# NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry

NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry plays a pivotal role in the CBSE class 12 Chemistry examination. Chemistry Class 12 solutions Chapter 3 is a comprehensive material that has answers to the textbook questions, important questions from previous papers. By studying chemistry Class 12 NCERT solutions Chapter 3, you will be able to solve different kinds of questions you can expect to appear in the main examination and entrance examinations.

Electrochemistry is the branch of chemistry that deals with the relationship between chemical energy and electrical energy produced in a redox reaction and how they can be converted into each other. The solutions are prepared by the best subject experts. In essence, these solutions can be useful for those students preparing for class 12 board exams and also for JEE advance and other medical entrance exams. Students can successfully answer the numerical problems based on electrochemistry by downloading the free pdf.

## Class 12 NCERT Solutions for Electrochemistry

The NCERT solutions for Chapter 3 – Electrochemistry has mainly been designed to help the students in preparing well and score good marks in CBSE class 12 Chemistry paper. Further, the solutions consist of well thought or structured questions along with detailed explanations to help students learn and remember concepts easily.

### Subtopics for Class 12 Chemistry Chapter 3 – Electrochemistry

1. Electrochemical Cells
2. Galvanic Cells
1. Measurement of Electrode Potential
3. Nernst Equation
1. Equilibrium Constant from Nernst Equation
2. Electrochemical Cell and Gibbs Energy of Reaction
4. The conductance of Electrolytic Solutions
1. Measurement of the Conductivity of Ionic Solutions
2. Variation of Conductivity and Molar Conductivity with Concentration
5. Electrolytic Cells and Electrolysis
1. Products of Electrolysis
6. Batteries
1. Primary Batteries
2. Secondary Batteries
7. Fuel Cells
8. Corrosion

After studying electrochemistry class 12 important questions and solutions, you will be able to describe an electrochemical cell and differentiate between galvanic and electrolytic cells. You will study the application of the Nernst equation for calculating the emf of a galvanic cell and define the standard potential of the cell.

This chapter has derivations of the relation between the standard potential of the cell, Gibbs energy of cell reaction and its equilibrium constant. This solution will give the definition of resistivity (p), conductivity (K) and molar conductivity ( Am) of ionic solutions; differentiate between ionic (electrolytic) and electronic conductivity; describe the method for measurement of conductivity of electrolytic solutions and calculation of their molar conductivity; justify the variation of conductivity and molar conductivity of solutions with change in their concentration and define Aom (molar conductivity at zero concentration or infinite dilution); enunciate Kohlrausch law and learn its applications; understand quantitative aspects of electrolysis; describe the construction of some primary and secondary batteries and fuel cells and explain corrosion as an electrochemical process.

### Class 12 Chemistry NCERT Solutions (Electrochemistry) – Important Questions

Q 3.1:

Arrange the following metals in the order in which they displace each other from the solution of their salts. Al, Cu, Fe, Mg and Zn

According to their reactivity, the given metals replace the others from their salt solutions in the said order: Mg, Al, Zn, Fe, Cu.

Mg: Al: Zn: Fe: Cu

Q 3.2:

Given the standard electrode potentials,
K+/K = –2.93V,

Ag+/Ag = 0.80V,
Hg2+/Hg = 0.79V
Mg2+/Mg = –2.37 V, Cr3+/Cr = – 0.74V
Arrange these metals in their increasing order of reducing power.

Ans:

The reducing power increases with the lowering of reduction potential. In order of given standard electrode potential (increasing order) : K+/K < Mg2+/Mg < Cr3+/Cr < Hg2+/Hg < Ag+/Ag

Thus, in the order of reducing power, we can arrange the given metals as Ag< Hg < Cr < Mg < K

Q 3.3 :

Depict the galvanic cell in which the reaction
Zn(s)+2Ag+(aq) →Zn2+(aq)+2Ag(s) takes place. Further show:
(i) Which of the electrode is negatively charged?
(ii) The carriers of the current in the cell.
(iii) Individual reaction at each electrode.

