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Balbharti Maharashtra State Board 12th Chemistry Textbook Solutions Chapter 6 Chemical Kinetics Textbook Exercise Questions and Answers.

Chemical Kinetics Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Chemistry Chapter 6 Exercise Solutions Maharashtra Board

Chemistry Class 12 Chapter 6 Exercise Solutions

1. Choose the most correct option.

Question i.
The rate law for the reaction aA + bB → P is rate = k[A] [B]. The rate of reaction doubles if
a. concentrations of A and B are both doubled.
b. [A] is doubled and [B] is kept constant
c. [B] is doubled and [A] is halved
d. [A] is kept constant and [B] is halved.
Answer:
b. [A] is doubled and [B] is kept constant

Question ii.
The order of the reaction for which the units of rate constant are mol dm^-3 s^-1 is
a. 1
b. 3
c. 0
d. 2
Answer:
c. 0

Question iii.
The rate constant for the reaction 2N₂O₅(g) → 2N₂O₄(g) + O₂(g) is 4.98 × 10^-4 s^-1. The order of reaction is
a. 2
b. 1
c. 0
d. 3
Answer:
b. 1

Question iv.
Time required for 90 % completion of a certain first order reaction is t. The time required for 99.9 % completion will be
a. t
b. 2t
c. t/2
d. 3t
Answer:
d. 3t

Question v.
Slope of the graph ln[A]_t versus t for first order reaction is
a. -k
b. k
c. k/2. 303
d. -k/2. 303
Answer:
a. -k

Question vi.
What is the half life of a first order reaction if time required to decrease concentration of reactant from 0.8 M to 0.2 M is 12 h?
a. 12 h
b. 3 h
c. 1.5 h
d. 6 h
Answer:
d. 6 h

Question vii.
The reaction, 3ClO^3Θ ClO₃^Θ + 2 Cl^Θ occurs in two steps,
(i) 2 ClO^– → ClO₂^Θ
(ii) ClO₂^Θ + ClO^Θ → ClO₃^Θ + Cl^Θ

The reaction intermediate is
a. Cl^Θ
b. ClO₂^Θ
c. ClO₃^Θ
d. ClO^Θ
Answer:
b. ClO₂^Θ

Question viii.
The elementary reaction O₂(g) + O(g) → 2O₂(g) is
a. unimolecular and second order
b. bimolecular and first order
c. bimolecular and second order
d. unimolecular and first order
Answer:
c. bimolecular and second order

Question ix.
Rate law for the reaction, 2NO + Cl₂ → 2 NOCl is rate = k[NO₂]²[Cl₂]. Thus k would increase with
a. increase of temperature
b. increase of concentration of NO
c. increase of concentration of Cl₂
d. increase of concentrations of both Cl₂ and NO
Answer:
a. increase of temperature

Question x.
For an endothermic reaction, X ⇌ Y. If E f is activation energy of the forward reaction and Er that for reverse reaction, which of the following is correct?
a. E_f = E_r
b. E_f < E_r
c. E_f > E_r
d. ∆H = E_f – E_r is negative
Answer:
(c) Ef → Er

2. Answer the following in one or two sentences.

Question i.
For the reaction,
N₂(g) + 3 H₂(g) → 2NH₃(g), what is the relationship among \(\frac{\mathrm{d}\left[\mathrm{N}_{2}\right]}{\mathrm{dt}}\)\(\frac{\mathrm{d}\left[\mathrm{H}_{2}\right]}{\mathrm{dt}} \text { and } \frac{\mathrm{d}\left[\mathrm{NH}_{3}\right]}{\mathrm{dt}} ?\)
Answer:
N_2(g) + 3H_2(g) → 2NH_3(g)
From the above reaction, when 1 mole of N₂ reacts, 3 moles of H₂ are consumed and 2 moles of NH₃ are formed.

If the instantaneous rate R of the reaction is represented in terms of rate of the consumption of N₂ then, \(R=-\frac{d\left[\mathrm{~N}_{2}\right]}{d t}\)

Hence the rate of reaction in terms of concentration changes in N₂, H₂ and NH₃ may be represented as,

Question ii.
For the reaction,
CH₃Br(aq) + OH-(aq) → CH₃OH^Θ (aq) +Br^Θ (aq), rate law is rate = k[CH₃Br][OH^Θ]
a. How does reaction rate changes if [OH^Θ] is decreased by a factor of 5?
b. What is change in rate if concentrations of both reactants are doubled?
Solution :
Given :
(a) Rate = R = k [CH₃Br] x [OH^–]
If R₁ and R₂ are initial and final rates of reaction then,

Rate will be increased 4 time.

Question iii.
What is the relationship between coeffients of reactants in a balanced equation for an overall reaction and exponents in rate law. In what case the coeffients are the exponents?
Answer:
Explanation : Consider the following reaction, aA + bB → products

If the rate of the reaction depends on the concentrations of the reactants A and B, then, by rate law,
R α [A]^a [B]^b
∴ R = k [A]^a [B^b
where [A] = concentration of A and
[B] = concentration of B

The proportionality constant k is called the velocity constant, rate constant or specific rate of the reaction.

a and b are the exponents or the powers of the concentrations of the reactants A and B respectively when observed experimentally.

The exponents or powers may not be necessarily a and b but may be different x and y depending on experimental observations. Then the rate R will be,
R = k [A]^x [B]^y
For example, if x = 1 and y = 2, then,
R = k [A] x [B]²

Question iv.
Why all collisions between reactant molecules do not lead to a chemical reaction?
Answer:
(i) Collisions of reactant molecules : The basic re-quirements of a reaction is that the reacting species (atoms, ions or molecules) must come together and collide for a reaction to occur. Therefore the rate of the reaction shall depend on the rate and frequency of collisions which increase with the amount of reacting species and temperature.

However it is observed that the rate of reaction is very low as compared to the rate of collisions between reacting species in gaseous phase or liquid phase. This suggests that all the collisions are not fruitful leading to a reaction. Hence it is necessary to consider another factor like energy of colliding species along with collision frequency.

(ii) Energy requirement (Activation energy) : The colliding molecules must possess a certain mini-mum energy called activation energy required far breaking and making bonds resulting in the reaction. This implies that the colliding molecules must have energy equal to or greater than the activation energy. The colliding molecules with less energy do not react.

(iii) Orientation of reactant molecules : The concept of activation energy is satisfactory in case of simple molecules or ions but not in case of complex or higher polyatomic molecules. It is observed that the rates of reaction are less as compared to the rates of collisions between activated molecules with activation energy.

This suggests that in addition, the colliding molecules must have proper orientations relative to each other during collisions. For example, consider the reaction, A – B-l-C → A + B – C. For the reaction to occur, C must collide with B while collisions with A will not be fruitful. Since B has to bond with C.

Question v.
What is the activation energy of a reaction?
Answer:
Activation energy : The energy required to form activated complex or transition state from the reactant molecules is called activation energy.
OR
The height of energy barrier in the energy profile diagram is called activation energy.

Question vi.
What are the units for rate constants for zero order and second order reactions if time is expressed in seconds and concentration of reactants in mol/L?
Answer:
(a) For a zero order reaction, the rate constant has units, molL^-1s^-1.
(b) For second order reaction,
Rate = k x [Reactant]²

Question vii.
Write Arrhenius equation and explain the terms involved in it.
Answer:
Arrhenius equation is represented as k = A x e^-E_a/RT
where
k = Rate constant at absolute temperature T
E_a = Energy of activation R = Gas constant
A = Frequency factor or pre-exponential factor.

Question viii.
What is the rate determining step?
Answer:
Many chemical reactions take place in a series of elementary steps. Among many steps of the reaction, one of the steps is the slowest step compared to other steps.

Rate determining step : The slowest step in the reaction mechanism which involves many steps is called the rate-determining step.

Question ix.
Write the relationships between rate constant and half life of fist order and zeroth order reactions.
Answer:
(a) For first order reaction, half-life period t_1/2 is, \(t_{1 / 2}=\frac{0.693}{k}\) where k is the rate constant.
(b) For zeroth-order reaction, half half period (t_1/2) is, \(t_{1 / 2}=\frac{[\mathrm{A}]_{0}}{2 k}\) where k is the rate constant and [A]₀ is initial concentration of the reactant.

Question x.
How do half lives of the fist order and zero order reactions change with initial concentration of reactants?
Answer:
(A) For the first order reaction, half life, t_1/2 is given by, \(t_{1 / 2}=\frac{0.693}{k}\) where k is rate constant. Hence it is independent of initial concentration of the reactant.

(B) Zero order reaction,
\(t_{1 / 2}=\frac{[\mathrm{A}]_{0}}{2 k}\) where [A]₀ is initial concentration of the reactant.

Hence, half life period increases with the increase in concentration of the reactant.

3. Answer the following in brief.

Question i.
How instantaneous rate of reaction is determined?
Answer:
(1) The instantaneous rate is expressed as an infinite¬simal change in concentration (- dc) of the reactant with the infinitesimal change in time (dt).
For a reaction, A → B, let an infinitesimal change in A be – dc in time dt, then Rate \(=\frac{d[\mathrm{~A}]}{d t}\).

Hence, it is represented as,
∴ Instantaneous rate \(=-\frac{d[\mathrm{~A}]}{d t}\)

The negative sign indicates a decrease in the concentration of A.

It is obtained by drawing a tangent to the curve obtained by plotting the concentration against the time. Hence, the slope at a given point represents the instantaneous rate of the reaction.

