Find the emf of the cell in which the… - NEET UG 2022 Chemistry PYQ

NEET UG 2022 — Chemistry Physical Chemistry

2022mcqhard

Find the emf of the cell in which the following reaction takes place at $298K$

$\mathrm{Ni}(s)+2{\mathrm{Ag}}^{+}(0.001M)\rightarrow {\mathrm{Ni}}^{2+}(0.001M)+2\mathrm{Ag}(s)$

(Given that $E{^{\circ}}_{\mathrm{cell}}=1.05V,\frac{2.303\mathrm{RT}}{F}=0.059$ at $298K$ )

Official previous-year question

Held on 30 Apr 2022 · Verified 9 Jul 2026.

Options

A
$1.385V$
B
$0.9615V$
C
$1.05V$
D
$1.0385V$

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Solution

Given reaction: $\mathrm{Ni}(s)+2{\mathrm{Ag}}^{+}(0.001M)\rightarrow {\mathrm{Ni}}^{2+}(0.001M)+2\mathrm{Ag}(s)$

Applying Nernst equation we have:

${E}_{(\mathrm{cell})}={E}_{(\mathrm{cell})}^{0}-\frac{2.303\mathrm{RT}}{\mathrm{nF}}\mathrm{log}\frac{[{\mathrm{Ni}}^{2+}][\mathrm{Ag}]}{{[{\mathrm{Ag}}^{+}]}^{2}[\mathrm{Ni}]}$

Active mass of solid is taken to be unity so $[\mathrm{Ni}(s)]=[\mathrm{Ag}(s)]=1$

${E}_{(cell)}={E}_{(cell)}^{0}-\frac{0.059}{n}\mathrm{log}\frac{[N{i}^{2+}]}{{[A{g}^{+}]}^{2}}$

$=1.05-\frac{0.0591}{2}\mathrm{log}\frac{(0.001)}{{(0.001)}^{2}}$

$=1.05-0.0295\mathrm{log}(1\times {10}^{3})$

$=1.05-0.0295\times 3$

$=0.9615V$

Therefore, the emf of the cell is $0.9615V$ .

Note: Question is modified little bit for academic accuracy. $E{^{\circ}}_{\mathrm{cell}}=1.05V\mathrm{in}\mathrm{place}\mathrm{of}E{^{\circ}}_{\mathrm{cell}}=10.5V$ as given in NEET 2022 paper.

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