Chapter 2 Electrochemistry – NCERT Class 12 Chemistry

Question 2.1

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.

Analysis

A metal with a lower (more negative) standard reduction potential can displace a metal with a higher reduction potential from its salt solution. This follows the Electrochemical Series.

Standard Electrode Potentials ($E^\circ$):

Mg: -2.37 V
Al: -1.66 V
Zn: -0.76 V
Fe: -0.44 V
Cu: +0.34 V

Order of Displacement: Mg > Al > Zn > Fe > Cu

Question 2.2

Given standard electrode potentials: $K^+/K = -2.93V$, $Ag^+/Ag = 0.80V$, $Hg^{2+}/Hg = 0.79V$, $Mg^{2+}/Mg = -2.37V$, $Cr^{3+}/Cr = -0.74V$. Arrange in increasing order of reducing power.

Concept

Reducing power is the ability to lose electrons (get oxidized). A metal with a lower reduction potential has a higher tendency to oxidize, thus a higher reducing power.

Order of $E^\circ$ (Low to High):

K (-2.93 V) < Mg (-2.37 V) < Cr (-0.74 V) < Hg (0.79 V) < Ag (0.80 V)

Reducing Power: Ag < Hg < Cr < Mg < K

Question 2.3

Depict the galvanic cell in which the reaction $Zn(s) + 2Ag^+(aq) \to Zn^{2+}(aq) + 2Ag(s)$ takes place. Further show: (i) Negatively charged electrode, (ii) Current carriers, (iii) Reaction at each electrode.

Cell Representation

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

Answers

(i) Negatively charged electrode: The Anode (Zinc electrode) is negative.

(ii) Carriers of current: Electrons flow from Anode (Zn) to Cathode (Ag) in the external circuit. Ions flow in the internal circuit.

(iii) Reactions:
Anode (Oxidation): $Zn(s) \to Zn^{2+}(aq) + 2e^-$
Cathode (Reduction): $Ag^+(aq) + e^- \to Ag(s)$

Question 2.4

Calculate standard cell potentials, $\Delta_r G^\circ$, and equilibrium constant for:
(i) $2Cr(s) + 3Cd^{2+}(aq) \to 2Cr^{3+}(aq) + 3Cd(s)$
(ii) $Fe^{2+}(aq) + Ag^+(aq) \to Fe^{3+}(aq) + Ag(s)$

(i) Reaction 1

$E^\circ_{cell} = E^\circ_{cathode} – E^\circ_{anode} = E^\circ_{Cd} – E^\circ_{Cr}$
$= -0.40 – (-0.74) = 0.34 \text{ V}$.
$n = 6$ electrons exchanged.

$$\Delta_r G^\circ = -nFE^\circ = -6 \times 96487 \times 0.34 = -196833 \text{ J/mol} \approx -196.8 \text{ kJ/mol}$$ $$\log K_c = \frac{nE^\circ}{0.059} = \frac{6 \times 0.34}{0.059} \approx 34.57 \implies K_c \approx 3.7 \times 10^{34}$$

(ii) Reaction 2

$E^\circ_{cell} = E^\circ_{Ag} – E^\circ_{Fe^{3+}/Fe^{2+}} = 0.80 – 0.77 = 0.03 \text{ V}$.
$n = 1$.

$$\Delta_r G^\circ = -1 \times 96487 \times 0.03 = -2894.6 \text{ J/mol} \approx -2.89 \text{ kJ/mol}$$ $$\log K_c = \frac{1 \times 0.03}{0.059} \approx 0.508 \implies K_c \approx 3.22$$

Question 2.5

Write Nernst equation and emf of the following cells at 298 K:
(i) $Mg(s)|Mg^{2+}(0.001M)||Cu^{2+}(0.0001M)|Cu(s)$
(ii) $Fe(s)|Fe^{2+}(0.001M)||H^+(1M)|H_2(g)(1bar)|Pt(s)$

(i) Mg-Cu Cell

$E^\circ = 0.34 – (-2.37) = 2.71 \text{ V}$. $n=2$.

$$E = E^\circ – \frac{0.059}{2} \log \frac{[Mg^{2+}]}{[Cu^{2+}]}$$ $$E = 2.71 – 0.0295 \log \frac{10^{-3}}{10^{-4}} = 2.71 – 0.0295 \log(10)$$ $$E = 2.71 – 0.0295 = 2.68 \text{ V}$$

(ii) Fe-H Cell

$E^\circ = 0.00 – (-0.44) = 0.44 \text{ V}$. $n=2$.

$$E = 0.44 – \frac{0.059}{2} \log \frac{[Fe^{2+}]}{[H^+]^2}$$ $$E = 0.44 – 0.0295 \log \frac{10^{-3}}{1^2} = 0.44 – 0.0295(-3)$$ $$E = 0.44 + 0.0885 = 0.5285 \text{ V}$$

Similarly solve (iii) & (iv). Results: (i) 2.68V, (ii) 0.53V.

