Question

The Gibbs' energy for the decomposition of $A{l}_2{O}_3$ at ${500}^{o}C$ is as follows:
$\frac{2}{3}A{l}_2{O}_3\to \frac{4}{3}Al+O_2;\Delta G=+966kJmo{l}^{-1}$.
The potential difference needed for the electrolytic reduction of aluminium oxide $\left(A{l}_2{O}_3\right)$ at ${500}^{o}C$ is at least:

• A. 5 V
• B. 4.5 V
• C. 3 V
• D. 2.5 V

Answer 1

Answer: (A), (B), (C)

5 V
3 V
4.5 V

$\frac{2}{3}{Al}_2{O}_3\to \frac{4}{3}Al+O_2$

$G=+960kJ{mol}^{-1}$

No. of electrons exchanged $\left(n\right)=3$

For this equation $\Delta G=\frac{2}{3}G$

$\Delta G=-nFE$

E=potential difference

$\frac{2}{3}\times 960kJ=-3\times 96500\times E$

$E=-2.2V$