Calculate the emf of the cell $\mathrm{Cu}(s)\left|\mathrm{Cu}^{2+}(a q)ight|\left|\mathrm{Ag}^{+}(a q)ight|…

Calculate the emf of the cell $\mathrm{Cu}(s)\left|\mathrm{Cu}^{2+}(a q)ight|\left|\mathrm{Ag}^{+}(a q)ight| \mathrm{Ag}(s)$ Given $E_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^0=+0.34 \mathrm{~V}, E_{\mathrm{Ag}^{+} / \mathrm{Ag}}^0=0.80 \mathrm{~V}$
  1. $+0.46 \mathrm{~V}$
  2. $+1.14 \mathrm{~V}$
  3. $+0.57 \mathrm{~V}$
  4. $-0.46 \mathrm{~V}$

Solution

$\begin{aligned} E_{\text {cell }}^{\circ} & =E_{\text {red (cathode) }}^{\circ}-E_{\text {oxi (anode) }}^{\circ} \\ & =E_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{\circ}-E_{\mathrm{Cu}^{+} / \mathrm{Cu}^{2+}}^0 \\ & =0.80-(+0.34)=+0.46 \mathrm{~V}\end{aligned}$ *

Asked in: JEE-TOPICTESTS-CHEMISTRY

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