Question 2.6

In button cells: $Zn(s) + Ag_2O(s) + H_2O(l) \to Zn^{2+}(aq) + 2Ag(s) + 2OH^-(aq)$. Determine $\Delta_r G^\circ$ and $E^\circ$.

1. Standard E.M.F.

Anode: $Zn \to Zn^{2+} + 2e^-$ ($E^\circ_{Zn} = -0.76 \text{ V}$).
Cathode: $Ag_2O + H_2O + 2e^- \to 2Ag + 2OH^-$ ($E^\circ_{Ag} = 0.344 \text{ V}$).
$E^\circ_{cell} = 0.344 – (-0.76) = 1.104 \text{ V}$.

2. Gibbs Energy

$n = 2$.

$$\Delta_r G^\circ = -nFE^\circ = -2 \times 96487 \times 1.104$$ $$\Delta_r G^\circ = -213043 \text{ J/mol} \approx -2.13 \times 10^5 \text{ J/mol}$$

$E^\circ = 1.104 \text{ V}$, $\Delta G^\circ = -2.13 \times 10^5 \text{ J}$.

Question 2.7

Define conductivity and molar conductivity. Discuss their variation with concentration.

Conductivity ($\kappa$): Conductance of a solution of 1 cm length and 1 cm² cross-section. It decreases with decrease in concentration because the number of ions per unit volume decreases.
Molar Conductivity ($\Lambda_m$): Conducting power of all ions produced by 1 mole of electrolyte. It increases with decrease in concentration because the volume containing 1 mole increases, reducing interaction (weak) or increasing dissociation (weak).

Question 2.8

Conductivity of 0.20 M KCl is 0.0248 S/cm. Calculate molar conductivity.

Calculation

$\kappa = 0.0248 \text{ S cm}^{-1}$, $C = 0.20 \text{ M}$.

$$\Lambda_m = \frac{\kappa \times 1000}{C}$$ $$\Lambda_m = \frac{0.0248 \times 1000}{0.20} = \frac{24.8}{0.20} = 124 \text{ S cm}^2 \text{ mol}^{-1}$$

Result: $124 \text{ S cm}^2 \text{ mol}^{-1}$

Question 2.9

Resistance of 0.001 M KCl is $1500 \Omega$. Conductivity is $0.146 \times 10^{-3} \text{ S cm}^{-1}$. Calculate cell constant.

Formula

Cell Constant $G^* = \text{Conductivity} \times \text{Resistance}$ ($G^* = \kappa R$).

$$G^* = (0.146 \times 10^{-3}) \times 1500$$ $$G^* = 0.146 \times 1.5 = 0.219 \text{ cm}^{-1}$$

Cell Constant = 0.219 cm⁻¹

Question 2.10

The conductivity of NaCl at different concentrations is given. Calculate $\Lambda_m$ for all and find $\Lambda_m^\circ$ by plotting against $c^{1/2}$.

Method

1. Calculate $\Lambda_m = (\kappa \times 1000)/C$ for each data point (Note units: convert S/m to S/cm if needed or keep consistent).
2. Calculate $\sqrt{C}$ for each.
3. Plot $\Lambda_m$ (y-axis) vs $\sqrt{C}$ (x-axis).
4. Extrapolate the straight line to $C=0$ to find intercept $\Lambda_m^\circ$.

From plot extrapolation, $\Lambda_m^\circ \approx 126.4 \text{ S cm}^2 \text{ mol}^{-1}$.

Question 2.11

Conductivity of 0.00241 M acetic acid is $7.896 \times 10^{-5} \text{ S cm}^{-1}$. Calculate $\Lambda_m$. If $\Lambda_m^\circ = 390.5$, find dissociation constant.

1. Molar Conductivity

$$\Lambda_m = \frac{7.896 \times 10^{-5} \times 1000}{0.00241} = \frac{0.07896}{0.00241} \approx 32.76 \text{ S cm}^2 \text{ mol}^{-1}$$

2. Degree of Dissociation

$\alpha = \frac{\Lambda_m}{\Lambda_m^\circ} = \frac{32.76}{390.5} \approx 0.084$.

3. Dissociation Constant

$$K_a = \frac{C \alpha^2}{1-\alpha} = \frac{0.00241 \times (0.084)^2}{1 – 0.084}$$ $$K_a = \frac{0.00241 \times 0.007056}{0.916} \approx 1.86 \times 10^{-5} \text{ mol/L}$$

$K_a \approx 1.86 \times 10^{-5}$

Question 2.12

How much charge is required for: (i) 1 mol $Al^{3+} \to Al$, (ii) 1 mol $Cu^{2+} \to Cu$, (iii) 1 mol $MnO_4^- \to Mn^{2+}$?

