 12% GST and the amount received by cheque.
25 Paid for Telephone charges ₹ 90,000
27 Mrs. Varsha bought goods from us ₹ 90,000 at a 12% Trade Discount.
28 Purchased goods from Abhijeet & Sons ₹ 1,50,000 at 18% GST.
30 Paid to Abhijeet & Sons and received 10% Cash Discount by cheque.
31 Paid for Advertisement ₹ 8,000 and Brokerage ₹ 12,000.
Solution:
Journal of Varun

Question 8.
Journalise the following transactions in the books of Dhoni Auto Car Centre
2018 Nov
1 Sold 1,000 shares for ₹ 100 each and paid brokerage @ 1% and the amount credited to our account.
4 Purchased goods from Ashish & Co. of ₹ 2,00,000.
6 Sold goods to Virat & Co. ₹ 1,50,000.
8 Paid for Advertisement ₹ 30,000 to Times of India.
10 Paid for Printing and Stationery ₹ 7,000.
12 Purchased goods from Prakash & Co. ₹ 1,50,000 @ 18% GST.
15 Paid for Transport charges ₹ 10,000 @12% GST.
20 Purchased goods from Vikram & Sons ₹ 1,20,000 @ 18% GST and paid half the amount immediately.
25 Paid to Prakash & Co. less 10% discount.
30 Invoiced Goods to Sanjay ₹ 60,000.
31 Sanjay returned goods of ₹ 10,000.
31 Sanjay became insolvent and recovered only 50 paise in a rupee as a final settlement from him.
Solution:
Journal of Dhoni Auto Car Centre


Working Notes:
1. 2018, Nov. 1:
Amount credited to Bank A/c = Sales proceed – Brokerage @ 1%
= 1,000 × 100 – \(\frac{1}{100}\) × 1,000 × 100
= 1,00,00 – 1,000
= ₹ 99,000

2. Nov. 12:
Net Purchase price = Purchase price + 9% CGST + 9% SGST
= 1,50,000 + \(\frac{9}{100}\) × 1,50,000 + \(\frac{9}{100}\) × 1,50,000
= 1,50,000 + 13,500 + 13,500
= ₹ 1,77,000

3. Nov. 20:
Net Purchase price = Purchase price + 9% CGST + 9% SGST
= 1,20,000 + \(\frac{9}{100}\) × 1,20,000 + \(\frac{9}{100}\) × 1,20,000
= 1,20,000 + 10,800 + 10,800
= ₹ 1,41,600
Amount paid = \(\frac{1}{2}\) of 1,41,600 = ₹ 70,800 and credit purchased = ₹ 70,800.

4. Nov. 25:
Cash discount = Amount due to Prakash & Co × 10%
= 1,77,000 × \(\frac{10}{100}\)
= ₹ 17,700

5. Bad debts = Amount due from Sanjay – Amount recovered
= 50,000 – 50% of 50,000
= 50,000 – 25,000
= ₹ 25,000

Question 9.
Journalise the following transactions in the books of Hero Enterprises.
Balance on 1st April 2019
Cash at Bank ₹ 80,000, Sundry Debtors Ram ₹ 20,000, Rahim ₹ 30,000, Stock ₹ 55,000, Building ₹ 1,50,000.
Credit Balances on 1st April 2019
Sundry Creditors Swapna ₹ 20,000, Rohit ₹ 30,000, Bank Loan ₹ 50,000.
2019 April
1 Purchased goods worth ₹ 1,50,000 from Prashant & Co., less 10% Trade Discount.
4 Sold goods to Mr. Amit Sharma ₹ 70,000 at 10% Trade Discount on credit.
9 Purchased goods for cash ₹ 2,00,000 @ 28% GST amount paid by NEFT.
12 Sold Goods to Aditya Ray of ₹ 90,000 @ 28% GST.
15 Paid for Rent ₹ 5000 and Salary ₹ 18,000.
17 Paid for Proprietor’s house Rent ₹ 12,000.
20 Sold half of the goods purchased on 9th April at 20% Profit and 28% GST.
25 Paid for Wages ₹ 1,500.
25 Purchased Furniture ₹ 1,80,000 @ 12% GST and amount paid by RTGS.
28 Sold an old Furniture of ₹ 20,000 for ₹ 12,000.
30 Sold shares of ₹ 10,000 for ₹ 15,000 and the amount received by cheque.
Solution:
Journal of Hero Enterprises


Working Notes:
1. 2019 April, 20:
Cost of Goods sold = \(\frac{1}{2}\) of purchases on 9th April, 2019
= \(\frac{1}{2}\) × 2,00,000
= ₹ 1,00,000
Selling price of Goods sold = \(\frac{120}{100}\) × ₹ 1,00,000 = ₹ 1,20,000

2. April, 28:
Loss on sale of furniture = Cost of furniture – Selling price
= 20,000 -12,000
= ₹ 8,000

3. April, 30:
Profit on sale of shares = Selling price – Cost of shares
= 15,000 – 10,000
= ₹ 5,000

Question 10.
Journalise the following transactions in the books of Harbhajan & Co. for the month of 1st April 2019.
Balance on 1st April 2019
Cash in hand ₹ 35,000, Cash at Bank ₹ 25,000, Furniture ₹ 1,50,000, Laptop ₹ 1,00,000
Debtors: Sangita ₹ 40,000, Viru ₹ 30,000
Creditors: Ganesh ₹ 10,000, Garima ₹ 40,000, Bank loan ₹ 50,000.
2019 April
1 Purchased goods from Ajay Kumar worth ₹ 2,50,000 at 10% Trade discount @ 18% GST and paid 1/4 amount in Cash.
5 Purchased shares of Infosys Company ₹ 50,000 and ₹ 500 paid as a brokerage for Demat A/c.
8 Sold goods to Raj worth ₹ 90,000 at 10% Trade discount and 1/3 amount received by cash and 5% cash discount is allowed.
12 Paid house rent of proprietor ₹ 9,000 and office rent ₹ 5,000.
15 Purchased Laptop of ₹ 60,000 @ 18% GST and paid amount by cheque.
20 Paid transport charges on the above Laptop ₹ 1,000 @ 18% GST.
25 Paid Commission ₹ 20,000 to Ram.
26 Paid Telephone Charges ₹ 1,000.
28 Transferred from private Bank A/c of proprietor to business Bank A/c ₹ 40,000.
30 Bought goods for ₹ 1,50,000 @ 12% as GST by cheque.
30 Exchanged our Furniture of ₹ 30,000 against a Motor car of the same value for business.
Solution:
Journal of Harbhajan & Co.


