Question

$\Delta G$ for the reaction
$\frac{4}{3} Al + O _{2} \longrightarrow \frac{2}{3} Al _{2} O _{3}$
is $-772\, kJ\, mol ^{-1}$ of $O _{2}.$
Calculate the minimum EMF in volts required to carry out an electrolysis of $Al _{2} O _{3}$.

Answer 1

$Al \longrightarrow Al ^{3+}+3 e ^{-}$

$\frac{4}{3}$ mol of $Al =\frac{4}{3} \times 3$ mol $e ^{-}=4$ mol $e ^{-}$

$n=4$

$\Delta G =- nFE$

$-772 \times 1000 J =-4 \times 96500 \times E$

$\therefore E =\frac{772 \times 1000}{4 \times 96500}=2.0\, V.$