On the basis of the information available from the reaction

\(\frac{4}{3}Al + O_2 \longrightarrow \frac{2}{3}Al_2O_3,  \Delta G = -827\text{ kJ/mol}^{-1}\)

The minimum e.m.f. required to carry the electrolysis of \(Al_2O_3\) is \((F = 96500 \text{ C mol}^{-1})\)

2.14 V

Given,

\(\frac{4}{3}Al + O_2 \longrightarrow \frac{2}{3}Al_2O_3,  \Delta G = -827\text{ kJ/mol}^{-1}\)

\(n = 4\)

\(\Delta G = 872\text{ kJ/mol}^{-1}\) ['-' ve sign indicates exothermic raction]

or, \(\Delta G = 872000\text{ J/mol}^{-1}\)

\(F = 96500 C\)

We know that,

\(\Delta G= nFE\)

or, \(E = \frac{\Delta G}{nF}\)

or, \(E = \frac{872000}{4 × 96500}\)

or, \(E \approx 2.14 V\)