Class 12 Chemistry Chapter 2 Electrochemistry ncert solutions

Electrochemistry Class 12 ncert solutions: Chemistry Class 12 Chapter 2 exercise solutions

Textbook	NCERT
Class	Class 12
Subject	Chemistry
Chapter	Chapter 2
Chapter Name	Electrochemistry ncert solutions
Category	Ncert Solutions
Medium	English

Are you looking for Class 12 Chemistry Chapter 2 Electrochemistry ncert solutions? Now you can download Ncert Chemistry Class 12 Chapter 2 exercise solutions pdf from here.

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.

Solution 2.1: According to their reactivity, the given metals replace the others from their salt solutions in the said order: Mg, Al, Zn, Fe, and Cu.

Question 2.2: 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.

Solution 2.2: The reducing power increases with the lowering of the 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

Question 2.3: 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.

Solution 2.3: The galvanic cell in which the given reaction takes place is depicted as

Zn _(S)| Zn²⁺ _(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)  →   Zn²⁺(aq) + 2e-

Reaction at the anode is given by Ag⁺(aq)  + e^–  →  Ag(s)

Question 2.4: 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 ∆_rG° and equilibrium constant of the reactions.

Solution 2.4:

Write down the half-reactions for each reaction.

(i) Reaction: \(( 2 \text{Cr}(s) + 3 \text{Cd}^{2+}(aq) \rightarrow \)\(2 \text{Cr}^{3+}(aq) + 3 \text{Cd}(s) )\)

The half-reactions are:

• Oxidation (Cr): \( \text{Cr}(s) \rightarrow \text{Cr}^{3+}(aq) + 3e^- \)
• Reduction (Cd): \( \text{Cd}^{2+}(aq) + 2e^- \rightarrow \text{Cd}(s) \)

(ii) Reaction: \( \text{Fe}^{2+}(aq) + \text{Ag}^+(aq) \rightarrow\)\( \text{Fe}^{3+}(aq) + \text{Ag}(s) \)

The half-reactions are:

• Oxidation (Fe): \( \text{Fe}^{2+}(aq) \rightarrow \text{Fe}^{3+}(aq) + e^- \)
• Reduction (Ag): \( \text{Ag}^+(aq) + e^- \rightarrow \text{Ag}(s) \)

Use standard reduction potentials to calculate the standard cell potential \(( E^\circ_{\text{cell}} )\).

For each half-reaction, the standard reduction potentials \(( E^\circ )\) can be found in tables. Here are typical values:

• \( E^\circ(\text{Cr}^{3+}/\text{Cr}) = -0.74 \, V \)
• \( E^\circ(\text{Cd}^{2+}/\text{Cd}) = -0.40 \, V \)
• \( E^\circ(\text{Fe}^{3+}/\text{Fe}^{2+}) = +0.77 \, V \)
• \( E^\circ(\text{Ag}^+/\text{Ag}) = +0.80 \, V \)

The standard cell potential is calculated by subtracting the anode (oxidation) potential from the cathode (reduction) potential:

(i) E°_cell for the reaction \( 2\text{Cr} + 3\text{Cd}^{2+} \rightarrow\)\( 2\text{Cr}^{3+} + 3\text{Cd} \):

E°_cell = E°_cathode \((\text{Cd}^{2+}/\text{Cd}) – E^\circ_{\text{anode}}(\text{Cr}^{3+}/\text{Cr}) \)

\(= (-0.40 \, \text{V}) – (-0.74 \, \text{V}) = 0.34 \, \text{V}\)

(ii) E°_cell for the reaction \( \text{Fe}^{2+} + \text{Ag}^+ \rightarrow \text{Fe}^{3+} + \text{Ag} \):

E°_cell = E°_cathode \((\text{Ag}^+/\text{Ag}) – E^\circ_{\text{anode}}(\text{Fe}^{3+}/\text{Fe}^{2+}) \)

\(= 0.80 \, \text{V} – 0.77 \, \text{V} = 0.03 \, \text{V}\)

Calculate the Gibbs free energy change \(( \Delta_r G^\circ )\).

