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Q. The Gibbs energy for the decomposition of $Al_2O_3$ at $500\,{}^{\circ}C$ is as follows :
$2/3Al_{2}O_{3} \rightarrow 4/3Al +O_2$ ; $\Delta_{r}G=+966\,kj/mol$.
The potential difference needed for electrolytic reduction of $Al_2O_3$ at $500\,{}^{\circ}C$ is at least

Electrochemistry

Solution:

The ionic reactions are :
$2/3 \times \left(2Al^{3+}\right)+4e^{-}\rightarrow 4/3Al$
$2/3 \times \left(3O^{2-}\right)\rightarrow O_2+4e^-$
Thus, no. of electrons transferred, $n = 4$.
$\Delta G=-nFE=-4\times96500\times E$
or $966\times10^{3}$
$=-4\times96500\times E$
$\Rightarrow E=-\frac{966\times10^{3}}{4\times96500}$
$=-2.5\,V$