The Edison storage cell is represented as $Fe(s)/FeO(s)/KOH(aq)/Ni_2O_3(s)/Ni(s)$.The half cell reactions are $Ni_2O_3(s)+H_2O(l)+2e^-\rightarrow 2NiO(s)+2OH^-(aq)\Rightarrow E^0=0.4V$,$FeO(s)+H_2O(l)+2e^-\rightarrow Fe(s)+2OH^-(aq)\Rightarrow E^0=-0.87.V$.What will be the maximum amount of electrical energy that can be obtained from one mole of $Ni_2O_3$

Question

$\begin{array}{1 1}(a)\;2.45\times 10^2KJmole^{-1}\\(b)\;2.45\times 10KJmole^{-1}\\(c)\;2.45\times 10^3KJmole^{-1}\\(d)\;None\end{array}$