ChemistryCBSE Class 12 · Sample paper, session 2026-27 · 5 marks · Long answer

CBSE Class 12 Chemistry sample paper 2026-27: question 33(A)

From the official CBSE sample question paper for 2026-27, 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 answerCompetency-basedElectrochemistry

(I) A student investigates the effect of concentration on the cell potential of a Zn–Cu galvanic cell at 298 K.
The cell is: $Z n (s) ∣ Z n^{2 +} (a q) ∣∣ C u^{2 +} (a q) ∣ C u (s)$
standard electrode potentials:
$E^{0} (Z n^{2 +} / Z n) = - 0.76$ V, $E^{0} (C u^{2 +} / C u) = + 0.34$ V, $E_{ce l l}^{0} = 1.10$ V
The following experimental data were collected:

Experiment	$[Z n^{2 +}]$ (M)	$[C u^{2 +}]$ (M)	Measured EMF (V)
1	1.00	1.00	1.10
2	0.1	1.00	1.16
3	1.00	0.10	1.04
4	0.01	1.00	1.22
5	1.00	0.01	0.98

a) Why does experiment 4 show the maximum EMF?
b) For experiment 3, calculate the theoretical EMF using the Nernst equation. Suggest one possible reason why the measured EMF differs slightly from the calculated value. (1+2)

(II) If k is expressed in $S m^{- 1}$ and the concentration, c in $m o l m^{- 3}$ then the units of molar conductivity are in $S m^{2} m o l^{- 1}$ . However, if we use $S c m^{- 1}$ as the units for k and $m o l c m^{- 3}$ , the units of concentration, then the units of molar conductivity are in $S c m^{2} m o l^{- 1}$ . Show how $1 S m^{2} m o l^{- 1}$ is related to $1 S c m^{2} m o l^{- 1}$ (1)

(III) What happens to the specific gravity of electrolyte in lead storage battery during charging and discharging of battery? (1)

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2026-27.

Hints

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

Hint 1

Write the Nernst equation with $Q = [Z n^{2 +}] / [C u^{2 +}]$ and see how a small Q affects E.

Hint 2

For experiment 3, put $n = 2$ and the ratio 1.00/0.10 into the log term.

Hint 3

For the units, convert 1 m² into cm². For the battery, ask whether H₂SO₄ is used up or formed in each process.

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)(a) The cell reaction is $Z n + C u^{2 +} \to Z n^{2 +} + C u$ , so $E_{ce l l} = E_{ce l l}^{0} - \frac{0.059}{2} lo g \frac{[ Z n ^{2 +} ]}{[ C u ^{2 +} ]}$ .
In experiment 4, $[Z n^{2 +}] = 0.01$ M (lowest) and $[C u^{2 +}] = 1.00$ M (highest). This makes $Q_{c}$ the smallest, at $1 0^{- 2}$ . So $lo g Q_{c}$ is most negative, the subtracted term becomes the largest positive addition, and the EMF is the highest.
1 mark
(I)(b) Nernst equation for experiment 3 ( $n = 2$ ):
$E_{ce l l} = 1.10 - \frac{0.059}{2} lo g \frac{[ Z n ^{2 +} ]}{[ C u ^{2 +} ]} = 1.10 - 0.0295 lo g \frac{1.00}{0.10}$
½ mark
$E_{ce l l} = 1.10 - 0.0295 \times lo g 10 = 1.10 - 0.0295 = 1.0705 \approx 1.07$ V
½ mark
Reason for the difference: The measured value (1.04 V) is lower than the calculated one. When the cell is measured, some current flows, and the internal resistance of the electrolyte, electrodes and salt bridge causes a voltage drop. Non-ideal behaviour (activity is not equal to concentration) also contributes.
1 mark
(II) $Λ_{m} = κ / c$ .
$1 S m^{2} m o l^{- 1} = 1 S (100 c m)^{2} m o l^{- 1} = 1 0^{4} S c m^{2} m o l^{- 1}$
So $1 S m^{2} m o l^{- 1} = 1 0^{4} S c m^{2} m o l^{- 1}$ , or equivalently $1 S c m^{2} m o l^{- 1} = 1 0^{- 4} S m^{2} m o l^{- 1}$ .
1 mark
(III) The overall reaction is $P b + P b O_{2} + 2 H_{2} S O_{4} ⇌ 2 P b S O_{4} + 2 H_{2} O$ .
During charging, $H_{2} S O_{4}$ is regenerated, so the specific gravity of the electrolyte increases.
½ mark
During discharging, $H_{2} S O_{4}$ is consumed and water is formed, so the specific gravity decreases.
½ mark

Answer: (I)(a) Q is smallest (lowest [Zn²⁺], highest [Cu²⁺]), so E is highest; (b) $E = 1.07$ V, lower measured value due to internal resistance/non-ideality; (II) $1 S m^{2} m o l^{- 1} = 1 0^{4} S c m^{2} m o l^{- 1}$ ; (III) specific gravity increases on charging and decreases on discharging.

Where the marks are

Explains through the Nernst equation that the smallest $Q_{c}$ gives the largest EMF in experiment 4 (1 mark)
Writes the Nernst equation with correct substitution for experiment 3 (½ mark)
Obtains $E = 1.0705$ V (≈1.07 V) (½ mark)
Gives a valid reason such as internal resistance or IR drop for the lower measured EMF (1 mark)
Shows $1 S m^{2} m o l^{- 1} = 1 0^{4} S c m^{2} m o l^{- 1}$ (1 mark)
States that specific gravity increases on charging (½ mark) and decreases on discharging (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 2026-27: 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 sample paper 2026-27: question 33(A)

Hints

Step-by-step solution

Revise the chapter

More questions from Electrochemistry