Given :
$E^{\large\circ}_{Zn^{2+}/Zn}=-0.76V$
$E^{\large\circ}_{Zn(NH_{34}^{2+}/Zn^{\large\circ}}=-1.03V$
$\begin{array}{1 1}(A)\;3\times 10^{10}&(B)\;1.38\times 10^9\\(C)\;2.28\times 10^9&(D)\;9.1\times 10^6\end{array}$