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JEE MAIN 2022
29-07-2022 S2
Question
For a cell, $\mathrm{Cu}(\mathrm{s}) \mid \mathrm{Cu}^{2+}\left(0.001 \mathrm{M}| | \mathrm{Ag}^{+}(0.01 \mathrm{M}) \mid \mathrm{Ag}(\mathrm{s})\right.$ the cell potential is found to be 0.43 V at 298 K . The magnitude of standard electrode potential for $\mathrm{Cu}^{2+} / \mathrm{Cu}$ is $\_\_\_\_$ $\times 10^{-2} \mathrm{~V}$. $\left[\right.$
Given: $E_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{\Theta}=0.80 \mathrm{~V}$ and $\left.\frac{2.303 R T}{F}=0.06 \mathrm{~V}\right]$
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Solution
At anode $\mathrm{Cu} \rightarrow \mathrm{Cu}^{2+}+2 \mathrm{e}^{-}$
At cathode $2 \mathrm{Ag}^{+}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Ag}$
Cell reaction $\rightarrow \mathrm{Cu}+2 \mathrm{Ag}^{+} \rightarrow \mathrm{Cu}^{2+}+2 \mathrm{Ag}$ $$ \begin{aligned} & E_{\mathrm{cell}}=E_{\mathrm{cel}}^0-\frac{0.06}{2} \log \frac{\left[\mathrm{Cu}^{2+}\right]}{\left[\mathrm{Ag}^{+}\right]^2} \\ & 0.43=E_{\mathrm{cel}}^0-\frac{0.06}{2} \log \frac{[0.001]}{(0.01)^2} \\ & E_{\mathrm{cel}}^0=0.46 \\ & E_{\mathrm{cel}}^0=E_{\mathrm{Ag}^{\prime} / \mathrm{Ag}}^0-E_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^0 \\ & 0.46=0.80-E_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^0 \\ & E_{\mathrm{Cu}^{2-} / \mathrm{Cu}}^0=0.34 \mathrm{volt} \\ & E_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^0=34 \times 10^{-2} \end{aligned} $$
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