Working Notes:
1. 8 April 2019:
Net selling price = Selling price – 10% Trade discount
= 90,000 – 10% on 90,000
= 90,000 – 9,000
= ₹ 81,000
Cash received = \(\frac{1}{3}\) of 81,000
= \(\frac{1}{3}\) × 81,000
= ₹ 27,000
Net cash received = 27,000 – Cash discount @ 5%
= 27,000 – \(\frac{5}{100}\) × 27,000
= 27,000 – 1,350
= ₹ 25, 650
Credit sale = \(\frac{2}{3}\) of 81,000
= \(\frac{2}{3}\) × 81,000
= ₹ 54,000

Balbharti Maharashtra State Board 12th Chemistry Textbook Solutions Chapter 5 Electrochemistry Textbook Exercise Questions and Answers.

Maharashtra State Board Class 12 Chemistry Solutions Chapter 5 Electrochemistry

1. Choose the most correct option.

Question i.
Two solutions have the ratio of their concentrations 0.4 and ratio of their conductivities 0.216. The ratio of their molar conductivities will be
(a) 0.54
(b) 11.574
(c) 0.0864
(d) 1.852
Answer:
(a) 0.54

Question ii.
On diluting the solution of an electrolyte,
(a) both ∧ and κ increase
(b) both ∧ and κ decrease
(c) ∧ increases and κ decreases
(d) ∧ decreases and κ increases
Answer:
(c) ∧ increases and κ decreases

Question iii.
1 S m² mol^-1 is equal to
(a) 10^-4 S m² mol^-1
(b) 10⁴ Ω^-1 cm² mol^-1
(c) 10^-2 S cm² mol^-1
(d) 10² Ω^-1 cm² mol^-1
Answer:
(b) 10⁴Ω^-1 cm² mol^-1

Question iv.
The standard potential of the cell in which the following reaction occurs
H₂₊ (g, 1 atm) + Cu²⁺ (1 M) → 2H (1 M) + Cu_(s), (\(E_{\mathrm{Cu}}^{0}\) = 0.34 V) is
(a) – 0.34 V
(b) 0.34 V
(c) 0.17 V
(d) -0.17 V
Answer:
(b) 0.34 V

Question v.
For the cell, Pb_(s)|Pb²⁺ (1 M)|| Ag⁺ (1 M)|Ag_(s), if concentration of an ion in the anode compartment is increased by a factor of 10, the emf of the cell will
(a) increase by 10 V
(b) increase by 0.0296 V
(c) decrease by 10 V
(d) decrease by 0.0296 V
Answer:
(d) decrease by 0.0296 V

Question vi.
Consider the half reactions with standard potentials

The strongest oxidising and reducing agents respectively are
(a) Ag and Fe²⁺
(b) Ag⁺ and Fe
(c) Pb²⁺ and I^–
(d) I₂ and Fe²⁺
Answer:
(b) Ag⁺ and Fe

Question vii.
For the reaction
Ni_(s) + Cu²⁺ (1 M) → Ni²⁺ (1 M) + Cu_(s), \(E_{\text {cell }}^{0}\) = 0.57 V. Hence ΔG⁰ of the reaction is
(a) 110 kJ
(b) -110 kJ
(c) 55 kJ
(d) -55 kJ
Answer:
(b) -110 kJ

Question viii.
Which of the following is not correct ?
(a) Gibbs energy is an extensive property
(b) Electrode potential or cell potential is an intensive property.
(c) Electrical work = -ΔG
(d) If half reaction is multiplied by a numerical factor, the corresponding E⁰ value is also multiplied by the same factor.
Answer:
(d) If half reaction is multiplied by a numerical factor, the corresponding E⁰ value is also multiplied by the same factor.

Question ix.
The oxidation reaction that takes place in lead storage battery during discharge is

Answer:
(c) \(\mathrm{Pb}_{(\mathrm{s})}+\mathrm{SO}_{4(\mathrm{aq})}{ }^{2-} \longrightarrow \mathrm{PbSO}_{4(\mathrm{~s})}+2 \mathrm{e}^{-}\)

Question x.
Which of the following expressions represent molar conductivity of Al₂(SO₄)₃ ?

Answer:
(b) \(2 \lambda_{\mathrm{Al}^{3+}}^{0}+3 \lambda_{\mathrm{SO}_{4}^{2-}}^{0}\)

2. Answer the following in one or two sentences.

Question i.
What is a cell constant ?
Answer:
(A) Cell constant of a conductivity cell is defined as the ratio of the distance between the electrodes divided by the area of cross section of the electrodes.

In SI units it is expressed as m-1.

Question ii.
Write the relationship between conductivity and molar conductivity and hence unit of molar conductivity.
Answer:
If k is conductivity and ∧_m is molar conductivity then, ∧_m = \(\frac{\kappa \times 1000}{C}\)
Unit of molar conductivity is, Ω^-1 cm² mol^-1 or S cm² mol^-1.

Question iii.
Write the electrode reactions during electrolysis of molten KCl.
Answer:

Question iv.
Write any two functions of salt bridge.
Answer:
The functions of a salt bridge are :

• It maintains the electrical contact between the two electrode solutions of the half cells.
• It prevents the mixing of electrode solutions.
• It maintains the electrical neutrality in both the solutions of two half cells by a flow of ions.
• It eliminates the liquid junction potential.