The relationship between the Gibbs free energy change and the cell potential is given by the equation:
\(\Delta_r G^\circ = -nFE^\circ_{\text{cell}}\)

where:

• n is the number of moles of electrons transferred.
• F is the Faraday constant \( (96,485 \, \text{C/mol}) \).
• \( E^\circ_{\text{cell}} \) is the standard cell potential.

(i) For \( 2\text{Cr} + 3\text{Cd}^{2+} \):

• n = 6 (as 2 moles of Cr lose 6 electrons and 3 moles of Cd gain 6 electrons).
• \( E^\circ_{\text{cell}} = 0.34 \, \text{V} \).

\(\Delta_r G^\circ = -6 \times 96,485 \times 0.34\)

\( = -196,864 \, \text{J/mol} = -196.86 \, \text{kJ/mol}\)

(ii) For \( \text{Fe}^{2+} + \text{Ag}^+ \):

• n = 1 (as 1 mole of Fe loses 1 electron and 1 mole of Ag gains 1 electron).
• \( E^\circ_{\text{cell}} = 0.03 \, \text{V} \).

\(\Delta_r G^\circ = -1 \times 96,485 \times 0.03 \)

\(= -2,894.55 \, \text{J/mol} = -2.895 \, \text{kJ/mol}\)

Calculate the equilibrium constant ( K ).

The equilibrium constant is related to the Gibbs free energy by the equation:
\(\Delta_r G^\circ = -RT \ln K\)

where:

• R is the gas constant \( (8.314 \, \text{J/mol K}) \),
• T is the temperature in Kelvin (assume \( 298 \, \text{K} \) unless stated otherwise).

Rearranging to solve for ( K ):
\(K = e^{-\Delta_r G^\circ / (RT)}\)

(i) For \( 2\text{Cr} + 3\text{Cd}^{2+} \):

\(K = e^{-(-196,860) / (8.314 \times 298)}\)

\(= e^{79.56} \approx 3.124 \times 10^{34}\)

(ii) For \( \text{Fe}^{2+} + \text{Ag}^+ \):

\(K = e^{-(-2,895) / (8.314 \times 298)} \)

\(= e^{1.17} \approx 3.2\)

Final Answer:

(i) For \( 2\text{Cr} + 3\text{Cd}^{2+} \rightarrow \)\(2\text{Cr}^{3+} + 3\text{Cd} \):

• \( E^\circ_{\text{cell}} = 0.34 \, \text{V} \)
• \( \Delta_r G^\circ = -196.86 \, \text{kJ/mol} \)
• \( K = 3.124 \times 10^{34} \)

(ii) For \( \text{Fe}^{2+} + \text{Ag}^+ \rightarrow\)\( \text{Fe}^{3+} + \text{Ag} \):

• \( E^\circ_{\text{cell}} = 0.03 \, \text{V} \)
• \( \Delta_r G^\circ = -2.895 \, \text{kJ/mol} \)
• \( K = 3.2 \)

Question 2.5: 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).

Solution 2.5: To calculate the EMF (electromotive force) of each cell, we use the Nernst equation:

\(E_{\text{cell}} = E^\circ_{\text{cell}} – \frac{0.0591}{n} \log Q\)

Where:

• \( E^\circ_{\text{cell}} \) is the standard cell potential.
• n is the number of moles of electrons transferred.
• Q is the reaction quotient, which depends on the concentrations of the ions involved.

(i) Cell: Mg(s) | Mg²⁺ (0.001 M) || Cu²⁺ (0.0001 M) | Cu(s)

Half-reactions:

• \( \text{Mg}(s) \rightarrow \text{Mg}^{2+}(aq) + 2e^- ) \)\((oxidation, ( E^\circ = -2.37 \, \text{V} )\)
• \( \text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s) ) \)\((reduction, ( E^\circ = +0.34 \, \text{V} )\)

Standard cell potential:
\(E^\circ_{\text{cell}} = 0.34 \, \text{V} – (-2.37 \, \text{V}) = 2.71 \, \text{V}\)

