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Q. The Gibbs' energy for the decomposition of $Al_2O_3 $ at $500^{\circ} C$ is as follows:
$ \frac{2}{3} Al_2 O_3 \rightarrow \frac{4}{3} Al + O_2$ ;
$\Delta_r G = + 960 \, kJ \, mol^{-1} $
The potential difference needed for the electrolytic reduction of aluminium oxide $(Al_2O_3)$ at $500^{\circ} C$ is at least

AIEEEAIEEE 2010Electrochemistry

Solution:

$\Delta G^{\circ} = - nFE^{\circ}$
$960 \times 1000 = - 4 \times 96500 \times E^{\circ}$
$E^{\circ} = - 2.48 V$
So difference = $2.5\, V$