Question

The Gibbs' energy for the decomposition of ${Al}_2{O}_3$ at ${500}^{o}C$ is as follows:

$\frac{2}{3}{Al}_2{O}_3\to \frac{4}{3}Al+O_2;{\Delta }_{r}G=+960kJ{mol}^{-1}$

The potential difference needed for the electrolytic reduction of aluminium oxide $\left({Al}_2{O}_3\right)$ at ${500}^{o}C$ is at least:

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

Answer 1

Answer: (B)

$2.5V$

The relationship between the Gibb's free energy change and the cell potential is as given below.
$\Delta G^o=-nF{E}^o$

where n is the number of electrons transferred

Substitute values in the above expression.
$960\times 1000=-4\times 96500\times E^o$
$E^o=-2.48V$
So difference $=2.5V$

Thus, the potential difference needed for the electrolytic reduction of aluminium oxide $\left({Al}_2{O}_3\right)$ at ${500}^{o}C$ is at least $2.5V$.