Reaction quotient ( Q ):
\(Q = \frac{[\text{Mg}^{2+}]}{[\text{Cu}^{2+}]} = \frac{0.001}{0.0001} = 10\)

Using the Nernst equation:
\(E_{\text{cell}} = 2.71 – \frac{0.0591}{2} \log 10 \)

\(= 2.71 – 0.02955 = 2.68 \, \text{V}\)

(ii) Cell: Fe(s) | Fe²⁺ (0.001 M) || H⁺ (1 M) | H₂(g)(1 bar) | Pt(s)

Half-reactions:

• \( \text{Fe}(s) \rightarrow \text{Fe}^{2+}(aq) + 2e^- )\)\( (oxidation, ( E^\circ = -0.44 \, \text{V} )\)
• \( 2\text{H}^+(aq) + 2e^- \rightarrow \text{H}_2(g) ) \)\((reduction, ( E^\circ = 0.00 \, \text{V} )\)

Standard cell potential:
\(E^\circ_{\text{cell}} = 0.00 \, \text{V} – (-0.44 \, \text{V}) = 0.44 \, \text{V}\)

Reaction quotient ( Q ):
\(Q = \frac{[\text{Fe}^{2+}]}{[\text{H}^+]^2} = \frac{0.001}{1^2} = 0.001\)

Using the Nernst equation:
\(E_{\text{cell}} = 0.44 – \frac{0.0591}{2} \log 0.001 \)

\(= 0.44 – 0.08865 = 0.53 \, \text{V}\)

(iii) Cell: Sn(s) | Sn²⁺ (0.050 M) || H⁺ (0.020 M) | H₂(g)(1 bar) | Pt(s)

Half-reactions:

\( \text{Sn}(s) \rightarrow \text{Sn}^{2+}(aq) + 2e^- )\)\( (oxidation, ( E^\circ = -0.14 \, \text{V} )\)

\( 2\text{H}^+(aq) + 2e^- \rightarrow \text{H}_2(g) )\)\( (reduction, ( E^\circ = 0.00 \, \text{V} )\)

Standard cell potential:
\(E^\circ_{\text{cell}} = 0.00 \, \text{V} – (-0.14 \, \text{V}) = 0.14 \, \text{V}\)

Reaction quotient ( Q ):
\(Q = \frac{[\text{Sn}^{2+}]}{[\text{H}^+]^2} = \frac{0.050}{(0.020)^2} = 125\)

Using the Nernst equation:
\(E_{\text{cell}} = 0.14 – \frac{0.0591}{2} \log 125 \)

\(= 0.14 – 0.0584 = 0.08 \, \text{V}\)

(iv) Cell: Pt(s) | Br^– (0.010 M) | Br₂(l) || H⁺ (0.030 M) | H₂(g)(1 bar) | Pt(s)

Half-reactions:

\( \text{Br}_2(l) + 2e^- \rightarrow 2\text{Br}^-(aq) )\)\( (reduction, ( E^\circ = +1.07 \, \text{V} )\)

\( 2\text{H}^+(aq) + 2e^- \rightarrow \text{H}_2(g) ) \)\((oxidation, ( E^\circ = 0.00 \, \text{V} )\)

Standard cell potential:
\(E^\circ_{\text{cell}} = 0.00 \, \text{V} – 1.07 \, \text{V} = -1.07 \, \text{V}\)

Reaction quotient ( Q ):
\(Q = \frac{[\text{Br}^-]^2}{[\text{H}^+]^2} \)

\(= \frac{(0.010)^2}{(0.030)^2} = 0.1111\)

Using the Nernst equation:
\(E_{\text{cell}} = -1.07 – \frac{0.0591}{2} \log 0.1111 \)

\(= -1.07 – 0.228 = -1.298 \, \text{V}\)

Final Answers:

(i) \( E_{\text{cell}} = 2.68 \, \text{V} \)
(ii) \( E_{\text{cell}} = 0.53 \, \text{V} \)
(iii) \( E_{\text{cell}} = 0.08 \, \text{V} \)
(iv) \( E_{\text{cell}} = -1.298 \, \text{V} \)

Question 2.6: In the button cells widely used in watches and other devices the following reaction takes place: Zn(s) + Ag₂O(s) + H₂O(l) → Zn²⁺(aq) + 2Ag(s) + 2OH⁻(aq) Determine ∆_rG° and E° for the reaction.

