(i) State Faraday's first law of electrolysis. How much charge, in terms of Faraday, is required for the reduction of 1 mol Cu²+ to Cu ? (ii) Calculate emf of the following cell at 298 K for Mg(s) Mg²+(0·1 M) \ Cu²+(0·01 M) Cu(s) [E°cell = +2·71 V, 1 F = 96500 C mol^-1, log 10 = 1]

ChemistryCBSE Class 12 · Board exam 2023, set 56/1/1 · 5 marks · Long answer

CBSE Class 12 Chemistry 2023: question 33(b)

Set in the CBSE Class 12 board exam of 2023, paper 56/1/1, section E, question 33(b), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q33(b) · 5 marks · Long answerElectrochemistry

(i) State Faraday's first law of electrolysis. How much charge, in terms of Faraday, is required for the reduction of 1 mol $C u^{2 +}$ to Cu ?
(ii) Calculate emf of the following cell at 298 K for
$Mg (s) ∣M g^{2 +} (0 \cdot 1 M) ∥C u^{2 +} (0 \cdot 01 M) ∣Cu (s)$
[ $E_{ce l l}^{\circ} = + 2 \cdot 71$ V, 1 F = 96500 C mol $^{- 1}$ , log 10 = 1]

Source: official question paper from cbse.gov.in (PDF). See it in the full 2023 board paper, set 56/1/1.

Hints

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

Hint 1

Faraday's first law relates mass deposited to charge passed.

Hint 2

How many electrons does one Cu²⁺ ion need to become Cu?

Hint 3

Write the cell reaction, find n, then put the ion concentrations into the Nernst equation.

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) Faraday's first law: The mass of a substance deposited or liberated at an electrode is directly proportional to the quantity of electricity passed through the electrolyte: $w = Z I t$ .
1 mark
$C u^{2 +} + 2 e^{-} \to Cu$ . One mole of Cu²⁺ needs 2 mol of electrons, so the charge required is 2 F (= 193000 C).
1 mark
(ii) Cell reaction: $Mg + C u^{2 +} \to M g^{2 +} + Cu$ , $n = 2$ .
Nernst equation: $E_{ce l l} = E_{ce l l}^{\circ} - \frac{0 \cdot 059}{n} lo g \frac{[ M g ^{2 +} ]}{[ C u ^{2 +} ]}$
1 mark
$E_{ce l l} = 2 \cdot 71 - \frac{0 \cdot 059}{2} lo g \frac{0 \cdot 1}{0 \cdot 01} = 2 \cdot 71 - 0 \cdot 0295 lo g 10$
1 mark
$E_{ce l l} = 2 \cdot 71 - 0 \cdot 0295 = 2 \cdot 68$ V
1 mark

Answer: (i) w ∝ Q; 2 F required; (ii) E_cell ≈ 2.68 V

Where the marks are

Statement of Faraday's first law (1 mark)
2 F charge for 1 mol Cu²⁺ (1 mark)
Correct Nernst equation with n = 2 (1 mark)
Correct substitution (1 mark)
E_cell = 2.68 V with unit; ½ deducted if the unit is missing (1 mark)

Revise the chapter

Chapter: Electrochemistry (notes, formulas, where marks are lost and every past question)
Topic: Electrolysis and Faraday's Laws

Practise this chapterBoard-style questions from Electrochemistry, one at a time, marked against the answer key and marking points.Back to the full paper2023 board paper, set 56/1/1: every question, by section.

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.

CBSE Class 12 Chemistry 2023: question 33(b)

Hints

Step-by-step solution

Revise the chapter

More questions from Electrochemistry