Question v.
What is standard cell potential for the reaction
3Ni_(s) + 2Al³⁺ (1M) → 3NI²⁺ (1M) + 2Al_(s)
if \(\boldsymbol{E}_{\mathrm{Ni}}^{0}\) = – 0.25 V and \(\boldsymbol{E}_{\mathrm{Al}}^{0}\) = -1.66V?
Solution :
Given : E⁰_Ni²⁺/Ni = -0.25 V
E⁰_Al³⁺/Al = – 1.66 V; E⁰_cell = ?
Since Ni is oxidised and Al³⁺ is reduced,
\(E_{\text {cell }}^{0}=E_{\mathrm{Al}^{3+} / \mathrm{Al}}^{0}-E_{\mathrm{Ni}^{2+} / \mathrm{Ni}}^{0}\)
= – 1.66 – (-0.25)
= – 1.41 V
Ans. \(E_{\text {cell }}^{0}\) = -1.41 V
[Note : Since \(E_{\text {cell }}^{0}\) is negative, the given reaction is not possible but reverse reaction is possible.]

Question vi.
Write Nerst equation. What part of it represents the correction factor for nonstandard state conditions ?
Answer:
(1) Nernst equation for cell potential is,

(2) The part of equation namely,

represents the correction factor for nonstandard state conditions.

Question vii.
Under what conditions the cell potential is called standard cell potential ?
Answer:
In the standard cell, the active masses of the substances taking part in the electrochemical reaction have unit value, i.e., 1 M solution or ions and 1 atm gas.

Question viii.
Formulate a cell from the following electrode reactions :
\(\mathbf{A u}_{(\mathrm{aq})}^{3+}+\mathbf{3 e}^{-} \longrightarrow \mathbf{A} \mathbf{u}_{(\mathrm{s})}\)
\(\mathbf{M g}_{(\mathbf{s})} \longrightarrow \mathbf{M g}_{(\mathrm{aq})}^{2+}+\mathbf{2 e}^{-}\)
Answer:
An electrochemical cell from above electrode reactions is,

Question ix.
How many electrons would have a total charge of 1 coulomb ?
Answer:
Given : 1 Faraday = charge on 1 mol of electrons
= 6.022 × 10²³ electrons and 1 Faraday = 96500 C
∵ 96500 C = 6.022 × 10²³ electrons 6 022 × 10²³
∴ 1 C ≡ \(\frac{6.022 \times 10^{23}}{96500}\) = 6.24 × 10¹⁸ electrons
Ans. Number of electrons = 6.24 × 10¹⁸

Question x.
What is the significance of the single vertical line and double vertical line in the formulation galvanic cell.
Answer:
(i) Consider representation of Daniell cell,

Single vertical line represents separation of two phases, solid Zn_(s) and solution of ions.
(ii) Double vertical lines represent a salt bridge.

3. Answer the following in brief

Question i.
Explain the effect of dilution of solution on conductivity ?
Answer:

• The conductance of a solution is due to the presence of ions in the solution. More the ions, higher is the conductance of the solution.
• Conductivity or the specific conductance is the conductance of unit volume (1 cm³) of the electrolytic solution.
• The conductivity of the electrolytic solution always decreases with the decrease in the concentration of the electrolyte or the increase in dilution of the solution.
• On dilution, the concentration of the solution decreases, hence the number of (current carrying) ions per unit volume decreases. Therefore the conductivity of the solution decreases, with the decrease concentration or increase in dilution. (It is to be noted here that, molar conductivity increases with dilution.)

Question ii.
What is a salt bridge ?
Answer:
A salt bridge is a U-shaped glass tube containing a saturated solution of a strong electrolyte, like KCl, NH₄NO₃, Na₂SO₄ in a solidified agar-agar gel. A hot saturated solution of these electrolytes in 5% agar solution is filled in the U-shaped tube and allowed it to cool and solidify forming a gel.

Fig. 5.9 : Salt bridge
It is used to connect two half cells or electrodes forming a galvanic or voltaic cell.

Question iii.
Write electrode reactions for the electrolysis of aqueous NaCl.
Answer:
Reactions in electrolytic cell :
(i) Reduction half reaction at cathode : There are Na⁺ and H⁺ions but since H⁺ are more reducible than Na⁺, they undergo reduction liberating hydrogen and Na⁺ are left in the solution.
2H₂O_(l) + 2e^– → H_2(g) + 2OH^–_(aq) (reduction) E⁰ = -0.83 V

(ii) Oxidation half reaction at anode : At anode there are Cl^– and OH^–. But Cl^– ions are preferably oxidised due to less decomposition potential.

Net cell reaction : Since two electrons are gained at cathode and two electrons are released at anode for each redox step, the electrical neutrality is maintained. Hence we can write,

Since Na⁺ and OH^– are left in the solution, they form NaOH_(aq).

Question iv.
How many moles of electrons are passed when 0.8 ampere current is passed for 1 hour through molten CaCl₂ ?
Answer:
Given : I = 0.8 A; t = 1 × 60 × 60 = 3600 s
Number of moles of electrons = ?
Q = I × t
= 0.8 × 3600
= 2880 C
1 Faraday = 1 mol electrons
1 Faraday = 96500 C
∵ 96500 C = 1 mol electrons
∴ 2880 C ≡ \(\frac{2880}{96500}\)
= 0.02984 mol electrons
Ans. Number of moles of electrons = 0.02984

Question v.
Construct a galvanic cell from the electrodes Co³⁺|Co and Mn²⁺|Mn. \(\boldsymbol{E}_{\mathrm{Co}}^{0}\) = 1.82 V,
\(\boldsymbol{E}_{\mathrm{Mn}}^{0}\) = – 1.18V. Calculate \(\boldsymbol{E}_{\text {cell }}^{0}\).
Answer:

Question vi.
Using the relationsip between ∆G⁰ of cell reaction and the standard potential associated with it, how will you show that the electrical potential is an intensive property ?
Answer:
(1) For an electrochemical cell involving n number of electrons in the overall cell reaction,
ΔG⁰ = -nF\(E_{\text {cell }}^{0}\)
where ΔG⁰ is standard Gibbs energy change and \(E_{\text {cell }}^{0}\) is a standard cell potential.
(2) ∴ \(E_{\mathrm{cell}}^{0}=\frac{-\Delta G^{0}}{n F}\)
Since ΔG⁰ changes according to number of moles of electrons involved in the cell reaction, the ratio, ΔG⁰/nF remains constant.
(3) Therefore \(E_{\text {cell }}^{0}\) is independent of the amount of substance and it represents the intensive property.

