$E^o_{Zn^{+2}/Zn}=-0.76V$.So $E^o_{Zn/Zn^{+2}}=$?

Question

$\begin{array}{1 1}(a)\;-0.076-\large\frac{0.0591}{2}\normalsize\log\large\frac{1}{Zn^{+2}}\\(b)\;-0.76\\(c)\;0.76\\(d)\;0\end{array}$