(2) The instantaneous rate can also be expressed as an infinitesimal change (or increase) in the concentration of the product with the infinitesimal change in time (dt).

Let dB be an infinitesimal change in the concentration of product B in time dt, then Rate \(=\frac{d[\mathrm{~B}]}{d t}=\frac{d x}{d t}\).

Hence,
Instantaneous rate \(=\frac{d x}{d t}\)

It is obtained from the slope of the curve obtained by plotting the concentration of the product against time.

The instantaneous rate is more useful in obtaining the rate law integrated equations.

Question ii.
Distinguish between order and molecularity of a reaction.
Answer:

Order	Molecularity
1. It is the sum of the exponents to which the concentration terms in rate law expression are raised.	1. ¡t is the number of molecules (or atoms or ions) of the reaCtants taking part in the elementary reaction.
2. It is experimentally determined and indicates the dependence of the reaction rate on the concentration of particular reactants.	2. It is the oretical property and indicátes the number of molecules of reactant in each step of the reaction.
3. It may have values that are integer, fractional, or zero.	3. It is always an integer.
4. Its value depends upon experimental conditions.	4. Its value does not depend upon experimental conditions.
5. It is the property of elementary and complex reactions.	5. It is the property of elementary reactions only.
6. Rate law expression describes the order of the reaction.	6. Rate law does not describe molecularity.

Question iii.
A reaction takes place in two steps,
1. NO(g) + Cl₂(g) NOCl₂(g)
2. NOCl₂(g) + NO(g) → 2NOCl(g)
a. Write the overall reaction.
b. Identify reaction intermediate.
c. What is the molecularity of each step?
Solution :
Given :
(1) NO_(g) + Cl_2(g) → NOCl_2(g)
(2) NOCl_2(g) + NO_(g) → 2NOCl_(g)

(a) Overall reaction is obtained by adding both the reactions
2NO_(g) + Cl_2(g) → 2NOCl_2(g)
(b) The reaction intermediate is NOCl₂, since it is formed in first step and consumed in the second step.
(c) Since the first step is a slow and rate determin­ing step, the molecularity is two.

Since the second step is a fast step its molecularity is not considered.

Question iv.
Obtain the relationship between the rate constant and half-life of a fist order reaction.
Answer:
Consider the following reaction,

If [A]₀ and [A]t are the concentrations of A at start and after time t, then [A]₀ = a and [A]_t = a – x.

The velocity constant or the specific rate constant k for the first order reaction can be represented as,

where, a is the initial concentration of the reactant A, x is the concentration of the product B after time t, so that (a – x) is the concentration of the reactant A after time t.

Half-life of a reaction : The time required to reduce the concentration of the reactant to half of its initial value is called the half-life period or the half-life of the reaction.

If t_1/2 is the half-life of a reaction, then at t = t_1/2, x = a/2, hence a – x = a – a/2 = a/2

Hence, for a first order reaction, the half-life of the reaction is independent of the initial concentration of the reactant.

Question v.
How will you represent zeroth-order reaction graphically?
Answer:
(1) A graph of concentration against time : In case of a zero-order reaction, the rate of reaction is independent of the concentration of the reactant. The concentration [A]_t of the reactant at a time t is given by
[A]_t = – kt + [A]₀ (y = – mx + c)
where [A]₀ is the initial concentration of the reactant and k is a rate constant.

Hence in case of zero order reaction, when the concentration of the reactant is plotted against time, a straight line with the slope equal to – k is obtained. The concentration of the reactants de-crease with time. The intercept on the concentration axis gives the initial concentration, [A]₀.

(2) A graph of rate of a reaction against the concen-tration of the reactant: Rate of a zero order reaction is independent of the concentration of the reactant.

Rate, R = k [A]⁰ = k

Hence even if the concentration of the reactant decreases, the rate of the reaction remains constant.

Therefore if the rate of a zero order reaction is plotted against concentration, then a straight line with zero slope is obtained indicating, no change in the rate of the reaction with a change in the concentration of the reactants.

(3) A graph of half-life period against concentration : The half-life period of a zero-order reaction is given by, \(t_{1 / 2}=\frac{[\mathrm{A}]_{0}}{2 k}\) where [A]₀ is initial con-centration of the reactant and k is the rate constant. Hence the half-life period is directly proportional to the concentration.

When a graph of t_1/2 is plotted against concentration, a straight line passing through origin is obtained, and the slope gives \(\frac{1}{2 k}\), where k is the rate constant.

Question vi.
What are pseudo-fist order reactions? Give one example and explain why it is pseudo-fist order.
Answer:
Pseudo-first-order reaction : A reaction which has higher-order true rate law but is experimentally found to behave as first order is called pseudo first order reaction.
Explanation : Consider an acid hydrolysis reaction of an ester like methyl acetate.
CH₃COOCH_3(aq) + H₂O₍₁₎ \(\stackrel{\mathrm{H}_{(\mathrm{aq})}^{+}}{\longrightarrow}\) CH₃COOH_(aq) + CH₃OH_(aq)
Since the reaction involves two substances, ester and water, it is a bimolecular reaction and the true rate law should be, Rate = k’ [CH₃COOCH₃] x [H₂O]

Hence the reaction is expected to follow second order kinetics. However experimentally it is found that the reaction follows first order kinetics.

This is because solvent water being in a large excess, its concentration remains constant. Hence, [H₂O] = constant = k”
Rate = k [CH₃COOCH₃] x [H₂O]
= k [CH₃COOCH₃] x k”
= k’ x k” x [CH₃COOCH₃]
If k’ x k” = k, then Rate = k [CH₃COOCH₃],

This indicates that second-order true rate law is forced into first order rate law. Therefore this bimolecular reaction which appears of second order is called pseudo first order reaction.

Question vii.
What are the requirements for the colliding reactant molecules to lead to products?
Answer:
Collisions of reactant molecules : The basic re­quirements of a reaction is that the reacting species (atoms, ions or molecules) must come together and collide for a reaction to occur. Therefore the rate of the reaction shall depend on the rate and frequency of collisions which increase with the amount of reacting species and temperature.

However it is observed that the rate of reaction is very low as compared to the rate of collisions between reacting species in gaseous phase or liquid phase. This suggests that all the collisions are not fruitful leading to a reaction. Hence it is necessary to consider another factor like energy of colliding species along with collision frequency.

Energy requirement (Activation energy) : The colliding molecules must possess a certain mini­mum energy called activation energy required far breaking and making bonds resulting in the reac­tion. This implies that the colliding molecules must have energy equal to or greater than the activation energy. The colliding molecules with less energy do not react.

Orientation of reactant molecules : The concept of activation energy is satisfactory in case of simple molecules or ions but not in case of complex or higher polyatomic molecules. It is observed that the rates of reaction are less as compared to the rates of collisions between activated molecules with activa­tion energy.

This suggests that in addition, the colliding mole­cules must have proper orientations relative to each other during collisions. For example, consider the reaction, A – B + C → A + B – C. For the reaction to occur, C must collide with B while collisions with A will not be fruitful. Since B has to bond with C.

Question viii.
How catalyst increases the rate of reaction? Explain with the help of a potential energy diagram for catalyzed and uncatalyzed reactions.
Answer:
(i) A catalyst is a substance, when added to the reactants, increases the rate of the reaction without being consumed. For example, the decomposition of KClO₃ in the presence of small amount of MnO₂ is very fast but very slow in the absence of MnO₂.

2KClO_3(s) \(\frac{\mathrm{MnO}_{2}}{\Delta}\) 2KCl_(s) + 3O_2(g)

(ii) The phenomenon of catalysed reaction is called catalysis and depends on nature of the catalyst. In heterogeneous catalysis, the reactant molecules are adsorbed on the solid catalyst surface while in case of homogeneous catalysis, the catalyst reacts with one of the reactants, forms intermediate and decomposes reforming original catalyst and the products.

(iii) The catalyst provides alternative and lower energy path or mechanism for the reaction.

(iv) In the presence of the catalyst, the activation energy of the reaction is lowered. The height of activation energy barrier is less than that in the uncatalysed reaction.

(v) Due to lowering of energy of activation, (E_a) the number of molecules possessing E_a increases, hence the rate of the reaction increases.

(vi) The rate constant = k = A x e^-E_a/RT where A is a frequency factor and hence the rates of the catalysed reaction are higher than those of un-catalysed reactions.

(vii) The catalyst does not change the extent of the reaction but hastens the reaction.

(viii) The catalyst enters the reaction but does not appear in the balanced equation since it is consumed in one step and regenerated in the another.

Question ix.
Explain with the help of the Arrhenius equation, how does the rate of reaction changes with (a) temperature and (b) activation energy.
Answer:
(a) By Arrhenius equation, k = Ax e^-E_a/RT where k is rate constant, A is a frequency factor and Ed is energy of activation at temperature T. As E_a increases, the rate constant and rate of the reaction decreases.

(b) As temperature increases E_a/RT decreases but due to negative sign, k and rate increase with the increase in temperature.

Question x.
Derive the integrated rate law for first order reaction.
Answer:
Consider following gas phase reaction,

Let initial pressure of A be P₀ at t = 0. If after time t the pressure of a A decreases by jc then the partial pressures of the substances will be, P_A = P_Q – x; P_B = x and P_c = x

Total pressure will be,

P_T + P₀ – x + x + x = Po + x
∴ x = P_T – P_n

The partial pressures at time t will be,

Question xi.
How will you represent first-order reactions graphically.
Answer:
(1) A graph of rate of a reaction and concentra­tion : The differential rate law for first-order reac­tion, A → Products is represented as, Rate = [/latex]-\frac{d[\mathrm{~A}]}{d t}=k[\mathrm{~A}]\(

∴ Rate = k x [A]_t (y = mx). When the rate of a first order reaction is plotted against concentration, [A]_t, a straight line graph is obtained.