Ans :

The galvanic cell in which the given reaction takes place is depicted as:

$Zn_{ ( s ) } | Zn^{ 2+ }_{( aq )}||Ag^{ + }_{( aq )}|Ag_{( s )}$

(i) The negatively charged electrode is the Zn electrode (anode)

(ii) The current carriers in the cell are ions. Current flows to zinc from silver in the external circuit.

(iii) Reaction at the anode is given by :

$Zn_{ ( s ) }\rightarrow Zn^{ 2+ }_{( aq )} + 2 e^-$

Reaction at the anode is given by :

$Ag^{+}_{ ( aq ) } + e^- \rightarrow Ag_{( s )}$

Q 3.4:

Calculate the standard cell potentials of galvanic cell in which the following
reactions take place:
(i) 2Cr(s) + 3Cd2+(aq) → 2Cr3+(aq) + 3Cd
(ii) Fe2+(aq) + Ag+(aq) → Fe3+(aq) + Ag(s)
Calculate the ∆rGJ and equilibrium constant of the reactions.

Ans :

(i) $E^{\Theta}_{Cr^{3+}/Cr}$ = 0.74 V

$E^{\Theta}_{Cd^{2+}/Cd}$ = -0.40 V

The galvanic cell of the given reaction is depicted as :

$Cr_{ ( s ) }|Cr^{ 3+ }_{ ( aq ) }||Cd^{ 2+ }_{ aq }|Cd_{ ( s ) }$

Now, the standard cell potential is

$E^{\Theta }_{cell} = E^{\Theta }_{g}-E^{\Theta }_{L}$

= – 0.40 – ( -0.74 )

= + 0.34 V

In the given equation, n = 6

F = 96487 C mol−1

$E^{\Theta }_{cell}$ = + 0.34 V

Then, $\Delta_rG^{\Theta}$ = −6 × 96487 C mol−1 × 0.34 V

= −196833.48 CV mol−1

= −196833.48 J mol−1

= −196.83 kJ mol−1

Again,

$\Delta_rG^{\Theta} = – R T ln K$ $\Delta_rG^{\Theta} = – 2.303 R T ln K$ $log {k} = \frac{\Delta_rG}{ 2.303 R T }$ $= \frac{-196.83\times10^{3}}{ 2.303 \times8.314\times 298 }$

= 34.496

K = antilog (34.496) = 3.13 × 1034

The galvanic cell of the given reaction is depicted as:

$Fe^{ 2+ }_{( aq )}|Fe^{ 3+ }_{( aq )}|| Ag^+_{( aq )}|Ag_{( s )}$

Now, the standard cell potential is

$E^{\Theta }_{cell} = E^{\Theta }_{g}-E^{\Theta }_{L}$

Here, n = 1.

Then, $\Delta_t G^0 = -nFE^0_{cell}$

= −1 × 96487 C mol−1 × 0.03 V

= −2894.61 J mol−1

= −2.89 kJ mol−1

Again,  $\Delta_t G^0 = -2.303 RT \; ln K$ $ln K = \frac{\Delta_t G}{ 2.303 RT }$ $= \frac{-2894.61 }{ 2.303 \times 8.314 \times 298 }$

= 0.5073

K = antilog (0.5073)

= 3.2 (approximately)

Q 3.5:

Write the Nernst equation and emf of the following cells at 298 K:
(i) Mg(s)|Mg2+(0.001M)||Cu2+(0.0001 M)|Cu(s)
(ii) Fe(s)|Fe2+(0.001M)||H+(1M)|H2(g)(1bar)| Pt(s)
(iii) Sn(s)|Sn2+(0.050 M)||H+(0.020 M)|H2(g) (1 bar)|Pt(s)
(iv) Pt(s)|Br–(0.010 M)|Br2(l )||H+(0.030 M)| H2(g) (1 bar)|Pt(s).

