The cell potential for the given cell at… - JEE Main 2022 Chemistry PYQ

JEE Main 2022 — Chemistry Physical Chemistry

2022integermedium

The cell potential for the given cell at $298K\mathrm{Pt}|{H}_{2}(g,1\mathrm{bar})||{H}^{+}(\mathrm{aq})\|{\mathrm{Cu}}^{2+}(\mathrm{aq})|\mathrm{Cu}(s)$ is $0.31V$ . The $\mathrm{pH}$ of the acidic solution is found to be $3$ , whereas the concentration of ${\mathrm{Cu}}^{2+}$ is ${10}^{x}M$ . The value of $x$ is _________.

(Given: ${E}_{{\mathrm{Cu}}^{2+}/\mathrm{Cu}}^{\Theta }=0.34V$ and $\frac{2.303\mathrm{RT}}{F}=0.06V$ )

Official previous-year question

Held on 29 Jun 2022 · Verified 6 Jul 2026.

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Solution

${H}_{2}(g)+{\mathrm{Cu}}^{2+}(\mathrm{aq}.)\rightarrow 2{H}^{+}(\mathrm{aq}.)+\mathrm{Cu}(s)$

Applying nernst equation

${E}_{\mathrm{cell}}={E}_{\mathrm{cell}}^{0}-\frac{0.06}{n}\mathrm{log}\frac{[\mathrm{products}]}{[\mathrm{Reactants}]}$

${E}_{\mathrm{cell}}=\mathrm{Cell}\mathrm{potential}=0.31V {E}_{\mathrm{cell}}^{0}=0.34 {p}^{H}=3,\mathrm{Hence},[{H}^{+}]={10}^{-3}$

$0.31=0.34-\frac{0.06}{2}\mathrm{log}\frac{{[{H}^{+}]}^{2}}{[{\mathrm{Cu}}^{2+}]}$

$[{\mathrm{Cu}}^{2+}]={10}^{-7}M$

$x=7$

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Solution

${H}_{2}(g)+{\mathrm{Cu}}^{2+}(\mathrm{aq}.)\rightarrow 2{H}^{+}(\mathrm{aq}.)+\mathrm{Cu}(s)$

Applying nernst equation

${E}_{\mathrm{cell}}={E}_{\mathrm{cell}}^{0}-\frac{0.06}{n}\mathrm{log}\frac{[\mathrm{products}]}{[\mathrm{Reactants}]}$

${E}_{\mathrm{cell}}=\mathrm{Cell}\mathrm{potential}=0.31V {E}_{\mathrm{cell}}^{0}=0.34 {p}^{H}=3,\mathrm{Hence},[{H}^{+}]={10}^{-3}$

$0.31=0.34-\frac{0.06}{2}\mathrm{log}\frac{{[{H}^{+}]}^{2}}{[{\mathrm{Cu}}^{2+}]}$

$[{\mathrm{Cu}}^{2+}]={10}^{-7}M$

$x=7$

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Solution

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Solution

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JEE Main Physical Chemistry in other years

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