With the increase in the concentration [A]_t, rate R, increases. The slope of the line gives the value of rate constant k.

(2) A graph of concentration against time : When the concentration of the reactant is plotted against time t, a curve is obtained. The concentration [A], of the reactant decreases exponentially with time. The variation in the concentration can be represented as,

where [A]₀ and [A]_t are initial and final concentra­tions the reactant and k is the rate constant. The time required to complete the first order reaction is infinity.

(3) A graph of log₁₀ (a – x) against time t :

When log₁₀(a – x) is plotted against time t, a straight line with negative slope is obtained, from which the velocity constant k can be calculated.

(4) A graph of half-life period and concentration : The half-life period, t_1/2 of a first order reaction is given by, where k is the rate constant.

For the given reaction at a constant temperature, t_1/2 is constant and independent of the concentration of the reactant.

Hence when a graph of t_1/2 is plotted against concentration, a straight line parallel to the concen­tration axis (slope = zero) is obtained.

(5) A graph of log₁₀ [latex]\left(\frac{a}{a-x}\right)\) against time : The rate constant, for a first order reaction is represented as, where [A₀] and [A]_t are the respective initial and final concentrations of the reactant after time t.


When \(\log _{10}\left(\frac{a}{a-x}\right)\) is plotted against time t, a straight line graph passing through the origin is obtained and the slope gives the value of k/2.303. From this slope, the rate constant can be calculated.

Question xii.
Derive the integrated rate law for the first order reaction, A(g) → B(g) + C(g) in terms of pressure.
Answer:
Consider following gas phase reaction,

Let initial pressure of A be P₀ at t = 0. If after time t the pressure of a A decreases by jc then the partial pressures of the substances will be, P_A = P_Q – x; P_B = x and P_c = x

Total pressure will be,

P_T + P₀ – x + x + x = Po + x
∴ x = P_T – P_n

The partial pressures at time t will be,

Question xiii.
What is zeroth-order reaction? Derive its integrated rate law. What are the units of rate constant?
Answer:
Definition : Zero order reaction : A reaction in which the rate of the reaction does not depend on the concentration of any reactant taking part in the reaction is called zero order reaction.

Consider a zero order reaction, A → Products
The rate of the reaction is, Rate \(=\frac{-d[\mathrm{~A}]}{d t}\)

By rate law,
Rate = k x [A]⁰ = k
∴ – d[A] = k x dt

If [A]₀ is the initial concentration of the reactant A at t = 0 and [A]_t is the concentration of A present after time t, then by integrating above equation,

This is the integrated rate law expression for rate constant for zero order reaction.
∴ k x t = [A]₀ – [A]_t
∴ [A]_t = – kt + A₀

For a zero order reaction :
The rate of reaction is R = k [A]⁰ = k
Hence, the velocity constant k has the unit of the rate of the reaction, i.e., mol dm^-3 s^-1.

Question xiv.
How will you determine activation energy: (a) graphically using Arrhenius equation (b) from rate constants at two different temperatures?
Answer:
(a) By Arrhenius equation,
Rate constant = = A x e^-E_a/RT where A is a fre-quency factor.

When log₁₀k is plotted against 1/T a straight line with negative slope is obtained. From the slope of the graph, energy of activation E_a, is obtained as follows :
Slope = \(\frac{E_{\mathrm{a}}}{2.303 R}\)
∴ Ea = 2303R x sloPe

(b) For the given reaction, rate constants k₁ and k₂ are measured at two different temperatures T₁ and T₂ respectively. Then \(\log _{10} \frac{k_{2}}{k_{1}}=\frac{E_{\mathrm{a}}\left(T_{2}-T_{1}\right)}{2.303 R \times T_{1} \times T_{2}}\) where E_a is the energy of activation.

Hence by substituting appropriate values, energy of activation E_a for the reaction is determined.

Question xv.
Explain graphically the effect of temperature on the rate of reaction.
Answer:
(i) It has been observed that the rates of chemical reactions increase with the increase in temperature.
(ii) The kinetic energy of the molecules increases with the increase in temperature. The fraction of molecules possessing minimum energy barrier,
i. e. activation energy E_a increases with increase in temperature.
(iii) Hence the fraction of colliding molecules that possess kinetic energy (E_a) also increases, hence the rate of the reaction increases with increase in temperature.

(iv) The above figure shows that the area that represents the fraction of molecules with kinetic energy exceeding E_a is greater at higher temperature T₂ than at lower temperature T₁. This explains that the rate of the reaction increases at higher temperature.
(v) The shaded area to the right of activation energy E_a represents fraction of collisions of activated molecules having energy E_a or greater.

Question xvi.
Explain graphically the effect of catalyst on the rate of reaction.
Answer:
(i) The phenomenon of catalysed reaction is called catalysis and depends on nature of the catalyst. In heterogeneous catalysis, the reactant molecules are adsorbed on the solid catalyst surface while in case of homogeneous catalysis, the catalyst reacts with one of the reactants, forms intermediate and decomposes reforming original catalyst and the products.
(ii) The catalyst provides alternative and lower energy path or mechanism for the reaction.
(iii) In the presence of the catalyst, the activation energy of the reaction is lowered. The height of activation energy barrier is less than that in the uncatalysed reaction.

Question xvii.
For the reaction 2A + B → products, find the rate law from the following data.

[A]/M	[A]/M	rate/M s-1
0.3	0.05	0.15
0.6	0.05	0.30
0.6	0.2	1.20

Solution :
Given : 2A + B → Products
Rates : R₁ = 0.15 Ms^-1 R₂ = 0.3 Ms^-1
[A]₁ = 0.3 M [A]₂ = 0.6 M
[B]₁ = 0.05 M [B]₂ = 0.05 M
(i) If order of the reaction in A is x and in B is y then, by rate law,

∴ y = 1. Hence the reaction has order one in B.
The order of overall reaction = n = n_A + n_B = 1 + 1 = 2

Answer:
(i) Rate law : Rate = fc [A] x [B]
Rate constant = k = 10M^-1s^-1
Order of the reaction = 2

4. Solve

Question i.
In a first order reaction, the concentration of reactant decreases from 20 mmol dm^-3 to 8 mmol dm^-3 in 38 minutes. What is the half life of reaction? (28.7 min)
Solution :
Given: [A]₀ =20 mmol dm^-3;
[A]_t=8 mmol dm^-3; t=38 mm;

Answer:
Half life period = 28.74 min

Question ii.
The half life of a first order reaction is 1.7 hours. How long will it take for 20% of the reactant to react? (32.9 min)
Solution :
Given : t_1/2 = 1.7 hr; [A]₀ = 100;
[A]_t = 100 – 20 = 80; t =?
\(t_{1 / 2}=\frac{0.693}{k}\)

Answer:
Time required = t = 32.86 min

Question iii.
The energy of activation for a first order reaction is 104 kJ/mol. The rate constant at 25 0C is 3.7 × 10^-5 s^-1. What is the rate constant at 30⁰C? (R = 8.314 J/K mol) (7.4 × 10^-5)
Solution :

Answer:
k₂ = 7.382 x 10^-5 s^-1

Question iv.
What is the energy of activation of a reaction whose rate constant doubles when the temperature changes from 303 K to 313 K? (54.66 kJ/mol)
Solution :
Given : k₂ = 2k_t, T₁ = 303 K; T₂ = 313 K; E_a = ?

Answer:
Energy of activation = E_a = 54.66 kJ

Question v.
The rate constant of a reaction at 5000C is 1.6 × 10³ M^-1 s^-1. What is the frequency factor of the reaction if its activation energy is 56 kJ/mol. (9.72 × 106 M^-1 s^-1)
Solution :

Answer:
Frequency factor = A = 9.727 x 10⁶ M^-1s^-1

Question vi.
Show that time required for 99.9% completion of a first order reaction is three times the time required for 90% completion.
Solution :
Given : For 99.9 % completion, if [A]₀ = 100,

If t₁ and t₂ are the times required for 99.9 % and 90 % completion of reaction respectively, then

Answer:
Time required for 99.9 % completion of a first order reaction is three time the time required for 90 % completion of the reaction.

Question vii.
A first order reaction takes 40 minutes for 30% decomposition. Calculate its half life. (77.66 min)
Solution :

Answer:
Half life period = 77.70 min.

Question viii.
The rate constant for the first order reaction is given by log₁₀ k = 14.34 – 1.25 × 10⁴ T. Calculate activation energy of the reaction. (239.3 kJ/mol)
Solution :
Given : log₁₀ k = 14.34 – \(\frac{1.25 \times 10^{4}}{T}\) ……………………. (1)
From Arrhenius equation we can write,
\(\log _{10} k=\log _{10} A-\frac{E_{\mathrm{a}}}{2.303 R \times T}\) ……………………. (2)
By comparing equations (1) and (2),
\(\frac{E_{\mathrm{a}}}{2.303 \times R}\) = 1.25 x 10⁴
∴ Ea = 1.25 x 10⁴ x 2.303 x R
= 1.25 x 10⁴ x 2.303 x 8.314
= 23.93 x 10⁴ = 239.3 kJ mol^-1

[Note : Frequency factor A may also be calculated as follows : log₁₀ A = 14.34
∴ A = Antilog 14.34 = 2.188 x 10⁴
Answer:
Energy of activation = E_a = 239.3 kJ mol^-1.

