Chemistry - Electrochemistry Question with Solution | TestHub

On the basis of information available for the reaction.
$\frac{\text{4}}{\text{3}}\text{Al}\text{+}{\text{O}}_{\text{2}}\to \frac{\text{2}}{\text{3}}{\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}\text{;}\text{ΔG}\text{=}\text{–827}\text{kJ/mol}\text{of}{\text{O}}_{\text{2}}$
The minimum emf required to carry out an electrolysis of$\mathrm{A}{\mathrm{l}}_2{\mathrm{O}}_3$is:
$\left(\text{Given:}\text{One Faraday}\text{=}\text{96500}\text{C}\right)$