The equilibrium constant of cell reaction : Sn^4+(aq) + Al(s) → Al³+ + Sn²+(aq) is 4.617 × 10^184, at 25 °C a. Calculate the standard emf of the cell. (Given: 4.617 × 10^184 = 184.6644) (2) b. What will be the E° of the half cell Al³+/Al, if E° of half cell Sn^4+/Sn²+ is 0.15 V. (1)

ChemistryCBSE Class 12 · Sample paper, session 2024-25 · 3 marks · Short answer

CBSE Class 12 Chemistry sample paper 2024-25: question 28

From the official CBSE sample question paper for 2024-25, section C, question 28, for 3 marks. Try it first; the hints and our step-by-step solution are below.

Q28 · 3 marks · Short answerElectrochemistry

The equilibrium constant of cell reaction :
$S n^{4 +} (a q) + A l (s) \to A l^{3 +} + S n^{2 +} (a q)$ is $4.617 \times 1 0^{184}$ , at 25 $^{\circ}$ C
a. Calculate the standard emf of the cell. (Given: $lo g 4.617 \times 1 0^{184} = 184.6644$ ) (2)
b. What will be the $E^{\circ}$ of the half cell $A l^{3 +} / A l$ , if $E^{\circ}$ of half cell $S n^{4 +} / S n^{2 +}$ is 0.15 V. (1)

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2024-25.

Hints

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

Hint 1

Balance the electrons: Sn⁴⁺ gains 2e⁻ and Al loses 3e⁻. What is n?

Hint 2

At equilibrium E_cell = 0; use the Nernst equation to relate E° and K.

Hint 3

For (b), Sn⁴⁺/Sn²⁺ is the cathode and Al³⁺/Al the anode.

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) Balanced cell reaction: $3 S n^{4 +} + 2 A l \to 2 A l^{3 +} + 3 S n^{2 +}$ , so $n = 6$ .
½ mark
At equilibrium $E_{ce l l} = 0$ , so $E_{ce l l}^{\circ} = \frac{0.059}{n} lo g K_{c}$ at 298 K.
½ mark
$E_{ce l l}^{\circ} = \frac{0.059}{6} \times lo g (4.617 \times 1 0^{184}) = 0.00983 \times 184.6644$
½ mark
$E_{ce l l}^{\circ} = 1.815 V$
½ mark
(b) $E_{ce l l}^{\circ} = E_{c a t h o d e}^{\circ} - E_{an o d e}^{\circ} = E_{S n^{4 +} / S n^{2 +}}^{\circ} - E_{A l^{3 +} / A l}^{\circ}$
$1.815 = 0.15 - E_{A l^{3 +} / A l}^{\circ}$
½ mark
$E_{A l^{3 +} / A l}^{\circ} = 0.15 - 1.815 = - 1.66 V$
½ mark

Answer: (a) E°cell ≈ 1.815 V; (b) E°(Al³⁺/Al) = −1.66 V

Where the marks are

n = 6 from the balanced reaction (½ mark)
Uses E°cell = (0.059/n) log K (½ mark)
Correct substitution (½ mark)
E°cell = 1.815 V (½ mark)
E°cell = E°cathode − E°anode with values (½ mark)
E°(Al³⁺/Al) = −1.66 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 paperSample paper 2024-25: every question, by section.

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CBSE Class 12 Chemistry sample paper 2024-25: question 28

Hints

Step-by-step solution

Revise the chapter

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