Question vii.
Derive the relationship between standard cell potential and equilibrium constant of cell reaction.
Answer:
For any galvanic cell, the overall cell reaction at equilibrium can be represented as,
Reactants ⇌ Products.
[For example for Daniell cell,
\(\mathrm{Zn}_{(s)}+\mathrm{Cu}_{(\mathrm{aq})}^{2+} \rightleftharpoons \mathrm{Zn}_{(\mathrm{aq})}^{2+}+\mathrm{Cu}_{(\mathrm{s})}\) ]
The equilibrium constant, K for the reversible reaction will be, \(K=\frac{[\text { Products }]}{[\text { Reactants }]}\)

The equilibrium constant is related to the stan-dard free energy change Δ G⁰, as follows,
ΔG⁰ = -RTlnK
If \(E_{\text {cell }}^{0}\) is the standard cell potential (or emf) of the galvanic cell, then ΔG⁰ = -nFE⁰_cell
By comparing above equations,

Question viii.
It is impossible to measure the potential of a single electrode. Comment.
Answer:
(1)

Fig 5.12(a) : Measurement of single electrode potential

Fig 5.12(b) : Measurement of cell potential
According to Nemst theory, electrode potential is the potential difference between the metal and ionic layer around it at equilibrium, i.e. the potential across the electric double layer.

(2) For measuring the single electrode potential, one part of the double layer, that is metallic layer can be connected to the potentiometer but not the ionic layer. Hence, single electrode potential can’t be measured experimentally.

(3) When an electrochemical cell is developed by combining two half cells or electrodes, they can be connected to the potentiometer and the potential difference or cell potential can be measured.
E_cell = E₂ – E₁
where E₁ and E₂ are reduction potentials of two electrodes.

Question ix.
Why do the cell potential of lead accumulators decrease when it generates electricity ? How the cell potential can be increased ?
Answer:
Working of a lead accumulator :
(1) Discharging : When the electric current is withdrawn from lead accumulator, the following reactions take place :

(2) Net cell reaction :
(i) Thus, the overall cell reaction during discharging is

OR
Pb_(s) + PbO_2(s) + 2H₂SO_4(aq) → 2PbSO_4(s) + 2H₂O_(l)
The cell potential or emf of the cell depends upon the concentration of sulphuric acid. During the operation, the acid is consumed and its concentration decreases and specific gravity decreases from 1.28 to 1.17. As a result, the emf of the cell decreases. The emf of a fully charged cell is about 2.0 V.

(ii) Recharging of the cell : When the discharged battery is connected to external electric source and a higher external potential is applied the cell reaction gets reversed generating H₂SO₄.
Reduction at the -ve electrode or cathode :
PbSO4_(s) + 2e^– → Pb_(s) + \(\mathrm{SO}_{4(\mathrm{aq})}^{2-}\)
Oxidation at the + ve electrode or anode :

The emf of the accumulator depends only on the concentration of H₂SO₄.

Question x.
Write the electrode reactions and net cell reaction in NICAD battery.
Answer:
Reactions in the cell :
(i) Oxidation at cadmium anode :
Cd_(s) + 2OH^–_(aq) → Cd(OH)_2(s) + 2e^–
(ii) Reduction at NiO_2(s) cathode :
NiO_2(s) + 2H₂O_(l) + 2e^– → Ni(OH)_2(s) + 2OH^–_(aq)
The overall cell reaction is the combination of above two reactions.
Cd_(s) + NiO_2(s) + 2H₂O_(l) → Cd(OH)_2(s) + Ni(OH)_2(s)

4. Answer the following :

Question i.
What is Kohrausch law of independent migration of ions? How is it useful in obtaining molar conductivity at zero concentration of a weak electrolyte ? Explain with an example.
Answer:
(A) Statement of Kohlrausch’s law : This states that at infinite dilution of the solution, each ion of an electrolyte migrates independently of its co-ions and contributes independently to the total molar conductivity of the electrolyte, irrespective of the nature of other ions present in the solution.

(B) Explanation : Both the ions, cation and anion of the electrolyte make a definite contribution to the molar conductivity of the electrolyte at infinite dilution or zero concentration (∧₀).
If \(\lambda_{+}^{0}\) and \(\lambda_{-}^{0}\) are the molar ionic conductivities of cation and anion respectively at infinite dilution, then
∧₀ = \(\lambda_{+}^{0}\) + \(\lambda_{-}^{0}\).
This is known as Kohlrausch’s law of independent migration of ions.
For an electrolyte, B_xA_y giving x number of cations and y number of anions,
∧₀ = x\(\lambda_{+}^{0}\) + y\(\lambda_{-}^{0}\).

(C) Applications of Kohlrausch’s law :
(1) With this law, the molar conductivity of a strong electrolyte at zero concentration can be determined. For example,
\(\wedge_{0(\mathrm{KCl})}=\lambda_{\mathrm{K}^{+}}^{0}-\lambda_{\mathrm{Cl}^{-}}^{0}\)
(2) ∧₀ values of weak electrolyte with those of strong electrolytes can be obtained. For example,

Molar conductivity of a weak electrolyte at infinite dilution or zero concentration cannot be measured experimentally.
Consider the molar conductivity (∧₀) of a weak acid, CH₃COOH at zero concentration. By Kohlrausch s law,

where λ⁰CH₃COO^– and λ⁰_H⁺ are the molar ionic conductivities of CH₃COO^– and H⁺ ions respectively.
If ∧_0CH3COONa, ∧_0HCl and ∧_0NaCl are molar conductivities of CH₃COONa, HCl and NaCl respectively at zero concentration, then by
Kohlrausch’s law,

Hence, from ∧₀ values of strong electrolytes, ∧₀ of a weak electrolyte CH₃COOH, at infinite dilution can be calculated.

