Calculate the emf of the following cell at 298 K : Fe(s) Fe²+ (0.01M) \ H^+(1M) H2(g) (1\ bar), Pt(s) Given E°Cell = 0.44 V.

ChemistryCBSE Class 12 · Board exam 2023, set 56/5/1 · 2 marks · Very short answer

CBSE Class 12 Chemistry 2023: question 19(b)

Set in the CBSE Class 12 board exam of 2023, paper 56/5/1, section B, question 19(b), for 2 marks. Try it first; the hints and our step-by-step solution are below.

Q19(b) · 2 marks · Very short answerElectrochemistry

Calculate the emf of the following cell at 298 K :
$F e_{(s)} ∣F e^{2 +} (0.01M) ∥ H_{(1M)}^{+} ∣ H_{2 (g)} (1 bar), P t_{(s)}$
Given $E_{C e l l}^{\circ} = 0.44$ V.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2023 board paper, set 56/5/1.

Hints

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

Hint 1

Write the overall cell reaction first to find how many electrons are transferred.

Hint 2

Use the Nernst equation; the reaction quotient is $[F e^{2 +}] / [H^{+}]^{2}$ (solids and gas at 1 bar are omitted).

Hint 3

Remember $lo g 1 0^{- 2} = - 2$ .

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.

Cell reaction: $Fe (s) + 2 H^{+} (aq) \to F e^{2 +} (aq) + H_{2} (g)$ , so $n = 2$ .
Nernst equation at 298 K:
$E_{ce l l} = E_{ce l l}^{\circ} - \frac{0.059}{n} lo g \frac{[ F e ^{2 +} ]}{[ H ^{+} ] ^{2}}$
½ mark
Substitute the values:
$E_{ce l l} = 0.44 - \frac{0.059}{2} lo g \frac{0.01}{( 1 ) ^{2}} = 0.44 - 0.0295 lo g 1 0^{- 2}$
1 mark
$E_{ce l l} = 0.44 - 0.0295 \times (- 2) = 0.44 + 0.059 = 0.499 V$
½ mark

Answer: $E_{ce l l} = 0.499$ V

Where the marks are

Correct Nernst equation with $n = 2$ (½ mark)
Correct substitution of concentrations (1 mark)
Final answer 0.499 V (½ 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 paper2023 board paper, set 56/5/1: every question, by section.

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CBSE Class 12 Chemistry 2023: question 19(b)

Hints

Step-by-step solution

Revise the chapter

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