Calculate the equilibrium constant for the reaction, 
   $\mathrm{Zn}\left(\mathrm{s}\right)+{\mathrm{Cu}}^{2+}\left(\mathrm{aq}\right)⟶{\mathrm{Zn}}^{2+}\left(\mathrm{aq}\right)+\mathrm{Cu}\left(\mathrm{s}\right)$ 
                                                         $\left[\text{ Given, }{\mathrm{E}}_{\text{cell }}^{\circ }=1.1\mathrm{V}\right]$

$\mathrm{Zn}\left(\mathrm{s}\right)+{\mathrm{Cu}}^{2+}\left(\mathrm{aq}\right)⟶{\mathrm{Zn}}^{2+}\left(\mathrm{aq}\right)+\mathrm{Cu}\left(\mathrm{s}\right)$
$0={\mathrm{E}}_{\mathrm{cell}}={\mathrm{E}}_{\mathrm{cell}}^{\circ }-\frac{2.303\mathrm{RT}}{2\mathrm{F}}\log \frac{\left[{\mathrm{Zn}}^{2+}\right]}{\left[{\mathrm{Cu}}^{2+}\right]}$
or     ${\mathrm{E}}_{\mathrm{cell}}^{\circ }=\frac{2.303\mathrm{RT}}{2\mathrm{F}}\log \frac{\left[{\mathrm{Zn}}^{2+}\right]}{\left[{\mathrm{Cu}}^{2+}\right]}$
But at equilibrium $\frac{\left[{\mathrm{Zn}}^{2+}\right]}{\left[{\mathrm{Cu}}^{2+}\right]}={\mathrm{K}}_{\mathrm{C}}$
          $\begin{array}{l} {\mathrm{E}}_{\text{cell }}^{\circ }=\frac{0.059\mathrm{V}}{2}\log {\mathrm{K}}_{\mathrm{C}}=1.1\mathrm{V}\\ \log {\mathrm{K}}_{\mathrm{C}}=\frac{1.1\mathrm{V}\times 2}{0.059\mathrm{V}}=37.288\\ {\mathrm{K}}_{\mathrm{C}}=2\times {10}^{37}\text{ at }298\mathrm{K}\end{array}$