Question ii.
Explain electrolysis of molten NaCl.
Answer:
(1) Construction of an electrolytic cell : It consists of a vessel containing molten (fused) NaCl. Two graphite (carbon) inert electrodes are dipped in it, and connected to an external source of direct electric current (battery). The electrode connected to a negative terminal of the battery is a cathode and that connected to a positive terminal is an anode.

(2) Working of the cell :
(A) In the external circuit, the electrons flow through the wires from anode to cathode of the cell.
(B) The fused NaCl dissociates to form cations (Na⁺) and anions (Cl^–).
\(\mathrm{NaCl}_{\text {(fused) }} \longrightarrow \mathrm{Na}_{(\mathrm{l})}^{+}+\mathrm{Cl}_{(\mathrm{l})}^{-}\)
Na⁺ migrate towards cathode and Cl^– migrate towards anode.

Fig. 5.7 : Electrolysis of fused sodium chloride

(C) Reactions in electrolytic cell :
(i) Reduction half reaction at cathode : The Na⁺ ions get reduced by accepting electrons from a cathode supplied by a battery and form metallic sodium.
\(\mathrm{Na}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Na}_{(\mathrm{s})} \text { (reduction) }\)

(ii) Oxidation half reaction at anode : The Cl^– ions get oxidised by giving up electrons to the anode forming neutral Cl atoms in the primary process, and these Cl atoms combine forming Cl₂ gas in the secondary process.

The released electrons in the anodic oxidation half reaction return to battery through the metallic wires.

Net cell reaction : In order to maintain the electrical neutrality, the number of electrons gained at cathode must be equal to the number of electrons released at anode. Hence the reduction half reaction is multiplied by 2 and both reactions, oxidation half reaction and reduction half reaction are added to obtain a net cell reaction.

Results of electrolysis :

• A molten silvery white Na is formed at cathode which floats on the surface of molten NaCl.
• A pale green Cl₂ gas is liberated at anode.

Question iii.
What current strength in amperes will be required to produce 2.4g of Cu from CuSO₄ solution in 1 hour ? Molar mass of Cu = 63.5 g mol^-1.
Answer:
Given : W_Cu = 2.4 g; t = 1 hr = 1 × 60 × 60 s
M_Cu = 63.5 g mol^-1; I = ?

Ans. Current strength = I = 2.026 A

Question iv.
Equilibrium constant of the reaction,
2Cu⁺_(aq) → Cu²⁺_(aq) + Cu_(s)
is 1.2 × 10⁶. What is the standard potential of the cell in which the reaction takes place ?
Answer:
For the cell reaction, n = 1

Question v.
Calculate emf of the cell
Zn_(s)|Zn²⁺ (0.2M)||H⁺(1.6M)|H₂(g, 1.8 atm)|Pt at 25°C.
Answer:
Given : Zn_(s)|Zn²⁺(0.2M)||H⁺(1.6M)|H₂(g, 1.8 atm)|Pt


= 0.763 – 0.0296 × (- 0.8521)
= 0.763 + 0.02522
= 0.7882
Ans. \(E_{\text {cell }}^{0}\) = 0.7882 V

Question vi.
Calculate emf of the following cell at 25°C.
Zn_(s)| Zn²⁺(0.08M)||Cr³⁺(0.1M)|Cr
E⁰_Zn = – 0.76 V, E⁰_Cr = – 0.74 V
Answer:

Question vii.
What is a cell constant ? What are its units? How is it determined experimentally?
Answer:
(A) Cell constant of a conductivity cell is defined as the ratio of the distance between the electrodes divided by the area of cross section of the electrodes.

In SI units it is expected as m^-1.

The resistance of an electrolytic solution is measured by using a conductivity cell and Wheatstone

Fig. 5.6 : Measurement of conductance
The measurement of molar conductivity of a solution involves two steps as follows :
Step I : Determination of cell constant of the conductivity cell :
KCl solution (0.01 M) whose conductivity is accurately known (κ = 0.00141 Ω^-1 cm^-1) is taken in a beaker and the conductivity cell is dipped. The two electrodes of the cell are connected to one arm while the variable known resistance (R) is placed in another arm of Wheatstone bridge.

A current detector D’ which is a head phone or a magic eye is used. J is the sliding jockey (contact) that slides on the arm AB which is a wire of uniform cross section. A source of A.C. power (alternating power) is used to avoid electrolysis of the solution.

By sliding the jockey on wire AB, a balance point (null point) is obtained at C. Let AC and BC be the lengths of wire.

If R_solution is the resistance of KCl solution and R_x is the known resistance then by Wheatstone’s bridge principle,
\(\frac{R_{\text {solution }}}{\mathrm{BC}}=\frac{R_{x}}{\mathrm{AC}}\)
∴ \(R_{\text {solution }}=\mathrm{BC} \times \frac{R_{x}}{\mathrm{AC}}\)
Then the cell constant ‘ b ’ of the conductivity cell is obtained by, b = κ_Kcl × R_solution.

Step II : Determination of conductivity of the given solution :
KCl solution is replaced by the given electrolytic solution and its resistance (R_s) is measured by Wheatstone bridge method by similar manner by obtaining a null point at D.
The conductivity (κ) of the given solution is,
κ = \(\frac{\text { cell constant }}{R_{\mathrm{s}}}=\frac{b}{R_{\mathrm{s}}}\)

Step III: Calculation of molar conductivity :
The molar conductivity (∧_m) is given by,

Since the concentration of the solution is known, ∧_m can be calculated.

