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

CBSE Class 12 Chemistry 2025: question 31(B)

Set in the CBSE Class 12 board exam of 2025, paper 56/1/1, section E, question 31(B), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q31(B) · 5 marks · Long answerElectrochemistry

(a) Write the anode and cathode reactions and the overall cell reaction occurring in a lead storage battery during its use. (2 + 3 = 5)
(b) Calculate the potential for half-cell containing 0.01 M $K_{2} C r_{2} O_{7} (aq)$ , 0.01 M $C r^{3 +} (aq)$ and $1.0 \times 1 0^{- 4}$ M $H^{+} (aq)$ .
The half cell reaction is
$C r_{2} O_{7}^{2 -} (aq) + 14 H^{+} (aq) + 6 e^{-} ⟶ 2 C r^{3 +} (aq) + 7 H_{2} O (l)$
and the standard electrode potential is given as E° = 1.33 V.
[Given : log 10 = 1]

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

Hints

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

Hint 1

During discharge, both electrodes end up as PbSO₄.

Hint 2

Write Q with products over reactants, including [H⁺]¹⁴; [Cr₂O₇²⁻] = [K₂Cr₂O₇].

Hint 3

Combine the powers of 10 carefully: (10⁻²)²/(10⁻²·10⁻⁵⁶).

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.

(a) Anode: $Pb (s) + S O_{4}^{2 -} (aq) \to PbS O_{4} (s) + 2 e^{-}$
½ mark
Cathode: $Pb O_{2} (s) + S O_{4}^{2 -} (aq) + 4 H^{+} (aq) + 2 e^{-} \to PbS O_{4} (s) + 2 H_{2} O (l)$
½ mark
Overall: $Pb (s) + Pb O_{2} (s) + 2 H_{2} S O_{4} (aq) \to 2 PbS O_{4} (s) + 2 H_{2} O (l)$
1 mark
(b) Nernst equation for the half cell (n = 6):
$E = E^{\circ} - \frac{0.059}{6} lo g \frac{[ C r ^{3 +} ] ^{2}}{[ C r _{2} O _{7}^{2 -} ] [ H ^{+} ] ^{14}}$
1 mark
$E = 1.33 - \frac{0.059}{6} lo g \frac{( 1 0 ^{- 2} ) ^{2}}{( 1 0 ^{- 2} ) ( 1 0 ^{- 4} ) ^{14}} = 1.33 - \frac{0.059}{6} lo g 1 0^{54}$
1 mark
$E = 1.33 - \frac{0.059 \times 54}{6} = 1.33 - 0.531 = 0.799$ V
1 mark

Answer: Overall: Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O; E = 0.799 V

Where the marks are

Anode reaction (½ mark)
Cathode reaction (½ mark)
Overall reaction (1 mark)
Correct Nernst expression with n = 6 (1 mark)
Correct substitution giving log 10⁵⁴ (1 mark)
E = 0.799 V (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 56/1/1: every question, by section.

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CBSE Class 12 Chemistry 2025: question 31(B)

Hints

Step-by-step solution

Revise the chapter

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