For the cell Cu(s)| Cu^2+( aq)(0.1M)… - JEE Main 2021 Chemistry PYQ

JEE Main 2021 — Chemistry Physical Chemistry

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For the cell $\mathrm{Cu}(s)|{\mathrm{Cu}}^{2+}(\mathrm{aq})(0.1M)\|{\mathrm{Ag}}^{+}(\mathrm{aq})(0.01M)|\mathrm{Ag}(s)$ the cell potential ${E}_{1}=0.3095V$ . For the cell $\mathrm{Cu}(s)|{\mathrm{Cu}}^{2+}(\mathrm{aq})(0.01M)\|{\mathrm{Ag}}^{+}(\mathrm{aq})(0.001M)|\mathrm{Ag}(s)$ the cell potential $=x\times {10}^{-2}V.$ Find value of x (Round off the Nearest Integer).

[ Use $:\frac{2.303\mathrm{RT}}{F}=0.059J$ ]

Official previous-year question

Held on 27 Jul 2021 · Verified 6 Jul 2026.

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Solution

Cell reaction is :

$\mathrm{Cu}(s)+2{\mathrm{Ag}}^{+}(\mathrm{aq})\rightarrow {\mathrm{Cu}}^{2+}(\mathrm{aq})+2\mathrm{Ag}(s)$

Now, ${E}_{\mathrm{cell}}={E}_{\mathrm{Cell}}^{o}-\frac{0.059}{2}\mathrm{log}\frac{[{\mathrm{Cu}}^{2+}]}{{[{\mathrm{Ag}}^{+}]}^{2}}$

${E}_{1}=0.3095={E}_{\mathrm{Cell}}^{o}-\frac{0.059}{2}\cdot \mathrm{log}\frac{0.1}{(0.01{)}^{2}}$ ....... $(1)$

${E}_{2}={E}_{\mathrm{Cell}}^{o}-\frac{0.059}{2}\cdot \mathrm{log}\frac{0.01}{(0.001{)}^{2}}..........(2)$

From $(1)$ and $(2),{E}_{2}=0.28V=28\times {10}^{-2}V$

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