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Calculate the emf of the cell
$\mathrm{Cu}(s)\left|\mathrm{Cu}^{2+}(a q)\right|\left|\mathrm{Ag}^{+}(a q)\right| \mathrm{Ag}(s)$
Given
$E_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^0=+0.34 \mathrm{~V}, E_{\mathrm{Ag}^{+} / \mathrm{Ag}}^0=0.80 \mathrm{~V}$
Options:
$\mathrm{Cu}(s)\left|\mathrm{Cu}^{2+}(a q)\right|\left|\mathrm{Ag}^{+}(a q)\right| \mathrm{Ag}(s)$
Given
$E_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^0=+0.34 \mathrm{~V}, E_{\mathrm{Ag}^{+} / \mathrm{Ag}}^0=0.80 \mathrm{~V}$
Solution:
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Verified Answer
The correct answer is:
$+0.46 \mathrm{~V}$
$\begin{aligned} E_{\text {cell }}^{\circ} & =E_{\text {red (cathode) }}^{\circ}-E_{\text {oxi (anode) }}^{\circ} \\ & =E_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{\circ}-E_{\mathrm{Cu}^{+} / \mathrm{Cu}^{2+}}^0 \\ & =0.80-(+0.34)=+0.46 \mathrm{~V}\end{aligned}$
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