In the reaction 32Al2O3⟶34Al+O2 For the oxidation half-reaction Al3++3e−⟶Al no. of electron transfered (n)=3 $\begin{aligned}
& \Delta \mathrm{G}^{\circ}=-n \mathrm{FE}^{\circ} \
& 940=3 \times 96500 \times \mathrm{E}^{\circ}
\end{aligned}\begin{aligned}
\mathrm{E}^{\circ} & =\frac{940 \times 10^3 \mathrm{J}}{3 \times 96500} \
& =3.24 \approx 3 \mathrm{V}
\end{aligned}$