(i) For the given reaction, the Nernst equation can be given as:

$E_{cell} = E^0_{cell} – \frac{0.591}{n}log\frac{[Mg^{2+}]}{[Cu^{2+}]}$ $= 0.34 – (-2.36) – \frac{0.0591}{2} log \frac{0.001}{0.0001}$ $2.7 -\frac{0.0591}{2}log10$

= 2.7 − 0.02955

= 2.67 V (approximately)

(ii) For the given reaction, the Nernst equation can be given as:

$E_{cell} = E^0_{cell} – \frac{0.591}{n}log\frac{[Fe^{2+}]}{[H ^{+}]^2}$

= 0 – ( – 0.14) – $\frac{0.0591}{n}log\frac{0.050}{(0.020)^{2}}$

= 0.52865 V

= 0.53 V (approximately)

(iii) For the given reaction, the Nernst equation can be given as:

$E_{cell} = E^0_{cell} – \frac{0.591}{n}log\frac{[Sn^{2+}]}{[H ^{+}]^2}$

= 0 – ( – 0.14) – $\frac{0.591}{2}log\frac{0.050}{(0.020)^2}$

= 0.14 − 0.0295 × log125

= 0.14 − 0.062

= 0.078 V

= 0.08 V (approximately)

(iv) For the given reaction, the Nernst equation can be given as:

$E_{cell} = E^0_{cell} – \frac{0.591}{n}log\frac{1}{[Br^{-}]^2[H ^{+}]^2}$

= 0 – 1.09 – $\frac{0.591}{2}log\frac{1}{(0.010)^2(0.030)^2}$

= -1.09 – 0.02955 x $log\frac{1}{0.00000009}$

= -1.09 – 0.02955 x $log\frac{1}{9\times 10^{-8}}$

= -1.09 – 0.02955 x $log{ (1.11 \times 10^{7} )}$

= -1.09 – 0.02955 x (0.0453 + 7)

= -1.09 – 0.208

= -1.298 V

Q 3.6:

In the button cells widely used in watches and other devices the following reaction takes place:

Determine ∆r GJ and EJ for the reaction.

Ans:

$E^0$ = 1.104 V

We know that,

$\Delta_r G^\Theta = -nFE^\Theta$

= −2 × 96487 × 1.04

= −213043.296 J

= −213.04 kJ

Q 3.7:

Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.

The conductivity of a solution is defined as the conductance of a solution of 1 cm in length and area of cross-section 1 sq. cm. Specific conductance is the inverse of resistivity and it is represented by the symbol κ. If ρ is resistivity, then we can write:

$k = \frac{1}{\rho}$

At any given concentration, the conductivity of a solution is defined as the unit volume of solution kept between two platinum electrodes with the unit area of the cross-section at a distance of unit length.

$G = k \frac{a}{l} = k \times 1 = k$     [Since a = 1 , l = 1]

When concentration decreases there will a decrease in Conductivity. It is applicable for both weak and strong electrolyte. This is because the number of ions per unit volume that carry the current in a solution decreases with a decrease in concentration.

Molar conductivity –

Molar conductivity of a solution at a given concentration is the conductance of volume V of a solution containing 1 mole of the electrolyte kept between two electrodes with the area of cross-section A and distance of unit length.

$\Lambda_m = k \frac{A}{l}$

Now, l = 1 and A = V (volume containing 1 mole of the electrolyte).

$\Lambda_m = k V$

Molar conductivity increases with a decrease in concentration. This is because the total volume V of the solution containing one mole of the electrolyte increases on dilution. The variation of $\Lambda_m$ with $\sqrt{c}$ for strong and weak electrolytes is shown in the following plot :

Q  3.8:

The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S cm–1. Calculate its molar conductivity

Ans :

Given, κ = 0.0248 S cm−1 c

c = 0.20 M

Molar conductivity, $\Lambda_m = \frac{k \times 1000}{c}$ $= \frac{0.0248 \times 1000}{0.2}$

= 124 Scm2mol-1

Q 3.9:

The resistance of a conductivity cell containing 0.001M KCl solution at 298 K is 1500 Ω. What is the cell constant if the conductivity of 0.001M KCl solution at 298 K is 0.146 × 10–3 S cm–1

Given,

Conductivity, k = 0.146 × 10−3 S cm−1

Resistance, R = 1500 Ω

Cell constant = k × R

= 0.146 × 10−3 × 1500

= 0.219 cm−1

Q 3.10:

The conductivity of sodium chloride at 298 K has been determined at different concentrations and the results are given below:

Concentration/M            0.001     0.010     0.020     0.050     0.100

102 × k/S m−1                      1.237     11.85     23.15     55.53     106.74

Calculate Λm for all concentrations and draw a plot between Λm and c½. Find the value of 0 Λ m.