Question ix.
What fraction of molecules in a gas at 300 K collide with an energy equal to activation energy of 50 kJ/mol? (2 × 10^-9)
Solution :
Given : T = 300 K; E_a = 50 kJ mol^-1
= 50 x 10³ mol^-1
The fraction of molecules undergoing fruitful collisions is

Answer:
Fraction of molecules undergoing collision = 2 x 10^-9

Activity :
1. If you wish to determine the reaction order and rate constant for the reaction, 2AB₂ → A₂ + 2B₂.
a) What data would you collect?
b) How would you use these data to determine whether the reaction is zeroth or first order?

2. The activation energy for two reactions are E_a and E’_a with E_a > E’_a. If the temperature of reacting system increases from T₁ to T₂, predict which of the following is correct?

k values are rate constants at lower temperatures and k values at higher temperatures.

12th Chemistry Digest Chapter 6 Chemical Kinetics Intext Questions and Answers

(Textbook Page No 121)

Question 1.
Write the expressions for rates of reaction for :
2N₂O_5(g) → 4NO_2(g) + O_2(g)?
Answer:
For the given reaction, Rate of reaction =
\(=R=-\frac{1}{2} \frac{d\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right]}{d t}\)
\(\begin{aligned}
&=+\frac{1}{4} \frac{d\left[\mathrm{NO}_{2}\right]}{d t} \\
&=+\frac{d\left[\mathrm{O}_{2}\right]}{d t}
\end{aligned}\)

Problem 6.1: (Textbook Page No 121)

Question 1.
For the reaction,
\(\mathbf{3 I}_{(a q)}^{-}+\mathbf{S}_{2} \mathbf{O}_{8(a q)}^{2-} \longrightarrow \mathbf{I}_{3(\text { (aq) }}^{-}+2 \mathbf{S O}_{4(\mathrm{aq})}^{2-}\)
Calculate (a) the rate of formation of I₃^–,
(b) the rates of consumption of 1 and S₂O and (c) the overall rate of reaction if the rate of formation of \(\mathrm{SO}_{4}^{2-}\) is 0.O22 moles dm^-3 sec^-1.
Answer:


∴ (a) Rate of formation of \(\mathrm{I}_{3}^{-}\) = 0.011 mol dm^-3 s^-1
(b) Rate of consumption of I^– = 0.033 mol dm^-3 s^-1
(c) Rate of consumption of \(\mathrm{S}_{2} \mathrm{O}_{8}^{2-}\) = 0.011 mol dm^-3 s^-1
(d) Overall rate of reaction = Rate of consumption of reactant = Rate of formation of product

Try this….. (Textbook Page No 122)

Question 1.
For the reaction :
NO_2(g) + CO_(g) → NO_(g) + CO_2(g), the rate of reaction is experimentally found to be proportional to the square of the concentration of NO2 and independent that of CO. Write the rate law.
Answer:
Since the rate of the reaction is proportional to [NO₂]² and [CO]⁰, the rate law is R = k[NO₂]² x [CO]⁰
∴ R = k[NO₂]².

Try this….. (Textbook Page No 124)

Question 1.
The reaction,
CHCl_3(g) + Cl_2(g) → CCl_4(g) + HCl_(g) is first order in CHCl₃ and 1/2 order in Cl₂. Write the rate law and overall order of reaction.
Answer:
Since the reaction is first order in CHCl₃ and 1/2 order in Cl₂, the rate law for the reaction will be, Rate = k[CHCl₃] X [Cl₂]^1/2
The overall order (n) of the reaction will be, n = l + = \(\frac{1}{2}=\frac{3}{2}\)

Use your brain power! (Textbook Page No 124)

Question 1.
The rate of the reaction 2A + B → 2C + D is 6 x 10^-4 mol dm^-3 s^-1 when [A] =[B] = O.3 mol dm^-3 If the reaction is of first order in A and zeroth order in B, what is the rate constant?
Answer:
For the reaction,
2A + B → 2C + D,

(Problem 6.7) (Textbook Page No 126)

Question 1.
A reaction occurs in the following steps :
(i) NO_2(g) + F_2(g) → NO₂F_(g) + F_(g) (slow)
(ii) F_(g) + NO_2(g) → NO₂F_(g) (fast)
(a) Write the equation of overall reaction.
(b) Write down rate law.
(c) Identify the reaction intermediate.
Solution :
(a) The addition of two steps gives the overall reaction as
2NO_2(g) + F_2(g) → 2NO₂ F_(g)
(b) Step (i) is slow. The rate law of the reaction is predicted from its stoichiometry. Thus, rate = k [NO₂] [F₂]
(c) F is produced in step (i) and consumed in step (ii) hence F is the reaction intermediate.

Try this….. (Textbook Page No 126)

Question 1.
A complex reaction takes place in two steps :
(i) NO_(g) + O_3(g) → NO_3(g) + O_(g)
(ii) NO_3(g) + O_(g) → NO_2(g) + O_2(g)
The predicted rate law is rate = k [NO] [O₃]. Identify the rate-determining step. Write the overall reaction. Which is the reaction inter-mediate? Why?
Answer:
(i) NO_(g) + O_3(g) → NO_3(g) + O_(g)
(ii) NO_3(g) + O_(g) → NO_2(g) + O_2(g)
(a) The first step is slow and rate determining step since the rate depends on concentrations of NO_(g) and O_3(g). (Given : Rate = k [NO] x [O])
(b) The overall reaction is the combination of two steps.
NO_(g) ⁺ O_3(g) → NO_2(g) + O_2(g)
(c) NO_3(g) and O_(g) are reaction intermediates. They are formed in first step (i) and removed in the second step (ii).

Try this….. (Textbook Page No 129)

Question 1.
The half-life of a first-order reaction is 0.5 min. Calculate (a) time needed for the reactant to reduce to 20% and (b) the amount decomposed in 55 s.
Answer:

Try this….. (Textbook Page No 123)

Question 1.
For the reaction 2A + 2B → 2C + D, if concentration of A is doubled at constant [B] the rate increases by a factor of 4. If the concentration of B is doubled with [A] being constant the rate is doubled. Write the rate law of the reaction.
Answer:
Rate = R₁ = k[A]^x [B]^y
When concentration of A = [2A] and

Hence order with respect to A is 2 and with respect to B is 1. By rate law,
Rate = A: [A]² [B]

Question 2.
The rate law for the reaction A + B → C is found to be rate = k [A]² x [B]. The rate constant of the reaction at 25 °C is 6.25 M^-2 S^-1. What is the rate of reaction when [A] = 1.0 mol dm^-3 and [B] = 0.2 mol dm^-3?
Answer:
Rate = k x [A]² x [B]
= 6.25 x 1² x 0.2
Rate = 1.25 x 10² mol dm^-3 s^-1

12th Std Chemistry Questions And Answers:

• Solid State Class 12 Chemistry Questions And Answers
• Solutions Class 12 Chemistry Questions And Answers
• Ionic Equilibria Class 12 Chemistry Questions And Answers
• Chemical Thermodynamics Class 12 Chemistry Questions And Answers
• Electrochemistry Class 12 Chemistry Questions And Answers
• Chemical Kinetics Class 12 Chemistry Questions And Answers
• Elements of Groups 16, 17 and 18 Class 12 Chemistry Questions And Answers
• Transition and Inner Transition Elements Class 12 Chemistry Questions And Answers

Balbharti Maharashtra State Board 12th Chemistry Textbook Solutions Chapter 11 Alcohols, Phenols and Ethers Textbook Exercise Questions and Answers.

Alcohols, Phenols and Ethers Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Chemistry Chapter 11 Exercise Solutions Maharashtra Board

Chemistry Class 12 Chapter 11 Exercise Solutions

1. Choose the correct option.

Question i.
Which of the following represents the increasing order of boiling points of (1), (2) and (3)?
(1) CH₃ – CH₂ – CH₂ – CH₂ – OH
(2) (CH₃)₂ CH – O – CH₃
(3) (CH₃)₃COH
A. (1) < (2) < (3)
B. (2) < (1) < (3)
C. (3) < (2) < (1)
D. (2) < (3) < (1)
Answer:
(a) (1) < (2) < (3)

Question ii.
Which is the best reagent for carrying out following conversion ?

A. LiAlH₄
B. Conc. H₂SO₄, H₂O
C. H₂/Pd
D. B₂H₆, H₂O₂ – NaOH
Answer:
B. Conc. H₂SO₄, H₂O

Question iii.
Which of the following reaction will give ionic organic product on reaction ?
A. CH₃ – CH₂ – OH + Na
B. CH₃ – CH₂ – OH + SOCl₂
C. CH₃ – CH₂ – OH + PCl₅
D. CH₃ – CH₂ – OH + H₂SO₄
Answer:
C. CH₃ – CH₂ – OH + PCl₅

Question iv.
Which is the most resistant alcohol towards oxidation reaction among the follwoing ?

Answer:
(c)

Question v.
Resorcinol on distillation with zinc dust gives
A. Cyclohexane
B. Benzene
C. Toluene
D. Benzene-1, 3-diol
Answer:
(b) Benzene

Question vi.
Anisole on heating with concerntrated HI gives
A. Iodobenzene
B. Phenol + Methanol
C. Phenol + Iodomethane
D. Iodobenzene + methanol
Answer:
B. Phenol + Methanol

Question vii.
Which of the following is the least acidic compound ?

