$\begin{aligned}
& \mathrm{E}{\mathrm{cell}}=\mathrm{E}{\mathrm{cell}}^{\circ}-\frac{0.0591}{\mathrm{n}} \log \mathrm{Q} \
& \text { Where } \mathrm{Q}=\frac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{Cu}^{2+}\right]}
\end{aligned}Forcompletedischarge\mathrm{E}{\text {cell }}=0So\mathrm{E}{\text {cell }}^{\circ}=\frac{0.591}{2} \log \frac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{Cu}^{2+}\right]}\Rightarrow\left|\frac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{Cu}^{2+}\right]}\right|=10^{37.3}$ Hence, (C) is correct.