Ans:

Given,

κ = 1.237 × 10−2 S m−1, c = 0.001 M

Then, κ = 1.237 × 10−4 S cm−1, c1⁄2 = 0.0316 M1/2

$\Lambda_m =\frac{k}{c}$ $=\frac{1.237 \times 10^{ -4 } S\;cm^{-1} }{0.001 \; mol\; \; L^{ -1 }}\times \frac{1000\;cm^{-1}}{L}$

= 123.7 S cm2 mol−1

Given,

κ = 11.85 × 10−2 S m−1, c = 0.010M

Then, κ = 11.85 × 10−4 S cm−1, c1⁄2 = 0.1 M1/2

$\Lambda_m =\frac{k}{c}$ $=\frac{11.85 \times 10^{ -4 } S\;cm^{-1} }{0.010 \; mol\; \; L^{ -1 }}\times \frac{1000\;cm^{-1}}{L}$

= 118.5 S cm2 mol−1

Given,

κ = 23.15 × 10−2 S m−1, c = 0.020 M

Then, κ = 23.15 × 10−4 S cm−1, c1/2 = 0.1414 M1/2

$\Lambda_m =\frac{k}{c}$ $=\frac{23.15 \times 10^{ -4 } S\;cm^{-1} }{0.020 \; mol\; \; L^{ -1 }}\times \frac{1000\;cm^{-1}}{L}$

= 115.8 S cm2 mol−1

Given,

κ = 55.53 × 10−2 S m−1, c = 0.050 M

Then, κ = 55.53 × 10−4 S cm−1, c1/2 = 0.2236 M1/2

$\Lambda_m =\frac{k}{c}$ $=\frac{106.74 \times 10^{ -4 } S\;cm^{-1} }{0.050 \; mol\; \; L^{ -1 }}\times \frac{1000\;cm^{-1}}{L}$

= 111.1 1 S cm2 mol−1

Given,

κ = 106.74 × 10−2 S m−1, c = 0.100 M

Then, κ = 106.74 × 10−4 S cm−1, c1/2 = 0.3162 M1/2

$\Lambda_m =\frac{k}{c}$ $=\frac{106.74 \times 10^{ -4 } S\;cm^{-1} }{0.100 \; mol\; \; L^{ -1 }}\times \frac{1000\;cm^{-1}}{L}$

= 106.74 S cm2 mol−1

Now, we have the following data :

Since the line interrupts $\Lambda_m$ at 124.0 S cm2 mol−1, $\Lambda^0_m$ =  124.0 S cm2 mol−1

Q 3.11:

Conductivity of 0.00241 M acetic acid is 7.896 × 10–5 S cm–1. Calculate its molar conductivity. If 0 Λ m for acetic acid is 390.5 S cm2 mol–1, what is its dissociation constant?

Ans:

Given, κ = 7.896 × 10−5 S m−1 c

= 0.00241 mol L−1

Then, molar conductivity, $\Lambda_m = \frac{k}{c}$

= $\frac{7.896 \times 10^{-5} S cm^{-1}}{0.00241 \; mol \; L^{-1}}\times \frac{1000 cm^3}{L}$

= 32.76S cm2 mol−1

$\Lambda^0_m =$ 390.5 S cm2 mol−1

Again,

$\alpha =\frac{\Lambda_m }{\Lambda^0_m }$

= $= \frac{32.76 \; S\; cm^2 \; mol^{-1} }{390.5 \; S\; cm^2 \; mol^{-1} }$

Now,

= 0.084

Dissociation constant, $K_a = \frac{c\alpha^2}{(1-\alpha)}$

= $\frac{ ( 0.00241 \; mol \; L^{-1} )( 0.084 )^2}{ ( 1 – 0.084 ) }$

= 1.86 × 10−5 mol L−1

Q 3.12:

How much charge is required for the following reductions:
(i) 1 mol of Al3+ to Al?
(ii) 1 mol of Cu2+ to Cu?
(iii) 1 mol of MnO4– to Mn2+?