(i) $Al^{3+} + 3e^- \to Al$: Requires 3F charge. $3 \times 96487 = 289461 \text{ C}$.

(ii) $Cu^{2+} + 2e^- \to Cu$: Requires 2F charge. $2 \times 96487 = 192974 \text{ C}$.

(iii) $MnO_4^- (+7) \to Mn^{2+} (+2)$: Change is 5 units. Requires 5F. $5 \times 96487 = 482435 \text{ C}$.

Question 2.13

How much electricity in Faraday is required to produce: (i) 20g Ca from molten $CaCl_2$, (ii) 40g Al from molten $Al_2O_3$?

(i) Calcium

$Ca^{2+} + 2e^- \to Ca$. Molar mass = 40g.
40g requires 2F. So, 20g requires 1F.

(ii) Aluminium

$Al^{3+} + 3e^- \to Al$. Molar mass = 27g.
27g requires 3F.
40g requires $\frac{3}{27} \times 40 = \frac{40}{9} \approx 4.44 \text{ F}$.

Question 2.14

How much electricity (Coulombs) for oxidation of: (i) 1 mol $H_2O \to O_2$, (ii) 1 mol $FeO \to Fe_2O_3$?

(i) Water

$2H_2O \to O_2 + 4H^+ + 4e^-$.
2 moles water require 4F. So, 1 mol requires 2F.
$2 \times 96487 = 192974 \text{ C}$.

(ii) Ferrous to Ferric

$Fe^{2+} \to Fe^{3+} + e^-$.
1 mol requires 1F = 96487 C.

Question 2.15

Solution of $Ni(NO_3)_2$ electrolysed with 5A for 20 min. Mass of Ni deposited?

Calculation

$I = 5 \text{ A}, t = 20 \times 60 = 1200 \text{ s}$.
$Q = It = 5 \times 1200 = 6000 \text{ C}$.
Reaction: $Ni^{2+} + 2e^- \to Ni$. (Molar mass = 58.7 g/mol).
2F (192974 C) deposits 58.7 g.
6000 C deposits $\frac{58.7}{192974} \times 6000 \approx 1.825 \text{ g}$.

Mass of Nickel = 1.825 g

Question 2.16

Three cells A (ZnSO4), B (AgNO3), C (CuSO4) in series. 1.5 A current passed until 1.45 g Ag deposited. Find time and mass of Cu and Zn.

1. Time Calculation

Ag: $Ag^+ + e^- \to Ag$ (108 g/mol).
Charge $Q = \frac{1.45}{108} \times 96487 \approx 1295.6 \text{ C}$.
Time $t = Q/I = 1295.6 / 1.5 \approx 863.7 \text{ s}$ (approx 14.4 min).

2. Mass of Cu and Zn

Cu (Valency 2, M=63.5): Mass $= \frac{Q}{2F} \times 63.5 = \frac{1295.6}{2 \times 96487} \times 63.5 \approx 0.426 \text{ g}$.

Zn (Valency 2, M=65.3): Mass $= \frac{1295.6}{2 \times 96487} \times 65.3 \approx 0.438 \text{ g}$.

Question 2.17

Predict feasibility using standard potentials: (i) $Fe^{3+}$ and $I^-$, etc.

A reaction is feasible if $E^\circ_{cell} > 0$.

(i) $2Fe^{3+} + 2I^- \to 2Fe^{2+} + I_2$: $E^\circ = 0.77 – 0.54 = 0.23 \text{ V}$. Feasible.
(ii) $2Ag^+ + Cu \to 2Ag + Cu^{2+}$: $E^\circ = 0.80 – 0.34 = 0.46 \text{ V}$. Feasible.
(iii) $Fe^{3+} + Br^-$: $E^\circ = 0.77 – 1.09 = -0.32 \text{ V}$. Not Feasible.

Question 2.18

Predict products of electrolysis: (i) Aqueous $AgNO_3$ (Silver electrodes), (ii) Aqueous $AgNO_3$ (Platinum), etc.

(i) Ag electrodes: Cathode: Ag deposited. Anode: Ag dissolves (Oxidation of Ag electrode).
(ii) Pt electrodes: Cathode: Ag deposited. Anode: $O_2$ gas evolved (Water oxidation preferred over nitrate).
(iii) Dilute $H_2SO_4$: Cathode: $H_2$ gas. Anode: $O_2$ gas.
(iv) Aqueous $CuCl_2$: Cathode: Cu deposited. Anode: $Cl_2$ gas evolved.

NCERT Solutions