Answer:
(b)

Question viii.
The compound incapable of hydrogen bonding with water is ……

Answer:
(b)

Question ix.
Ethers are kept in air tight brown bottles because
A. Ethers absorb moisture
B. Ethers evaporate readily
C. Ethers oxidise to explosive peroxide
D. Ethers are inert
Answer:
C. Ethers oxidise to explosive peroxide

Question x.
Ethers reacts with cold and concentrated H₂SO₄ to form
A. oxonium salt
B. alkene
C. alkoxides
D. alcohols
Answer:
A. oxonium salt

2. Answer in one sentence/ word.

Question i.
Hydroboration-oxidation of propene gives…..
Answer:
n-propyl alcohol (CH₃ – CH₂ – CH₂ – OH)

Question ii.
Write the IUPAC name of alcohol having molecular formula C₄H₁₀O which is resistant towards oxidation.
Answer:

Question iii.
Write the structure of optically active alcohol having molecular formula C₄H₁₀O
Answer:

Question iv.
Write name of the electrophile used in Kolbe’s Reaction.
Answer:
Electrophile : Carbon dioxide (O = C = O)

3. Answer in brief.

Question i.
Why phenol is more acidic than ethyl alcohol ?
Answer:
(1) In ethyl alcohol, the -OH group is attached to sp³ – hybridised carbon while in phenols, it is attached to sp² – hybridised carbon.

(2) Due to higher electronegativity of sp² – hybridised carbon, electron density on oxygen decreases. This increases the polarity of O-H bond and results in more ionization of phenol than that of alcohols.

(3) Electron donating inductive effect (+1 effect) of the alkyl group destabilizes alkoxide ion. As a result alcohol does not ionize much in water, therefore alcohol is neutral compound in aqueous medium.

(4) In alkoxide ion, the negative charge is localized on oxygen, while in phenoxide ion the negative charge is delocalized. The delocalization of the negative charge (structure I to V) makes phenoxide ion more stable than that of phenol.

The delocalization of charge in phenol (structures VI to X), the resonating structures have charge separation (where oxygen atom of OH group to be positive and delocalization of negative charge over the ortho and para positions of aromatic ring) due to which phenol molecule is less stable than phenoxide ion. This favours ionization of phenol. Thus phenols are more acidic than ethyl alcohol.

Question ii.
Why p-nitrophenol is a stronger acid than phenol ?
Answer:
(1) In p-nitrophenol, nitro group (NO₂) is an electron withdrawing group present at para position which enhances the acidic strength (-1 effect). The O-H bond is under strain and release of proton (H⁺) becomes easy. Further p-nitrophenoxide ion is more stabilised due to resonance.

(2) Since the absence of electron withdrawing group (like – NO₂) in phenol at ortho and para position, the acidic strength of phenol is less than that of p-nitrophenol.

Question iii.
Write two points of difference between properties of phenol and ethyl alcohol.
Answer:

Question iv.
Give the reagents and conditions necessary to prepare phenol from
a. Chlorobenzene
b. Benzene sulfonic acid.
Answer:
(1) From chlorobenzene : Reagents required : NaOH and dil. HC1 Temperature : 623 K, Pressure : 150 atm
(2) From Benzene sulphonic acid : Reagents required : aq NaOH, caustic soda, dil. HC1 Temperature : 573 K

Question v.
Give the equations of the reactions for the preparation of phenol from isopropyl benzene.
Answer:
Preparation of phenol from cumene (isopropylbenzene) : This is the commercial method of preparation of phenol. When a stream of air is passed through cumene (isopropylbenzene) suspended in aqueous Na₂CO₃ solution in the presence of cobalt naphthenate catalyst, isopropyl benzene hydroperoxide or cumene hydroperoxide is formed. Isopropylbenzene hydroperoxide on warming with dil. H₂SO₄ gives phenol and acetone. Acetone is an important by-product of the reaction and is separated by distillation. The reaction is called auto oxidation.

Question vi.
Give a simple chemical test to distinguish between ethanol and ethyl bromide.
Answer:
When ethyl bromide is heated with aq NaOH; ethyl alcohol is formed whereas ethanol does not react with aq NaOH

4. An ether (A), C₅H₁₂O, when heated with excess of hot HI produce two alkyl halides which on hydrolysis form compound (B)and (C), oxidation of (B) gave and acid (D), whereas oxidation of (C) gave a ketone (E). Deduce the structural formula of (A), (B), (C), (D) and (E).
Answer:

5. Write structural formulae for

a. 3-Methoxyhexane
b. Methyl vinyl ether
c. 1-Ethylcyclohexanol
d. Pentane-1,4-diol
e. Cyclohex-2-en-1-ol
Answer:

6. Write IUPAC names of the following

Answer:

Activity :
• Collect information about production of ethanol as byproduct in sugar industry and its importance in fuel economy.
• Collect information about phenols used as antiseptics and polyphenols having antioxidant activity.

12th Chemistry Digest Chapter 11 Alcohols, Phenols and Ethers Intext Questions and Answers

Use your brain power! (Textbook Page No 235)

Question 1.
Classify the following alcohols as l0/2°/3° and allylic/benzylic

Answer:
(1) Ally lie alcohol (primary)
(2) Allylic alcohol (secondary)
(3) Allylic alcohol (tertiary)
(4) Benzylic alcohol (primary)
(5) Benzylic alcohol (secondary)

Use your brain power ….. (Textbook Page No 236)

Question 1.
Name t-butyl alcohol using carbinol system of nomenclature.
Answer:
Trimethyl carbinol.

Problem 11.1 (Textbook Page No 238)

Question 1.
Draw structures of following compounds:
(i) 2,5-DiethIphenoI
(ii) Prop-2-en-I-oI
(iii) 2-methoxypropane
(iv) Phenylmethanol
Solution :

Try this ….. (Textbook Page No 238)

Write IUPAC names ol (he following compounds.

Do you know (Textbook Page No 238)

Question 1.
The mechanism of hydration of ethylcnc to ethyl alcohol.
Answer:
The mechanism of hydration of ethylene involves three steps:

Step 1: Ethylene gets protonated to form carbocation by electrophilic attack of H₃O (Formation of carbocation intermediate).

Step 2 : Nucleophilic attack of water on carbocation

Step 3 : Deprotonation to form an alcohol

Problem 11.2 : (Textbook Page No 239)

Question 1.
Predict the products for the following reaction.

Solution:
The substrate (A) contains an isolated and an aldehyde group. H₂/Ni can reduce both these functional groups while LiAlH₄ can reduce only – CHO of the two, Hence

Try this ….. (Textbook page 240)

Question 1.
Arrange O – H, C – H and N – H bonds in increasing order of their bond polarity.
Answer:
Increasing order of polarity :C – H, N – H, O – H

Problem 11.3 : (Textbook Page No 241)

Question 1.
The boiling point of n-butyl alcohol, isobutyl alcohol, sec-butyl alcohol and tert-butyl alcohol are 118 °C, 108 °C. 99 °C and 82 °C respectively. Explain.
Solution:
As branching increases, intermolecular van der Waal’s force become weaker and the boiling point decreases. Therefore, n-butyl alcohol has highest boiling point 118 °C and tert-butyl alcohol has lowest boiling point 83 °C. Isobutyl alcohol is a primary alcohol and hence its boiling point is higher than that of sec-butyl alcohol.

Problem 11.4 : (Textbook Page No 242)

The solubility of o-nitrophenol and p-nitrophenol is 0.2 g and 1.7 g/100 g of H₂O respectively. Explain the difference.
Solution :

p-Nitrophenol has strong intermolecular hydrogen bonding with solvent water. On the other hand, o-nitrophenol has strong intramolecular hydrogen bonding and therefore the intermolecular attraction towards solvent water is weak. The stronger the intermolecular attraction between solute and solvent higher is the solubility. Hence p-nitrophenol has higher solubility in water than that of o-nitrophenol.

Problem 11.5 : (Textbook Page No 243 & 244)

Question 1.
Arrange the following compounds in decreasing order of acid strength and justify.
(1) CH₃ – CH₂ – OH
(2) (CH₃)₃ C – OH
(3) C₆H₅ – OH
(4) p-NO₂ – C₆H₄ – OH
Solution :
Compounds (3) and (4) are phenols and therefore are more acidic than the alcohols (1) and (2). The acidic strengths of compounds depend upon stabilization of the corresponding conjugate bases. Hence let us compare electronic effects in the conjugate bases of these compounds :

The conjugate base of the alcohol (1) is destabilized by + 1 effect of one alkyl group, whereas conjugate base of the alcohol (2) is destabilized by +1 effect of three alkyl groups. Hence (2) is weaker acid than (1)

Phenols : The conjugate base of p-nitrophenol (4) is better resonance stabilized due to six resonance structures compared to the five resonance structure of conjugate base of phenol (3). The resonance structure VI has – ve charge on only electronegative oxygens. Hence the phenol (4) is stronger acid than (3). Thus the decreasing order of acid strength is (4), (3), (1), (2).

Use your brain power (Textbook Page No 244)

Question 1.
What are the electronic effects exerted by – OCH₃ and – Cl? Predict the acid strength of

Answer:
The electronic effects exerted by – Cl and – O CH₃ are as follows :
(1) Cl being more electronegative atom it pulls the bonding electrons towards itself. This is known as negative inductive effect (- I).