Question viii.
How will you calculate the moles of electrons passed and mass of the substance produced during electrolysis of a salt solution using reaction stoichiometry.
Answer:
Calculation of moles of electrons passed : The charge carried by one mole of electrons is referred to as one faraday (F). If total charge passed is Q C, then moles of electrons passed = \(\frac{Q(\mathrm{C})}{F\left(\mathrm{C} / \mathrm{mol} \mathrm{e}^{-}\right)}\)

Calculation of mass of product : Mass, W of product formed is given by,
W = moles of product × molar mass of product (M)
= \(\frac{Q}{96500}\) × mole ratio × M
= \(\frac{I \times t}{96500}\) × mole ratio × M 96500
When two electrolytic cells containing different electrolytes are connected in series so that same quantity of electricity is passed through them, then the masses W₁ and W₂ of products produced are given by,

Question ix.
Write the electrode reactions when lead storage cell generates electricity. What are the anode and cathode and the electrode reactions during its recharging?
Answer:
Recharging of the cell : When the discharged battery is connected to external electric source and a higher external potential is applied the cell reaction gets reversed generating H₂SO₄.
Reduction at the – ve electrode or cathode :
\(\mathrm{PbSO}_{4(\mathrm{~s})}+2 \mathrm{e}^{-} \rightarrow \mathrm{Pb}^{(s)}+\mathrm{SO}_{4(\mathrm{aq})}^{2-}\)
Oxidation at the + ve electrode or anode :

The emf of the accumulator depends only on the concentration of H₂SO₄.

Question x.
What are anode and cathode of H₂-O₂ fuel cell ? Name the electrolyte used in it. Write electrode reactions and net cell reaction taking place in the fuel cell.
Answer:
Construction :
(i) In fuel cell the anode and cathode are porous electrodes with suitable catalyst like finely divided platinum.

(iii) H₂ is continuously bubbled through anode while O, gas is bubbled through cathode.

Working (cell reactions) :
(i) Oxidation at anode : At anode, hydrogen gas is oxidised to H₂O.
2H_2(g) + 4OH^–_(aq) → 4H₂O_(l) + 4e^– (oxidation half reaction)
(ii) Reduction at cathode : The electrons released at anode travel to cathode through external circuit and reduce oxygen gas to OH^–.
O_2(g) + 2H₂O_(l) + 4e^– → 4OH^–_(aq) (reduction half reaction)

(iii) Net cell reaction: Addition of both the above reactions at anode and cathode gives a net cell reaction.
2H_2(g) + O_2(g) → 2H₂O_(l) (overall cell reaction)

Question xi.
What are anode and cathode for Leclanche’ dry cell ? Write electrode reactions and overall cell reaction when it generates electricity.
Answer:
A dry cell has zinc vessel as anode and graphite rod as cathode and moist paste of ZnCl₂, MnO₂, NH₄Cl as electrolytes.
At anode :
Zn_(s) → \(\mathrm{Zn}_{(\mathrm{aq})}^{2+}\) + 2e^– (Oxidation half reaction)
At graphite (c) cathode :
\(2 \mathrm{NH}_{4(\mathrm{e})}^{+}\) + 2e^– → 2NH_3(aq) + H_2(g) (Reduction half reaction)
2MnO_2(s) + H₂ → Mn₂O_3(s) + H₂O_(l)
There is a side reaction inside the cell, between Zn²⁺ ions and aqueous NH₃.
\(\mathrm{Zn}_{(\mathrm{aq})}^{2+}+4 \mathrm{NH}_{3(\mathrm{aq})} \longrightarrow\left[\mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{4}\right]_{(\mathrm{aq})}^{2+}\)

Question xii.
Identify oxidising agents and arrange them in order of increasing strength under standard state conditions. The standard potentials are given in parenthesis.
Al(- 1.66 V), Cl₂ (1.36 V), Cd²⁺ (-0.4 V), Fe (-0.44 V), I₂ (0.54 V), Br^– (1.09 V).
Answer:
The oxidising agents are I₂, Br^– and Cl₂. The increasing strength is

(Note : Actually Br₂ acts as an oxidising agent but not Br^–.)

Question xiii.
Which of the following species are reducing agents? Arrange them in order of increasing strength under standard state conditions. The standard potentials are given in parenthesis.
K (-2.93V), Br₂(1.09V), Mg(-2.36V), Co³⁺(1.61V), Ti²⁺(-0.37V), Ag⁺(0.8V), Ni (-0.23V).
Answer:
Lower the standard reduction potential, higher is reducing power. The reducing agents are Ni, Mg and K. Their increasing strength is,

(Note : Cations don’t act as reducing agent since they are already in oxidised state.)

Question xiv.
Predict whether the following
reactions would occur spontaneously
under standard state conditions.
a. Ca_(s) + Cd²⁺_(aq) → Ca²⁺_(aq) + Cd_(s)
b. 2 Br-_(s) + Sn²⁺(aq) → Br_2(l) + Sn_(s)
c. 2Ag_(s) + Ni²⁺_(aq) → 2 Ag⁺_(aq) + Ni_(s)
(use information of Table 5.1)
Answer:

12th Chemistry Digest Chapter 5 Electrochemistry Intext Questions and Answers

Question 1.
How does electrical resistance depend on the dimensions of an electronic (metallic) conductor?
Answer:
The electrical resistance of an electronic conductor is linearly proportional to its length (l) and inversely proportional to its cross section area a.

Fig. 5.3 : Electronic conductor
Thus, R ∝ l; R ∝ \(\frac{1}{a}\)
∴ R ∝ \(\frac{l}{a}\) or R = ρ × \(\frac{l}{a}\)
where the proportionality constant p is called specific resistance. IUPAC recommends the term resistivity for specific resistance.

Question 2.
What are the units of resistivity ?
Answer:
For an electronic conductor of length l, and cross section area a, the resistance R is represented as
R = ρ × \(\frac{l}{a}\)
where ρ is the resistivity of the conductor.
∴ ρ = R × \(\frac{a}{l}\)
If l = 1 m, a = 1 m², ρ = R

Hence, resistivity is the resistance of a conductor of volume of 1 m³.
(In C.G.S. units, the units of ρ are ohm cm. Hence, ρ is the resistance of a conductor of unit volume or 1 cm³.)

