The minimum voltage required to electrolyse alumina in the Hall-Heroult process is [Given, ${\mathrm{\Delta G}}^{\mathrm{o}}\left({\mathrm{Al}}_2{\mathrm{O}}_3\right)=-1520$ kJ/mol and ${\mathrm{\Delta G}}^{\mathrm{o}}\left({\mathrm{CO}}_2\right)=-394$ kJ/mol]

In Hall-Heroult process the following reactions occur 

$\begin{array}{l}3\mathrm{C}+2{\mathrm{Al}}_2{\mathrm{O}}_3\to 4\mathrm{Al}+3{\mathrm{CO}}_2\\ {\mathrm{\Delta G}}^{\mathrm{o}}=3{\mathrm{\Delta }}_{\mathrm{f}}{\mathrm{G}}^{\mathrm{o}}\left({\mathrm{CO}}_2\right)-2{\mathrm{\Delta }}_{\mathrm{f}}{\mathrm{G}}^{\mathrm{o}}\left({\mathrm{Al}}_2{\mathrm{O}}_3\right)\end{array}$

$\begin{array}{l}=3(-394)-2(-1520)=1858\mathrm{kJ}\\ {\mathrm{\Delta G}}^{\mathrm{o}}=-{\mathrm{nFE}}^{\mathrm{o}}=\mathrm{or}=-{\mathrm{E}}^{\mathrm{o}}\\ =\frac{{\mathrm{\Delta G}}^{\mathrm{o}}}{\mathrm{nF}}=\frac{1858\times 1000}{12\times 96500}=1.60\mathrm{V}\end{array}$