(2) – OCH₃ is less electronegative group which repels the bonding electrons away from it. This is known as positive inductive effect ( + I).

(3) The relative to parent phenol, is more acidic than .

Problem 11.6 : (Textbook Page No 245)

Question 1.
Mechanism of acid catalyzed dehydration of ethanol to give ethene.
Answer:
The mechanism of dehydration of ethanol involves the following order :
Step 1 : Formation of protonated alcohols : Initially ethyl alcohol gets protonated to form ethyl oxonium ion.

Step 2 : Formation of carbocation : It is the slowest step and hence, the rate determining step of the reaction.

Steps 3: Formation of ethene: Removal of a proton (H⁺) from carbocation.

The acidused in step I is released in step 3, the equilibrium is shifted to the right, ethene is removed as it is formed.

Problem 11.6 : (Textbook Page No 245)

Question 1.
Write the reaction showing major and minor products formed on heating butan-2-ol with concentrrated sulphuric acid.
Solution :
In the reaction described butan-2-ol undergoes dehydration to give but-2-ene (major) and but-l-ene (minor) in accordance with Saytzeff rule.

Problem 11.7 : (Textbook Page No 246)

Question 1.
Write and explain reactions to convert propan-l-ol into propan-2-ol.
Solution :
The dehydration of propane-l-ol to propene is the first step. Markownikoff hydration of propene is the second step to get the product propan-2-ol. This is brought about by reaction with concemtrated H₂SO₄ followed by hydrolysis.

Problem 11.8 : (Textbook Page No 246)

Question 1.
An organic compound gives hydrogen on reaction with sodium metal. It forms an aldehyde having molecular formula C₂H₄O on oxidation with pyridinium chlorochromate. Name the compounds and give equations of these reactions.
Solution :
The given molecular formula C₂H₄O of aldehyde is written as CH₃ – CHO. Hence the formula of alcohol from which this is obtained by oxidation must be CH₃ – CH₂ – OH. The two reactions can, therefore, be represented as follows :

(Do you know? Textbook Page No 248)

Question 90.
Write the mechanism of dehydration of alcohol to give ether.
Answer:
Dehydration of alcohols to form ether is SN₂ reaction. The mechanism of dehydration of ethanol involves the following steps.

Step 1 (Protonation) : Initially ethyl alcohol gets protonated in the presence of acid to form ethyl oxonium ion.

Step 2 (SN² mechanism) : Protonated alcohol species undergoes a backside attack by second molecule of alcohol is a slow step.

Step 3 (Deprotonation) : Formation of diethyl ether by elimination of proton

Problem 11.9 : (Textbook Page No 249)

Question 1.
Ethyl isopropyl ether does not form on reaction of sodium ethoxide and isopropyl chloride.

(i) What would be the main product of this reaction?
(ii) Write another reaction suitable for the preparation of ethyl isopropyl ether.
Solution :
(i) Isopropyl chloride is a secondary chloride. On treating with sodium ethoxide it gives elimination reaction to form propene as the main product.

(ii) Ethyl isopropyl ether can be prepared as follows using ethyl chloride (1⁰ chloride) as substrate.

Do you know? (Textbook Page No 250)

Question 1.
The mechanism of the reaction of HI with methoxy ethane.
Answer:
The reaction mechanism takes place as follows :
Step 1 : Protonation of ether Initially the ether molecule (methoxy ethane) protonated by cone. HI to form oxonium ion.

Step 2 : Iodide is a good nucleophile. It attacks the least substituted carbon of the oxonium ion formed in step 1 and displaces an alcohol molecule by SN² mechanism.

For example :
• Use of excess HI converts the alcohol into alkyl iodide.
• In case of ether having one tertiary alkyl group the reaction with hot HI follows SN1 mechanism, and tertiary iodide is formed rather than tertiary alcohol.

Step 1 :

Step 2 :

12th Std Chemistry Questions And Answers:

• Coordination Compounds Class 12 Chemistry Questions And Answers
• Halogen Derivatives Class 12 Chemistry Questions And Answers
• Alcohols, Phenols and Ethers Class 12 Chemistry Questions And Answers
• Aldehydes, Ketones and Carboxylic Acids Class 12 Chemistry Questions And Answers
• Amines Class 12 Chemistry Questions And Answers
• Biomolecules Class 12 Chemistry Questions And Answers
• Introduction to Polymer Chemistry Class 12 Chemistry Questions And Answers
• Green Chemistry and Nanochemistry Class 12 Chemistry Questions And Answers

Balbharti Maharashtra State Board 12th Chemistry Textbook Solutions Chapter 15 Introduction to Polymer Chemistry Textbook Exercise Questions and Answers.

Introduction to Polymer Chemistry Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Chemistry Chapter 15 Exercise Solutions Maharashtra Board

Chemistry Class 12 Chapter 15 Exercise Solutions

1. Choose the correct option from the given alternatives.

Question i.
Nylon fibers are …………………………………..
A. Semisynthetic fibres
B. Polyamide fibres
C. Polyester fibres
D. Cellulose fibres
Answer:
B. polyamide fibres

Question ii.
Which of the following is naturally occurring polymer?
A. Telfon
B. Polyethylene
C. PVC
D. Protein
Answer:
D. Protein

Question iii.
Silk is a kind of …………………………………. fibre
A. Semisynthetic
B. Synthetic
C. Animal
D. Vegetable
Answer:
C. an animal

Question iv.
Dacron is another name of …………………………………. .
A. Nylon 6
B. Orlon
C. Novolac
D. Terylene
Answer:
D. Terylene

Question v.
Which of the following is made up of polyamides?
A. Dacron
B. Rayon
C. Nylon
D. Jute
Answer:
C. Nylon

Question vi.
The number of carbon atoms present in the ring of ε – caprolactam is
A. Five
B. Two
C. Seven
D. Six
Answer:
D. Six

Question vii.
Terylene is …………………………………. .
A. Polyamide fibre
B. Polyester fibre
C. Vegetable fibre
D. Protein fibre
Answer:
B. Polyester fibre

Question viii.
PET is formed by …………………………………. .
A. Addition
B. Condensation
C. Alkylation
D. Hydration
Answer:
D. Hydration

Question ix.
Chemically pure cotton is …………………………………. .
A. Acetate rayon
B. Viscose rayon
C. Cellulose nitrate
D. Cellulose
Answer:
D. Cellulose

Question x.
Teflon is chemically inert, due to presence of …………………………………. .
A. C-H bond
B. C-F bond
C. H- bond
D. C=C bond
Answer:
A. C-H bond

2. Answer the following in one sentence each.

Question i.
Identify ‘A’ and ‘B’ in the following reaction …………………………………. .

Answer:

Question ii.
Complete the following statements
a. Caprolactam is used to prepare …………………………………. .
b. Novolak is a copolymer of …………………………………. and …………………………………. .
c. Terylene is ………………………………….. polymer of terephthalic acid and ethylene glycol.
d. Benzoyl peroxide used in addtion polymerisation acts as …………………………………. .
e. Polyethene consists of polymerised …………………………………. .
Answer:
a. Nylon-6
b. Phenol, formaldehyde
c. polyester
d. initiator (catalyst)
e. linear or branched-chain

Question iii.
Draw the flow chart diagram to show the classification of polymers based on type of polymerisation.
Answer:

Question iv.
Write examples of Addition polymers and condensation polymers.
Answer:
Addition polymers : Polyvinyl chloride, polythene
Condensation polymers : Bakelite, terylene, Nylon-66

Question v.
Name some chain-growth polymers.
Answer:
Chain growth polymers : Polythene, polyacrylonitrile and polyvinyl chloride.

Question vi.
Define the terms :
1) Monomer
2) Vulcanisation
3) Synthetic fibres
Answer:
1. Monomer is a small and simple molecule and has a capacity to form two chemical bonds with other monomers. Examples : Ethene, Propylene.
2. The process by which a network of cross-links is introduced into an elastomer is called vulcanisation or it can also be defined as the process of heating natural rubber with sulphur to increase the tensile strength, toughness and elasticity of natural rubber is known as vulcanization of rubber.
3. The man-made fibres prepared by polymerization of one monomer or copolymerization of two or more monomers are called synthetic fibres.

Question vii.
What type of intermolecular force leads to high-density polymer?
Answer:
High density polymers have low degree of branching along the hydrocarbon chain. The molecules are closely packed together during crystallization. This closer packing means that the van der Waals attraction between the chains are greater and so the plastic (high density polymer) is stronger and has a melting point.

Question viii.
Give one example each of copolymer and homopolymer.
Answer:
Homopolymer : PVC, Nylon-6
Copolymer : Terylene, Buna-S

Question ix.
Identify Thermoplastic and Thermosetting Plastics from the following …………………………………. .
1. PET
2. Urea-formaldehyde resin
3. Polythene
4. Phenol formaldehyde
Answer:
Thermoplastic plastics : PET, Polythene
Thermosetting plastics : Urea formaldehyde resin, Phenol formaldehyde

3. Answer the following.

Question i.
Write the names of classes of polymers formed according to intermolecular forces and describe briefly their structural characteristics.
Answer:
Molecular forces bind the polymer chains either by hydrogen bonds or Vander Waal’s forces. These forces are called intermolecular forces. On the basis of magnitude of intcrmolccular forces, polymers are further classified as ebstomers, fibres, thermoplastic polymers. thermosetting polymers.

