Answer the following. (i) Calculate the emf of the following cell at 25°C : Zn(s) Zn²+(0.1 M) H^+(0.01 M) H_2(g)(1 bar), Pt(s) [ Given : E°Zn²+/Zn = – 0.76 V, E°2H^+/H_2 = 0.00 V, log 10 = 1 ] (ii) State Faraday’s second law of electrolysis. How much electricity is required in terms of Faraday for the reduction of 1 mol of Cr_2O_7²- to Cr³+ ?

ChemistryCBSE Class 12 · Board exam 2025, set 56/7/1 · 5 marks · Long answer

CBSE Class 12 Chemistry 2025: question 33(a)

Set in the CBSE Class 12 board exam of 2025, paper 56/7/1, section E, question 33(a), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q33(a) · 5 marks · Long answerElectrochemistry

Answer the following.

(i)
Calculate the emf of the following cell at 25 $^{\circ}$ C :
$Z n (s) ∣ Z n^{2 +} (0.1 M) ∣∣ H^{+} (0.01 M) ∣ H_{2} (g) (1 ba r), P t (s)$
[ Given : $E_{Z n^{2 +} / Z n}^{\circ}$ = – 0.76 V, $E_{2 H^{+} / H_{2}}^{\circ}$ = 0.00 V, log 10 = 1 ]
3 marks
(ii)
State Faraday’s second law of electrolysis. How much electricity is required in terms of Faraday for the reduction of 1 mol of $C r_{2} O_{7}^{2 -}$ to $C r^{3 +}$ ?
2 marks

Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 56/7/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

Write the overall cell reaction first to find n and the reaction quotient.

Hint 2

In Q, the concentration of $H^{+}$ is squared, so Q = $0.1/ (0.01)^{2}$ .

Hint 3

For (ii), find the change in oxidation state of Cr and multiply by the two Cr atoms.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

(i) Cell reaction: $Zn (s) + 2 H^{+} (aq) \to Z n^{2 +} (aq) + H_{2} (g)$ , with $n = 2$ .
Nernst equation at 298 K:
$E_{ce l l} = (E_{c a t h o d e}^{\circ} - E_{an o d e}^{\circ}) - \frac{0.059}{2} lo g \frac{[ Z n ^{2 +} ]}{[ H ^{+} ] ^{2}}$
1 mark
Substitute the values:
$E_{ce l l} = [0 - (- 0.76)] - \frac{0.059}{2} lo g \frac{0.1}{( 0.01 ) ^{2}} = 0.76 - 0.0295 lo g 1 0^{3}$
1 mark
$E_{ce l l} = 0.76 - 0.0295 \times 3 = 0.76 - 0.0885 = 0.6715 V \approx 0.67 V$
1 mark
(ii) Faraday's second law: when the same quantity of electricity is passed through different electrolytic solutions, the amounts of the substances liberated are proportional to their chemical equivalent weights.
1 mark
Half-reaction: $C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2C r^{3 +} + 7 H_{2} O$ .
Each Cr goes from +6 to +3, so 1 mol of $C r_{2} O_{7}^{2 -}$ needs 6 mol of electrons. Electricity required = 6 F (= $6 \times 96500 = 579000$ C).
1 mark

Answer: (i) $E_{ce l l} = 0.6715$ V ≈ 0.67 V; (ii) amounts liberated ∝ equivalent weights; 6 F

Where the marks are

Writes the Nernst equation for the cell with n = 2 (1 mark)
Correct substitution, giving $0.76 - 0.0295 lo g 1 0^{3}$ (1 mark)
$E_{ce l l} = 0.6715$ V with unit; ½ mark deducted for missing or wrong unit (1 mark)
Correct statement of Faraday's second law (1 mark)
6 F for reducing 1 mol $C r_{2} O_{7}^{2 -}$ to $C r^{3 +}$ (1 mark)

Revise the chapter

Chapter: Electrochemistry (notes, formulas, where marks are lost and every past question)
Topic: Nernst Equation, Gibbs Energy and Equilibrium Constant

Practise this chapterBoard-style questions from Electrochemistry, one at a time, marked against the answer key and marking points.Back to the full paper2025 board paper, set 56/7/1: every question, by section.

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CBSE Class 12 Chemistry 2025: question 33(a)

Hints

Step-by-step solution

Revise the chapter

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