Question 3.
Define resistivity. What are its units ?
Answer:
Resistivity (or specific resistance) : It is the resistance of a conductor that is 1 m in length and 1 m² in cross section area in SI units. (In C.G.S. units, it is the resistance of a conductor that is 1 cm in length and 1 cm² in cross section area.) Hence, the resistivity is the resistance of a conductor of unit volume. (In case of electrolytic solution, ρ is the resistivity i.e., resistance of a solution of unit volume.)
It has SI units, ohm m and C.G.S. units, ohm cm.

Question 4.
Why is alternating current used in the measurement of conductivity of the solution ?
Answer:
If direct current (D.C.) by battery is used, there will be electrolysis and the concentration of the solution is changed. Hence alternating current (A.C.) with high frequency is used.

Try this… (Textbook page No. 93)

Question 1.
What must be the concentration of a solution of silver nitrate to have the molar conductivity of 121.4 Ω^-1 cm² mol^-1 and the conductivity of 2.428 × 10^-3 Ω^-1 cm^-1 at 25 °C ?
Answer:

∴ Concentration of a Solution = 0.02 M

Try this… (Textbook page No. 96)

Question 1.
Obtain the expression for dissociation constant in terms of ∧_c and ∧₀ using Ostwald’s dilution law.
Answer:
Consider a solution of a weak electrolyte, BA having concentration C mol dm^-3. If α is the degree of dissociation, then by Ostwald’s theory of weak electrolytes,

If K is the dissociation constant of the weak electrolyte, then by Ostwald’s dilution law,
K = \(\frac{\alpha^{2} C}{(1-\alpha)}\)
If ∧_m is the molar conductivity of the electrolyte BA at the concentration C and ∧₀ is the molar conductivity at zero concentration or infinite dilution, then

Hence by measuring ∧_m at the concentration C and knowing ∧₀, the dissociation constant can be calculated.
If \(\lambda_{+}^{0}\) and \(\lambda_{-}^{0}\) are the ionic conductivities, then by Kohlrauseh’s law, ∧₀ = \(\lambda_{+}^{0}\) + \(\lambda_{-}^{0}\).

Learn this as well…

Question 1.
How is the cell constant of a conductivity cell determined?
Answer:
The cell constant of a given conductivity cell is obtained by measuring the resistance (R) (or the conductance) of a standard solution whose conductivity (f_c) is accurately known by using Wheatstone’s bridge (discussed in Q. 37). For this purpose, KCl solution of accurately known conductivity is used.
\(\kappa_{\mathrm{KCl}}=\frac{1}{R_{\mathrm{KCl}}} \times \frac{l}{a}\) where \(\frac{l}{a}\) is a cell constant, represented by b.
∴ \(\kappa_{\mathrm{KCl}}=\frac{b}{R_{\mathrm{KCl}}}\)
or b = κ_KCl × R_KCl
For example, the conductivity of 0.01 M KCl is 0.00141 Ω^-1 cm^-1 (S cm^-1). Hence by measuring R KCl the cell constant b can be obtained.

Try this… (Textbook page No. 95)

Question 1.
Calculate ∧₀ (CH₂ClCOOH) if ∧₀ values for HCl, KCl and CH₂ClCOOK are respectively, 4.261, 1.499 and 1.132 Ω^-1 m² mol^-1.
Solution :

Adding equations (i) and (ii) and subtracting equation (iii) we get equation (I).

Can you tell ? (Textbook page No. 103)

Question 1.
You have learnt Daniel cell in XIth standard. Write notations for anode and cathode. Write the cell formula.
Answer:
Daniel cell is represented as,

Try this… (Textbook page No. 104)

Question 1.
Write electrode reactions and overall cell reaction for Daniel cell you learnt in standard XI.
Answer:
Reactions for Daniell cell:

Question 1.
Describe different types of reversible electrodes with examples. (1 mark for each type)
Answer:
A reversible electrochemical cell or a galvanic cell consists of two reversible half cells or electrodes. There are four types of reversible electrodes according to their compositions.
(1) Metal-metal ion electrode : This electrode is set up by dipping a metal in a solution containing its own ions, e.g. Zn rod dipped into ZnSO₄ solution containing Zn⁺⁺ ions of concentration C.
It is represented as,
\(\mathrm{Zn}^{2+}{ }_{(\mathrm{aq})} \mid \mathrm{Zn}_{(\mathrm{s})}\)
The reduction reaction at the electrode is,
Zn⁺⁺_(aq) + 2e^– → Zn_(s)

(2) Metal-sparingly soluble salt electrode : This electrode consists of a metal coated with one of its sparingly soluble salts and immersed in a solution containing an electrolyte having a common anion as that of the salt. For example, silver electrode coated with sparingly soluble AgCl dipped in KCl solution with common anion Cl^–. This electrode is represented as,
Cl^–_(aq) | AgCl_(s) | Ag_(s)
The reduction reaction is,
AgCl_(s) + e → Ag_(s) + Cl^–_(aq)

(3) Gas electrode : This is developed by bubbling pure and dry gas around a platinised platinum foil dipped in the solution containing ions (of the gas) reversible with respect to the gas bubbled.
The gas is adsorbed on the surface of platinum foil and establishes an equilibrium with its ions in the solution. Pt electrode provides electrical contact and also acts as a catalyst.
Some of the gas electrodes are represented as follows :
(i) Hydrogen gas electrode :
H⁺_(aq) | H₂(g, P_H2) | Pt
Reduction reaction : H⁺_(aq) + e^– → \(\frac {1}{2}\)H_2(g)
(ii) Chlorine gas electrode :
Cl^–_(aq) | Cl₂(g, PCl₂) | Pt
Reduction reaction : \(\frac {1}{2}\)Cl_2(g) + e- → Cl^–_(aq)