Ans :

(i) $Al^{3+} + 3e^- \rightarrow Al$

Required charge = 3 F

= 3 × 96487 C

= 289461 C

(ii) $Cu^{2+} + 2e^- \rightarrow Cu$

Required charge = 2 F

= 2 × 96487 C

= 192974 C

(iii) $MnO^-_4 \rightarrow Mn^{2+}$

i.e $Mn^{7+} + 5e^-\rightarrow Mn^{2+}$

Required charge = 5 F

= 5 × 96487 C

= 482435 C

Q 3.13:

How much electricity in terms of Faraday is required to produce
(i) 20.0 g of Ca from molten CaCl2?
(ii) 40.0 g of Al from molten Al2O3?’

Ans:

(i)  From given data,

$Ca^{2+} + 2e^- \rightarrow Ca$

Electricity required to produce 40 g of calcium = 2 F

Therefore, electricity required to produce 20 g of calcium = (2 x 20 )/ 40 F

= 1 F

(ii) From given data,

$Al^{3+} + 3e^- \rightarrow Al$

Electricity required to produce 27 g of Al = 3 F

Therefore, electricity required to produce 40 g of Al = ( 3 x 40 )/27 F

= 4.44 F

Q 3.14:

How much electricity is required in coulomb for the oxidation of
(i) 1 mol of H2O to O2?
(ii) 1 mol of FeO to Fe2O3?

Ans :

(i) From given data,

$H_2O\rightarrow H_2 + \frac{1}{2}O_2$

We can say that :

$O^{2-}\rightarrow \frac{1}{2}O_2 + 2e^-$

Electricity required for the oxidation of 1 mol of H2O to O2 = 2 F

= 2 × 96487 C

= 192974 C

(ii) From given data,

$Fe^{2+}\rightarrow Fe^{3+} + e^-$

Electricity required for the oxidation of 1 mol of FeO to Fe2O3 = 1 F

= 96487 C

Q 3.15:

A solution of Ni(NO3)2 is electrolysed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode?

Ans :

Given,

Current = 5A

Time = 20 × 60 = 1200 s

Charge = current × time

= 5 × 1200

= 6000 C

According to the reaction,

$Ni^{2+} + 2e^-\rightarrow Ni_{ (s) } + e^-$

Nickel deposited by 2 × 96487 C = 58.71 g

Therefore, nickel deposited by 6000 C = $\frac{58.71 \times 6000}{2 \times 96487}g$

= 1.825 g

Hence, 1.825 g of nickel will be deposited at the cathode.

Q 3.16:

Three electrolytic cells A,B,C containing solutions of ZnSO4, AgNO3 and CuSO4, respectively are connected in series. A steady current of 1.5 amperes was passed through them until 1.45 g of silver deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited?

Ans :

According to the reaction:

$Ag^+_{(aq)} +e^- \rightarrow Ag_{(s)}$

i.e., 108 g of Ag is deposited by 96487 C.

Therefore, 1.45 g of Ag is deposited by = $\frac{96487\times 1.45}{107}C$

= 1295.43 C

Given,

Current = 1.5 A

Time = 1295.43/ 1.5 s

= 863.6 s

= 864 s

= 14.40 min

Again,

$Cu^{2+}_{(aq)} +2e^-\rightarrow Cu_{(s)}$

i.e., 2 × 96487 C of charge deposit = 63.5 g of Cu

Therefore, 1295.43 C of charge will deposit $\frac{63.5 \times 1295.43}{2 \times 96487}$

= 0.426 g of Cu

$Zn^{2+}_{(aq)} +2e^-\rightarrow Zn_{(s)}$

i.e., 2 × 96487 C of charge deposit = 65.4 g of Zn

Therefore, 1295.43 C of charge will deposit $\frac{65.4 \times 1295.43}{2 \times 96487}$

= 0.439 g of Zn

Q 3.17:

Using the standard electrode potentials given in Table 3.1, predict if the reaction between the following is feasible:
(i) Fe3+(aq) and I–(aq)
(ii) Ag+ (aq) and Cu(s)
(iii) Fe3+ (aq) and Br– (aq)
(iv) Ag(s) and Fe 3+ (aq)
(v) Br2 (aq) and Fe2+ (aq).

