In the electrochemical cell: Zn|… - NEET UG 2017 Chemistry PYQ

NEET UG 2017 — Chemistry Physical Chemistry

2017mcqhard

In the electrochemical cell:

$\mathrm{Zn}|{\mathrm{ZnSO}}_{4}(0.01M)||{\mathrm{CuSO}}_{4}(1.0 M)|\mathrm{Cu},$ the emf of this Daniel cell is ${E}_{1}$ . When the concentration ${\mathrm{ZnSO}}_{4}$ is changed to $1.0M$ and that of ${\mathrm{CuSO}}_{4}$ changed to $0.01M,$ the emf changes to ${E}_{2}$ . From the following, which one is the relationship between ${E}_{1}$ and ${E}_{2}$ ? (Given, $\frac{\mathrm{RT}}{F}=0.059$ )

Official previous-year question

Held on 30 Apr 2017 · Verified 9 Jul 2026.

Options

A
${E}_{1}={E}_{2}$
B
${E}_{1}<{E}_{2}$
C
${E}_{1}>{E}_{2}$
D
${E}_{2}=0 \neq {E}_{1}$

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Solution

$\mathrm{Zn}|{\mathrm{ZnSO}}_{4}(0.01 M)||{\mathrm{CuSO}}_{4}(1.0M)|\mathrm{Cu}$

$E={E}_{\mathrm{cell}}^{o}-\frac{2.303\mathrm{RT}}{2F}\mathrm{log}\frac{[{\mathrm{Zn}}^{+2}]}{[{\mathrm{Cu}}^{+2}]}$

$\therefore {E}_{1}={E}_{\mathrm{cell}}^{o}-\frac{2.303\mathrm{RT}}{2\times F}\times \mathrm{log}⁡\frac{(0.01)}{1}$

When concentrations are changed

$\therefore {E}_{2}={E}_{\mathrm{cell}}^{o}-\frac{2.303\mathrm{RT}}{2F}\times \mathrm{log}⁡\frac{1}{0.01}$

i.e., ${E}_{1}>{E}_{2}$

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