(4) Redox electrode (Oxidation reduction electrode) : This electrode consists of a platinum wire dipped in a solution containing the ions of the same metal (or a substance) in two different oxidation states, like Fe²⁺ – Fe³⁺, Sn²⁺ – Sn⁴⁺, Mn⁺⁺ – MnO^–₄, etc.
A platinum electrode which provides an electrical contact and acts as catalyst aquires an equilibrium between two ions in the solution, due to their tendency to undergo a change from one oxidation state to another. The electrodes are represented as,
Fe²⁺_(aq), Fe³⁺_(aq) | Pt
Reduction reaction : Fe³⁺_(aq) + e^– → Fe²⁺_(aq)
SnCl_2(aq), SnCl_4(aq) | Pt
Reduction reaction : Sn⁴⁺_(aq) + 2e^– →Sn²⁺_(aq)

Use your brain power! (Textbook page No. 98)

Question 1.
Distinguish between electrolytic and galvanic cells.
Answer:
Electrolytic cell:

1. This device is used to bring about a non-spontaneous chemical reaction by passing an electric current.
2. It is used to bring about a chemical reaction generally for the dissociation (electrolysis) of compounds.
3. In this cell, electrical energy is converted into chemical energy.
4. In this cell, the cathode is negative and the anode is positive.
5. Electrolytic cells are irreversible.
6. Oxidation takes place at the positive electrode and reduction at the negative electrode.
7. The electrons are supplied by the external source and enter through cathode and come out through anode.
8. It is used for electroplating, electrorefining, etc.

Electrochemical cell (Galvanic cell or Voltaic cell):

1. This device is used to produce electrical energy by a spontaneous chemical reaction.
2. It is used to generate electricity.
3. In this cell, chemical energy is converted into electrical energy.
4. In this cell, the cathode is positive and the anode is negative.
5. Electrochernical cells are reversible.
6. Oxidation takes place at the negative electrode and reduction at the positive electrode.
7. The electrons move from anode to cathode in the external circuit.
8. It is used as a source of electric current.

Try this… (Textbook page No. 107)

Question 1.
Write expressions to calculate equilibrium constant from
i. Concentration data
ii. Thermochemical data
iii. Electrochemical data
Answer:
(i) Consider following a reversible cell reaction.
aA + bB ⇌ cC + dD
If [A], [B], [C] and [D] represent concentrations of reactants and products then the equilibrium constant K is,
K = \(\frac{[\mathrm{C}]^{c} \times[\mathrm{D}]^{d}}{[\mathrm{~A}]^{a} \times[\mathrm{B}]^{b}}\)
(ii) If ΔG⁰ is the standard Gibbs free energy change at temperature T then,
ΔG⁰ = – RTlnK = – 2.303 RTlog₁₀K
(iii) From electrochemical data,
if \(E_{\text {cell }}^{0}\) is the standard cell potential and K is the equilibrium constant for the cell reaction at a temperature T, then,
\(E_{\text {cell }}^{0}=\frac{0.0592}{n} \log _{10} K\)

Learn this as well…

Question 1.
The construction and working of the calomel electrode.
Answer:
(1) Since standard hydrogen electrode (SHE) is not convenient for experimental use, a secondary reference electrode like calomel electrode is used.
(2) Construction : It consists of a glass vessel with side arm B for dipping in a desired solution of another electrode like, ZnSO_4(aq) for an electric contact. The vessel is filled with mercury, a paste of Hg and Hg₂Cl₂ (calomel) and saturated KCl solution.

Fig. 5.15 : Determination of standard electrode potential using calomel electrode
(3) The potential developed depends upon the concentration of Cl^– or KCl solution. When saturated KCl solution is used, its reduction potential is 0.242 V.
(4) Consider following cell :
Zn_(s) | ZnSO_4(aq) || KCl_(aq) | Hg₂Cl_2(s) | Hg
OR Zn_(s) | ZnSO_4(aq) || Calomel electrode
Reduction reaction for calomel electrode :
Hg₂Cl_2(s) + 2e^– → 2Hg_(l) + 2Cl^–_(aq)
Hence potential of calomel electrode depends on the concentration of Cl^– or KCl solution.

Can you tell ? (Textbook page No. 114)

Question 1.
In what ways are fuel cells and galvanic cells similar and in what ways are they different ?
Answer:
Similarity between fuel cells and galvanic cells :

• In both the cells, there is oxidation at anode and j reduction at cathode.
• The cell potential is developed due to net redox reactions.
• Both are galvanic cells.

Difference in fuel cells and galvanic cells :

• Fuel cells involve electrodes with large surface area while galvanic cells involve electrodes with j compact surface area.
• Fuel cells involve gaseous materials on a large scale while galvanic cells involve gaseous materials at a definite pressures along with electrolytes or there may not be gases.
• In fuel cells, the cell potential is developed due to exothermic combustion reactions while in galvanic cell, cell potential is developed due to normal redox reactions.
• In fuel cells gaseous electrode materials are continuously supplied from outside while in galvanic cells electrode materials have constant concentration or may change due to reactions.

Use your brain power (Textbook page No. 114)

Question 1.
Indentify the strongest and the weakest oxidizing agents from the electrochemical series.
Answer:
From the electrochemical series,
(a) The strongest oxidising agent is fluorine since it has the highest standard reduction potential (\(E_{\mathrm{F}_{2} / \mathrm{F}^{-}}^{0}\) = + 2.87 V).
(b) The weakest oxidising agent (or the strongest reducing agent) is lithium since it has the lowest standard reduction potential, (\(E_{\mathrm{Li}^{+} / \mathrm{Li}}^{0}\) = -3.045 V).

Use your brain power (Textbook page No. 115)

Question 1.
Identify the strongest and the weakest reducing agents from the electrochemical series.
Answer:
(a) From the electrochemical series, the strongest reducing agent is lithium since it has the lowest standard reduction potential (\(E_{\mathrm{Li}^{+} / \mathrm{Li}}^{0}\) = -3.045 V).
(b) The weakest reducing agent is fluorine since it has the highest standard reduction potential,
(\(E_{\mathrm{F}_{2} / \mathrm{F}^{-}}^{0}\) = +2.87 V).

Question 2.
From E° values given in Table 5.1, predict whether Sn can reduce I₂ or Ni²⁺.
Answer:
From electrochemical series,