Calculate emf and Δ G for the following cell at 298 K : Mg(s) / Mg²+(0.01\ M) // Ag^+(0.001\ M) / Ag(s) Given : E°Mg²+/Mg = -2.37 V, E°Ag^+/Ag = +0.80 V [1 F = 96500 C mol^-1, log 10 = 1]

ChemistryCBSE Class 12 · Board exam 2026, set 56/1/1 · 5 marks · Long answer

CBSE Class 12 Chemistry 2026: question 33(a)

Set in the CBSE Class 12 board exam of 2026, paper 56/1/1, 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

Calculate emf and $Δ G$ for the following cell at 298 K :
$Mg (s) / M g^{2 +} (0.01 M) // A g^{+} (0.001 M) / Ag (s)$
Given : $E_{M g^{2 +} / M g}^{\circ} = - 2.37$ V, $E_{A g^{+} / A g}^{\circ} = + 0.80$ V
[1 F = 96500 C mol $^{- 1}$ , log 10 = 1]

Source: official question paper from cbse.gov.in (PDF). See it in the full 2026 board paper, set 56/1/1.

Hints

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

Hint 1

First write the overall cell reaction and find how many electrons are transferred.

Hint 2

Use $E_{ce l l}^{\circ} = E_{c a t h o d e}^{\circ} - E_{an o d e}^{\circ}$ , then put the ion concentrations into the Nernst equation (remember the square on [Ag $^{+}$ ]).

Hint 3

Finally use $Δ G = - n F E_{ce l l}$ with the emf you found.

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.

The cell reaction is $Mg (s) + 2 A g^{+} (aq) \to M g^{2 +} (aq) + 2 Ag (s)$ , so $n = 2$ .
$E_{ce l l}^{\circ} = E_{c a t h o d e}^{\circ} - E_{an o d e}^{\circ} = 0.80 - (- 2.37) = 3.17 V$
1 mark
Nernst equation at 298 K:
$E_{ce l l} = E_{ce l l}^{\circ} - \frac{0.059}{n} lo g \frac{[ M g ^{2 +} ]}{[ A g ^{+} ] ^{2}}$
1 mark
Substitute the concentrations:
$E_{ce l l} = 3.17 - \frac{0.059}{2} lo g \frac{0.01}{( 0.001 ) ^{2}} = 3.17 - 0.0295 lo g 1 0^{4}$
$E_{ce l l} = 3.17 - 0.0295 \times 4 = 3.17 - 0.118 = 3.052 V$
1 mark
Gibbs energy change: $Δ G = - n F E_{ce l l}$
½ mark
$Δ G = - 2 \times 96500 \times 3.052$
½ mark
$Δ G = - 589036 J mol^{- 1} \approx - 589.04 kJ mol^{- 1}$
1 mark

Answer: $E_{ce l l} = 3.052$ V; $Δ G = - 589036$ J mol $^{- 1}$ (≈ −589.04 kJ mol $^{- 1}$ )

Where the marks are

Calculates $E_{ce l l}^{\circ} = 3.17$ V (1 mark)
Writes the Nernst equation with n = 2 (1 mark)
Substitutes values and obtains $E_{ce l l} = 3.052$ V (1 mark)
Writes $Δ G = - n F E_{ce l l}$ (½ mark)
Substitutes the values correctly (½ mark)
Obtains $Δ G \approx - 589$ kJ mol $^{- 1}$ (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 paper2026 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 2026: question 33(a)

Hints

Step-by-step solution

Revise the chapter

More questions from Electrochemistry