The equilibrium constant ${\mathrm{K}}_{\mathrm{C}}$ of the reaction: ${\mathrm{Cu}}_{\left(\mathrm{s}\right)}+2{\mathrm{Ag}}_{\left(\mathrm{aq}\right)}^{+}\to {\mathrm{Cu}}^{2+}\left(\mathrm{aq}\right)+2{\mathrm{Ag}}_{\left(\mathrm{s}\right)}$ is $10\times {10}^{15}$, calculate the ${\mathrm{E}}_{\text{cell }}$ of this reaction at $298 \mathrm{K}$

The equilibrium constant ${\mathrm{K}}_{\mathrm{C}}$ of the reaction: ${\mathrm{Cu}}_{\left(\mathrm{s}\right)}+2{\mathrm{Ag}}_{\left(\mathrm{aq}\right)}^{+}\to {\mathrm{Cu}}^{2+}\left(\mathrm{aq}\right)+2{\mathrm{Ag}}_{\left(\mathrm{s}\right)}$ is $10\times {10}^{15}$

${\mathrm{E}}_{\text{cell }}={\mathrm{E}}_{\text{cell }}°-\frac{0.059}{\mathrm{n}}\mathrm{logQ}$

${\mathrm{E}}_{\text{cell }}={\mathrm{E}}_{\text{cell }}°-\frac{0.059}{\mathrm{n}}\mathrm{logQ}$

At equilibrium,

${\mathrm{E}}_{\text{cell }}°=\frac{0.059}{\mathrm{n}}\mathrm{logK}$

${\mathrm{E}}_{\text{cell }}°=\frac{0.059}{2}\log \left(10\times {10}^{15}\right)$

${\mathrm{E}}_{\text{cell }}°=0.059\times 8$

${\mathrm{E}}_{\text{cell }}°=0.4736 \mathrm{V}$

Therefore, the correct option is (B).