Solution 2.6: To determine the Gibbs free energy change \(( \Delta_r G^\circ )\) and standard cell potential \(( E^\circ_{\text{cell}} )\) for the given reaction:

Reaction:

\(\text{Zn}(s) + \text{Ag}_2\text{O}(s) + \text{H}_2\text{O}(l) \rightarrow\)\( \text{Zn}^{2+}(aq) + 2\text{Ag}(s) + 2\text{OH}^-(aq)\)

Write down the half-reactions

Oxidation (Zn): \( \text{Zn}(s) \rightarrow \text{Zn}^{2+}(aq) + 2e^- \)

Standard reduction potential for \( \text{Zn}^{2+}/\text{Zn} ): ( E^\circ = -0.76 \, \text{V} \)

Reduction (Ag2O):
\( \text{Ag}_2\text{O}(s) + \text{H}_2\text{O}(l) + 2e^- \rightarrow\)\( 2\text{Ag}(s) + 2\text{OH}^-(aq) \)

Standard reduction potential for \( \text{Ag}_2\text{O}/\text{Ag} ): ( E^\circ = +0.34 \, \text{V} \)

Calculate the standard cell potential \( E^\circ_{\text{cell}} \)

The standard cell potential is calculated as:
\(E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} – E^\circ_{\text{anode}}\)

Where:

• \( E^\circ_{\text{cathode}} = +0.34 \, \text{V} ) (Ag2O/Ag\)
• \( E^\circ_{\text{anode}} = -0.76 \, \text{V} ) (Zn/Zn²⁺\)

\(E^\circ_{\text{cell}} = 0.34 \, \text{V} – (-0.76 \, \text{V}) \)

\(= 0.34 \, \text{V} + 0.76 \, \text{V} = 1.10 \, \text{V}\)

Calculate \( \Delta_r G^\circ \)

The relationship between \( \Delta_r G^\circ \) and the cell potential is given by:
\(\Delta_r G^\circ = -nFE^\circ_{\text{cell}}\)

Where:

• n = 2 (since 2 electrons are transferred)
• \( F = 96,485 \, \text{C/mol} \) (Faraday constant)
• \( E^\circ_{\text{cell}} = 1.10 \, \text{V} \)

Now, calculate \( \Delta_r G^\circ \):

\(\Delta_r G^\circ = -2 \times 96,485 \times 1.10 \)

\(= -212,267 \, \text{J/mol} = -212.27 \, \text{kJ/mol}\)

Final Answer:

• Standard cell potential: \( E^\circ_{\text{cell}} = 1.10 \, \text{V} \)
• Gibbs free energy change: \( \Delta_r G^\circ = -212.27 \, \text{kJ/mol} \)

Question 2.7: Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.

Solution 2.7: 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 be a decrease in Conductivity. It is applicable for both weak and strong electrolytes. 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 –

The 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 :

Question 2.8: The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S cm^–1. Calculate its molar conductivity.

Solution 2.8: 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

Question 2.9: The resistance of a conductivity cell containing 0.001M KCl solution at 298 K is 1500 Ω. What is the cell constant if conductivity of 0.001M KCl solution at 298 K is 0.146 × 10^–3 S cm^–1.
.

Solution 2.9: 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

Question 2.10: 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 Λ_m for all concentrations and draw a plot between Λ_m and c½. Find the value of 0 Λ_m.

