Given the equilibrium constant : $\mathrm{K}_{\mathrm{C}}$ 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 K $$ \left[2.303 \frac{\mathrm{RT}}{\mathrm{~F}} \text { at } 298 \mathrm{~K}=0.059 \mathrm{~V}\right] $$