(1) Elastomers: Weak van der Waals type of intermolecular forces of attraction between the polymer chains are observed in cbstomcrs. When polymer is stretched, the polymer chain stretches and when the strain is relieved the chain returns to its odginal position, Thus, polymer shows elasticity and is called elastomers. Elastomers, the elastic polymers, have weak van der Waals type of intermolecular forces which permit the polymer to be stretched. Lilastorners are soft and stretchy and used in making rubber bands. E.g.. neoprene, vulcanized rubber, buna.S, buna-N.

(2) Fibres : It consists of strong intermolecular forces of’ attraction due to hydrogen bonding and strong dipole-dipole forces. These polymers possess high tensile strength. Due to these strong intermolecular forces the fibres are crystalline in nature. They are used in textile industries, strung tyres. etc.. e.g., nylon, terylene.

(3) Thermoplastic polymers: These polymer possess moderately strong intermolecular forces of attraction between those of elastomers and fibres. These polymers arc called thermoplastic because they become soft on heating and hard on cooling. They are either linear or branched chain polymers. They can be remoulded and recycled. E.g. polyethenc, PVC, polystyrene.

(4) Thermosetting polymers: These polymers are cross linked or branched molecules and are rigid polymers. During their formation they have property of being shaped on heating. but they get hardened while hot. Once hardened these become infusible, cannot be softened by heating and therefore, cannot be remoulded and recycled.
This shows extensive cross linking by covalent bonds formed in the moulds during hardening/setting process while hot. E.g. Bakelite, urea formaldehyde resin.

Question ii.
Write reactions of formation of :
a. Nylon 6
b. Terylene
Answer:
Terylene is polyester fibre formed by the polymerization of terephthalic acid and ethylene glycol.

Terylene is obtained by condensation polymerization of ethylene glycol and terephthalic acid in presence of catalyst zinc acetate and antimony trioxide at high temperature.

Properties :

• Terylene has relatively high melting point (265 °C)
• It is resistant to chemicals and water.

Uses :

• It is used for making wrinkle free fabrics by blending with cotton (terycot) and wool (terywool), and also as glass reinforcing materials in safety helmets.
• PET is the most common thermoplastic which is another trade name of the polyester polyethylenetereph- thalate.
• It is used for making many articles like bottles, jams, packaging containers.

Question iii.
Write the structure of natural rubber and neoprene rubber along with the name and structure of thier monomers.
Answer:

Question iv.
Name the polymer type in which the following linkage is present.
Answer:
The linkage is present in terylene or dacron polymer.

Question v.
Write the structural formula of the following synthetic rubbers :
a. SBR rubber
b. Buna-N rubber
c. Neoprene rubber
Answer:

Question vi.
Match the following pairs :
Name of polymer – Monomer
1. Teflon – a. CH₂ = CH₂
2. PVC – b. CF₂ = CF₂
3. Polyester – c. CH₂ = CHCl
4. Polythene – d. C₆H₅OH and HCHO
5. Bakelite – e. Dicarboxylic acid and polyhydoxyglycol
Answer:

1. Teflon – CF₂ = CF₂
2. PVC – CH₂ = CHCI
3. Polyester-Dicarboxylic acid and polyhydoxyglycol
4. Polythene – CH₂ = CH₂
5. Bakelite – C₆H₅OH and HCHO

Question vii.
Draw the structures of polymers formed from the following monomers
1. Adipic acid + Hexamethylenediamine
2. e – Aminocaproic acid + Glycine
Answer:

Question viii.
Name and draw the structure of the repeating unit in natural rubber.
Answer:
Repeating unit of natural rubber (Basic unit : isoprene)

Question ix.
Classify the following polymers as natural and synthetic polymers
a. Cellulose
b. Polystyrene
c. Terylene
d. Starch
e. Protein
f. Silicones
g. Orlon (Polyacrylonitrile)
h. Phenol-formaldehyde resins
Answer:

Natural Polymers	1. Cellulose 4. Starch 5. Protein
Synthetic Polymers	2. Polystyrene 3. Terylene 6. Silicones 7. Orion (Polyacrylonitrile) 8. phenol-formaldehyde resin

Question x.
What are synthetic resins? Name some natural and synthetic resins.
Answer:
Synthetic resins are artificially synthesised high molecular weight polymers. They are the basic raw material of plastic. The main properties of plastic depend on the synthetic resin it is made from.

Examples of natural resins : Rosin, Damar, Copal, Sandarac, Amber, Manila
Examples of synthetic resins : Polyester resin, Phenolic resin, Alkyl resin, Polycarbonate resin, Polyamide resin, Polyurethane resin, silicone resin, Epoxy resin, Acrylic resin.

Question xi.
Distinguish between thermosetting and thermoplastic resins. Write example of both the classes.
Answer:

Thermosetting resin	Thermoplastic resin
(1) They harden when heated. Once hardened it no longer melts.	(1) They soften when heated and harden again when cooled.
(2) They cannot be re-shaped.	(2) They can be reshaped
(3) They are strong, hard.	(3) They are weak, soft.
(4) Thermosetting resin show cross-linking. Examples : Melamine resin Epoxy resins, Bake-lite.	(4) Thermoplastic molecules do not cross link, hence are flexible. Examples : Polythene, polypropylene, nylon, polyester.

Question xii.
Write name and formula of raw material from which bakelite is made.
Answer:
The raw material or monomers used to prepare bakelite are o-hydroxymethyl phenol and formaldehyde (HCHO)

4. Attempt the following :

Question i.
Identify condensation polymers and addition polymers from the following.

Answer:

Question ii.
Write the chemical reactions involved in the manufacture of Nylon 6, 6
Answer:
Nylon-6, 6 is a linear polyamide polymer formed by the condensation polymerisation reaction. The monomers used in the preparation of Nylon-6, 6 are :
(1) Adipic acid : HOOC-(CH₂)₄-COOH
(2) Hexamethylene diamine : H₂N-(CH₂)₆-NH₂

When equimolar aqueous solutions of adipic acid and hexamethylene diamine are mixed and heated, there is neutralization to form a nylon salt. During polymerisation at 553 k nylon salt loses a water molecule to form nylon 6, 6 polymer. Both monomers (hexamethylene diamine and adipic acid) contain six carbon atoms each, hence the polymer is termed as Nylon-6,6.

Properties and uses : Nylon 6,6 is high molecular mass (12000 – 50000 u) linear condensation polymer. It possesses high tensile strength. It does not soak in water. It is used for making sheets, bristles for brushes, surgical sutures, textile fabrics, etc.

Question iii.
Explain the vulcanisation of rubber. Which vulcanizing agents are used for the following synthetic rubber.
a. Neoprene
b. Buna-N
Answer:
The process by which a network of cross links is introduced into an elastomer is called vulcanization.

Vulcanization enhances the properties of natural rubber like tensile strength, stiffness, elasticity, toughness etc. Sulphur forms cross links between polyisoprene chains which results in improved properties of rubber.

• For neoprene vulcanizing agent is MgO.
• For Buna-N vulcanizing agent is sulphur.

Question iv.
Write reactions involved in the formation of …………………………………. .
1) Teflon
2) Bakelite
Answer:
The monomers phenol and formaldehyde undergo polymerisation in the presence of alkali or acid as catalyst.

Phenol reacts with formaldehyde to form ortho or p-hydroxy methyl phenols, which further reacts with phenol to form a linear polymer called Novolac. It is used in paints.

In the third stage, various articles are shaped from novolac by putting it in appropriate moulds and heating at high temperature (138 °C to 176 °C) and at high pressure forms rigid polymeric material called bakelite. Bakelite is insoluble and infusible and has high tensile strength.

Bakelite is used in making articles like telephone instrument, kitchenware, electric insulators frying pans, etc.

2. Teflon is polytetrafluoroethylene. The monomer used in preparation of teflon is tetrafluoroethylene, (CF₂ = CF₂) which is a gas at room temperature. Tetrafluoroethylene is polymerized by using free radical initiators such as hydrogen peroxide or ammonium persulphate at high pressure.

Properties:

• Telflon is tough, chemically inert and resistant to heat and attack by corrosive reagents.
• C – F bond is very difficult to break and remains unaffected by corrosive alkali, organic solvents.
  Uses : Telflon is used in making non-stick cookware, oil seals, gaskets, etc.

Question v.
What is meant by LDP and HDP? Mention the basic difference between the same with suitable examples.
Answer:

• LDP is low density polyethylene and HDP is high density polyethylene.
• LDP is a branched polymer with low density due to chains are loosely held and HDP is a linear polymer with density due to close packing.
• HDP is much stiffer than LDP and has high tensile strength and hardness.

LDP is mainly used in preparation of pipes for agriculture, irrigation and domestic water line connections. HDP is used in manufacture of toys and other household articles like bucket, bottles, etc.

Question vi.
Write preparation, properties and uses of Teflon.
Answer:
Teflon is polytetrafluoroethylene. The monomer used in preparation of teflon is tetrafluoroethylene, (CF₂ = CF₂) which is a gas at room temperature. Tetrafluoroethylene is polymerized by using free radical initiators such as hydrogen peroxide or ammonium persulphate at high pressure.

Properties:

• Telflon is tough, chemically inert and resistant to heat and attack by corrosive reagents.
• C – F bond is very difficult to break and remains unaffected by corrosive alkali, organic solvents.
  Uses : Telflon is used in making non-stick cookware, oil seals, gaskets, etc.