Solution 2.10: Given,

κ = 1.237 × 10⁻² S m−1, c = 0.001 M

Then, κ = 1.237 × 10⁻⁴ S cm⁻¹, c^1⁄2 = 0.0316 M^1/2

\(\Lambda_m =\frac{k}{c}\)

\(=\frac{1.237 \times 10^{ -4 } S\;cm^{-1} }{0.001 \; mol\; \; L^{ -1 }}\times \frac{1000\;cm^{-1}}{L}\)

= 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, c^1⁄2 = 0.1 M^1/2

\(\Lambda_m =\frac{k}{c}\)

\(=\frac{11.85 \times 10^{ -4 } S\;cm^{-1} }{0.010 \; mol\; \; L^{ -1 }}\times \frac{1000\;cm^{-1}}{L}\)

= 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, c^1/2 = 0.1414 M^1/2

\(\Lambda_m =\frac{k}{c}\)

\(=\frac{23.15 \times 10^{ -4 } S\;cm^{-1} }{0.020 \; mol\; \; L^{ -1 }}\times \frac{1000\;cm^{-1}}{L}\)

= 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, c^1/2 = 0.2236 M^1/2

\(\Lambda_m =\frac{k}{c}\)

\(=\frac{106.74 \times 10^{ -4 } S\;cm^{-1} }{0.050 \; mol\; \; L^{ -1 }}\times \frac{1000\;cm^{-1}}{L}\)

= 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, c^1/2 = 0.3162 M^1/2

\(\Lambda_m =\frac{k}{c}\)

\(=\frac{106.74 \times 10^{ -4 } S\;cm^{-1} }{0.100 \; mol\; \; L^{ -1 }}\times \frac{1000\;cm^{-1}}{L}\)

= 106.74 S cm² mol⁻¹

Now, we have the following data:

C1/2  / M1/2	0.0316	0.1	0.1414	0.2236	0.3162
λm (Scm2 mol-1)	123.7	118.5	115.8	111.1	106.74

Since the line interrupts

\(\Lambda_m\) at 124.0 S cm² mol⁻¹,

\(\Lambda^0_m\) =  124.0 S cm² mol⁻¹

Question 2.11: Conductivity of 0.00241 M acetic acid is 7.896 × 10^–5 S cm^–1. Calculate its molar conductivity. If \(\Lambda^0_m\) for acetic acid is 390.5 S cm² mol^–1, what is its dissociation constant?

Solution 2.11: 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

Again, \(\Lambda^0_m =\) 390.5 S cm² mol⁻¹

Now,

\(\alpha =\frac{\Lambda_m }{\Lambda^0_m }\)

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

= 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

Question 2.12: 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⁺?

Solution 2.12:

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

= Required charge = 5 F

= 5 × 96487 C

= 482435 C

Question 2.13: 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?

Solution 2.13:

(i)  From the 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 the 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

Question 2.14: 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?

Solution 2.14: (i) From the 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 the given data,

\(Fe^{2+}\rightarrow Fe^{3+} + e^-\)

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

= 96487 C

Question 2.15: 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?

Solution 2.15: 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.

Question 2.16: Three electrolytic cells, A, B, and 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 was deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited?

Solution 2.16: According to the reaction,

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,

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

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

Question 2.17: 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)

Solution 2.17:

Since Eº  for the overall reaction is positive, the reaction between Fe^3+₍aq) and I⁻(aq) is feasible.

Since Eº for the overall reaction is positive, the reaction between Ag⁺ _(aq) and Cu_(s) is feasible.

Since Eº for the overall reaction is negative, the reaction between Fe^3+_(aq) and Br⁻_(aq) is not feasible.

Since Eº for the overall reaction is negative, the reaction between Ag _(s) and Fe³⁺ _(aq) is not feasible.

Since Eº for the overall reaction is positive, the reaction between Br_2(aq) and Fe^2+_(aq) is feasible.

Question 2.18: 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

Solution 2.18: (i) At cathode:

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

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

At anode: 

The Ag anode is attacked by NO₃^–  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.

The reaction with a higher value of Eº 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 NO₃^–  ions. Therefore, OH⁻ or NO₃^–
ions can be oxidized at the anode. But OH⁻ ions having a lower discharge potential and get preference and decompose to liberate O₂.

OH^–  → OH + e^–

4OH^–  → 2H₂O +O₂

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

H+ (aq) + e-  → ½ H₂(g)

At the anode, the following processes are possible.

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.

The reaction with a higher value of  Eº 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.

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