Given the equilibrium constant : $K_{C}$ of the reaction : $\mathrm{Cu}(\mathrm{s})+2 \mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})$ is $10 \times 10^{15}$ calculate the $\mathrm{E}_{\text {cell }}^{0}$ of this reaction at $298 \mathrm{~K}$
$\left[2.303 \frac{\mathrm{RT}}{\mathrm{F}}\right.$ at $\left.298 \mathrm{~K}=0.059 \mathrm{~V}\right]$
Correct Option: , 3
$\mathrm{E}_{\mathrm{cell}}^{0}=\frac{2.303 \mathrm{RT}}{\mathrm{nF}} \log \mathrm{K}_{\mathrm{C}}$ or $\mathrm{E}_{\mathrm{cell}}^{0}=\frac{0.059 \mathrm{~V}}{\mathrm{n}} \log \mathrm{K}_{\mathrm{C}}$
$=\frac{0.059 \mathrm{~V}}{2} \log 10^{16}=0.4736 \mathrm{~V}$
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