Question vii.
Classify the following polymers as straight-chain, branched-chain and cross-linked polymers.

Answer:

5. Answer the following.

Question i.
How is polythene manufactured? Give their properties and uses.
Answer:
LDP means low density polyethylene. LDP is obtained by polymerization of ethylene under high pressure (1000 – 2000 atm) and temperature (350 – 570 K) in presence of traces of O₂ or peroxide as initiator.

The mechanism of this reaction involves free radical addition and H-atom abstraction. The latter results in branching. As a result the chains are loosely held and the polymer has low density.

Properties of LDP :

• LDP films are extremely flexible, but tough chemically inert and moisture resistant.
• It is poor conductor of electricity with melting point 110 °C.

Uses of LDP :

• LDP is mainly used in preparation of pipes for agriculture, irrigation, domestic water line connections as well as insulation to electric cables.
• It is also used in submarine cable insulation.
• It is used in producing extruded films, sheets, mainly for packaging and household uses like in preparation of squeeze bottles, attractive containers, etc.

HDP means high density polyethylene. It is a linear polymer with high density due to close packing.

HDP is obtained by polymerization of ethene in presence of Zieglar-Natta catalyst which is a combination of triethyl aluminium with titanium tetrachloride at a temperature of 333K to 343K and a pressure of 6-7 atm.

Properties of HDP :

• HDP is crystalline, melting point in the range of 144 – 150 °C.
• It is much stiffer than LDP and has high tensile strength and hardness.
• It is more resistant to chemicals than LDP.

Uses of HDP :

• HDP is used in manufacture of toys and other household articles like buckets, dustbins, bottles, pipes, etc.
• It is used to prepare laboratory wares and other objects where high tensile strength and stiffness is required.

Question ii.
Is synthetic rubber better than natural rubber? If so, in what respect?
Answer:
Yes. Synthetic rubber is more resistant to abrasion than natural rubber and is also superior in resistance to heat and the effects of aging (lasts longer). Many types of synthetic rubber are flame-resistant, so they can be used in insulation for electrical devices.

It also remains flexible at low temperatures and is resistant to grease and oil. It is resistant to heat, light and certain chemicals.

Question iii.
Write main specialities of Buna-S, Neoprene rubber?
Answer:
Buna-S is an elastomer and it is copolymer of styrene with butadiene. Its trade name is SBR. Buna-S is superior to natural rubber, because of its mechanical strength and abrasion resistance. It is used in tyre industry. It is vulcanized with sulphur. Neoprene is a synthetic rubber and it is a condensation polymer of chloroprene (2-chloro-l, 3-butadiene). Vulcanization of neoprene takes place in presence of MgO. It is resistant to petroleum, vegetable oils. Neoprene is used in making hose pipes for transport of gasoline and making gaskets.

Question iv.
Write the structure of isoprene and the polymer obtained from it.
Answer:

Question v.
Explain in detail the free radical mechanism involved during the preparation of the addition polymer.
Answer:
Polymerisation of ethylene is carried out at high temperature and at high pressure in presence of small amount of acetyl peroxide as initiator.

(1) Formation of free radicals : The first step involves clevage of acetyl peroxide to form two carboxy radicals. These carboxy radicals immediately undergo decarboxylation to give methyl initiator free radicals.

(2) Chain initiating step : The methyl radical thus formed adds to ethylene to form a new larger free radical.

(3) Chain propagation step : The larger free radical formed in the chain initiating step reacts with another molecule of ethene to form another big size free radicals and chain grows. This is called chain propagation step.

The chain reaction continues till thousands of ethylene molecules are added.

(4) Chain terminating step : The continuous chain reaction can be terminated by the combination of free radicals to form polyethene.

Activity :
i. Collect the information of the process like extrusion and moulding in Textile Industries.
ii. Make a list of polymers used to make the following articles
a. Photographic film
b. Frames of spectacles
c. Fountain pens
d. Moulded plastic chains
e. Terywool or Terycot fabric
iii. Prepare a report on factors responsible for degradation of polymers giving suitable example.
iv. Search and make a chart/note on silicones with reference to monomers, structure, properties and uses.
v. Collect the information and data about Rubber industry, plastic industry and synthetic fibre (rayon) industries running in India.

12th Chemistry Digest Chapter 15 Introduction to Polymer Chemistry Intext Questions and Answers

Use your brain power! (Textbook Page No 323)

Question 1.
Differentiate between natural and synthetic polymers.
Answer:

Natural polymers	Synthetic polymers
(1) The polymers are obtained either from plants or animals.	(1) The man made fibres prepared by polymerization of monomer or copolymerization of two or more monomers.
(2) They are further divided into two types : (i) plant polymers (ii) Animal polymers.Examples: Cotton, linen, latex	(2) They are further divided into three subtypes : (i) fibres (ii) synthetic rubbers (iii) plastics.Examples : Nylon, terylene Buna-S

Use your brain power! (Textbook Page No 325)

Question 1.
What is the type of polymerization in the following examples?

Answer:
(i) Addition polymerization
(ii) Condensation polymerization

Problem 15.1 : (Textbook Page No 326)

Question 1.
Refer to the following table listing for different polymers formed from respective monomers. Identify from the list whether it is copolymer or homopolymer.

Monomer	Polymers
Ethylene	Polyethene
Vinyl chloride	Polyvinyl chloride
Isobutylene	Polyisobutylene
Acrylonitrile	Polyacrylonitrile
Caprolactam	Nylon 6
Hexamethylene diammonium adipate	Nylon 6, 6
Butadiene + styrene	Buna-S

Solution :
In each of first five cases, there is only one monomer which gives corresponding homopolymer. In the sixth case hexamethylene diamine reacts with adipic acid to form the salt hexamethylene diammonium adipate which undergoes condensation to form Nylon 6, 6. Hence nylon 6, 6 is homopolymer. The polymer Buna-S is formed by polymerization of the monomers butadiene and styrene in presence of each other. The repeating units corresponding to the monomers butadiene and styrene are randomly arranged in the polymer. Hence Buna-S is copolymer.

Use your brain power! (Textbook Page No 328)

Question 1.
(1) From the cis-polyisoprene structure of natural rubber explain the low strength of van der Waals forces in it.
(2) Explain how the vulcanization of natural rubber improves its elasticity. (Hint : consider the intermolecular links.)
Answer:
(1) (i) Natural rubber is cis-polyisoprene. It is obtained by polymerization of isoprene units at 1, 4 positions. In rubber molecule, double bonds are located between C₂ and C₃ of each isoprene unit. These cis-double bonds do not allow the polymer chain to come closer. Therefore, only weak vander Waals’ forces are present. Since the chains are not linear, they can be stretched just like springs and exhibit elastic properties.

(ii) Cis-1, 4 polyisoprene (Natural rubber), due to this cis configuration about the double bonds, has the adjacent chain that do not fit together well (there is no close packing of adjacent chains). The only force that interact is the weak or low strength of van der Waals’ forces.

(iii) Cis-polyisoprene has a coiled structure in which the various polymer chains are held together by weak van der Waals’ forces.

(2) (i) Vulcanization of rubber is a process of improvement of the rubber elasticity and strength by heating it in the presence of sulphur, which results in three dimensional cross-linking of the chain rubber molecules (polyisoprene) bonded to each other by sulphur atoms.

(ii) Vulcanisation makes rubber more elastic and more stiffer. On vulcanisation, sulphur forms cross links at the reactive sites of double bonds and thus rubber get stiffened.

(iii) The improved properties of vulcanised rubber are (i) high elasticity (ii) low water absorption tendency,

(iii) resistance to oxidation.

Use your brain power! (Textbook Page No 334)

Question 1.
Write structural formulae of styrene and polybutadiene.
Answer:

(1) Classify the following polymers as addition or condensation.
(i) PVC (ii) Polyamides
(iii) Polystyrene
(iv) Polycarbonates
(v) Novolac
Answer:
Addition polymers: PVC, Polystyrene
(ondensatlon polymers: Polyamides. Polycarbonates, Novolac

Question 2.
Completed the following table :
Answer:

Use your brain power! (Textbook Page No 335)

(1) Represent the copolymerization reaction between glycine and e aminocaproic acid to form the copolymer nylon 2-nylon 6.
(2) What is the origin of the numbers 2 and 6 in the name of this polymer?
Answer:
(1) It is a copolymer and has polyamide linkages. The monomers glycine and e-amino caproic acid undergo condensation polymerisation to form nylon-2-nylon-6.

Nylon-2-nylon-6 is used in orthopaedic devices and implants.

(2) Monomer glycine contains two carbon atoms and e amino caproic acid contains six carbon atoms, hence the polymer is termed as nylon-2-nylon-6.

12th Std Chemistry Questions And Answers:

• Coordination Compounds Class 12 Chemistry Questions And Answers
• Halogen Derivatives Class 12 Chemistry Questions And Answers
• Alcohols, Phenols and Ethers Class 12 Chemistry Questions And Answers
• Aldehydes, Ketones and Carboxylic Acids Class 12 Chemistry Questions And Answers
• Amines Class 12 Chemistry Questions And Answers
• Biomolecules Class 12 Chemistry Questions And Answers
• Introduction to Polymer Chemistry Class 12 Chemistry Questions And Answers
• Green Chemistry and Nanochemistry Class 12 Chemistry Questions And Answers

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

Electrochemistry Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Chemistry Chapter 5 Exercise Solutions Maharashtra Board

Chemistry Class 12 Chapter 5 Exercise Solutions

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,

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