Ans :

(i)

(ii)

E0  is positive, hence reaction is feasible.

(iii)

E0  is negative, hence reaction is not feasible.

(iv)

E0  is negative, hence reaction is not feasible.

(v)

E0  is positive, hence reaction is feasible.

Q  3.18:

Predict the products of electrolysis in each of the following:
(i) An aqueous solution of AgNO3 with silver electrodes.
(ii) An aqueous solution of AgNO3 with platinum electrodes.
(iii) A dilute solution of H2SO4 with platinum electrodes.
(iv) An aqueous solution of CuCl2 with platinum electrodes.

Ans:

(i) At cathode:

The following reduction reactions compete to take place at the cathode.

$Ag^+_{(aq)}+e^- \rightarrow Ag_{(s)}$ ; E0 = 0.80 V

$H^+_{(aq)}+e^- \rightarrow \frac{1}{2}H_{2(g)}$ ;E0 = 0.00 V

The reaction with a higher value of E0 takes place at the cathode. Therefore, deposition of silver will take place at the cathode.

At anode:

The Ag anode is attacked by $NO^+_3$  ions. Therefore, the silver electrode at the anode dissolves in the solution to form Ag+.

(ii) At cathode:

The following reduction reactions compete to take place at the cathode.

$Ag^+_{(aq)}+e^- \rightarrow Ag_{(s)}$ ; E0 = 0.80 V

$H^+_{(aq)}+e^- \rightarrow \frac{1}{2}H_{2(g)}$ ;E0 = 0.00 V

The reaction with a higher value of E0 takes place at the cathode. Therefore, deposition of silver will take place at the cathode.

At anode:

Since Pt electrodes are inert, the anode is not attacked by $NO^+_3$ ions. Therefore, OH or $NO^+_3$ ions can be oxidized at the anode. But OH ions having a lower discharge potential  and get preference and decompose to liberate O2.

$OH^-\rightarrow OH + E^-$ $4OH^-\rightarrow 2H_2O + O_2$

(iii) At the cathode, the following reduction reaction occurs to produce H2 gas.

$H^+_{(aq)}+e^-\rightarrow \frac{1}{2}H_{2(g)}$

At the anode, the following processes are possible.

$2H_2O_{(l)}\rightarrow O_{2(g)} + 4H^+_{(aq)}+4e^-$ ; E0 = +1.23 V             —–(i)

$2SO^{2-}_{4(aq)}\rightarrow S_2O^{2-}_{6(aq)} + 2e^-$ ; E0 = +1.96 V          —–(ii)

For dilute sulphuric acid, reaction (i) is preferred to produce O2 gas. But for concentrated sulphuric acid, reaction (ii) occurs.

(iv) At cathode:

The following reduction reactions compete to take place at the cathode.

$Cu^{2+}_{(aq)}+2e^- \rightarrow Cu_{(s)}$ ; E0 = 0.34 V

$H^+_{(aq)}+e^- \rightarrow \frac{1}{2}H_{2(g)}$ ;E0 = 0.00 V

The reaction with a higher value of takes place at the cathode. Therefore, deposition of copper will take place at the cathode.

At anode:

The following oxidation reactions are possible at the anode.

$Cl^{-}_{(aq)} \rightarrow \frac{1}{2} Cl_{2(g)}+e^-$; E0 = 1.36 V

$2H_20_{(l)} \rightarrow O_{2(g)} + 4H^+_{(aq)} +e^-$; E0 = +1.23 V

At the anode, the reaction with a lower value of E0 is preferred. But due to the over potential of oxygen, Cl gets oxidized at the anode to produce Cl2 gas.

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