Question

The Gibb’s energy for the decomposition of $Al_{2}O_{3}$ at $500^{\circ}C$ is as follows :
$2/3 Al_{2}O_{3}\rightarrow 4/3 Al+O_{2'} \Delta_{r} G=+966 \,KJ\, mol^{-1}$
The potential difference needed for electrolytic reduction of $Al_{2}O_{3} $ at $500°C$ is at least

• A. 5.0 V
• B. 4.5 V
• C. 3.0 V
• D. 2.5 V

Answer 1

Answer: (D)

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