Question

The Gibb's energy for the decomposition of $A{l}_2{O}_3$ at ${500}^{0}C$ is as follows :

$\frac{2}{3}A{l}_2{O}_3⟶\frac{4}{3}Al+O_2$, ${△}_{r}G=+960kJmo{l}^{-1}$

The potential difference needed for the electrolytic reduction of aluminium oxide $\left(A{l}_2{O}_3\right)$ at ${500}^{0}C$ at least:

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

Answer 1

Answer: (C)

2.5 V

At Anode

$2O^{2-}+4e^{-}\to O_2]\times 3$

At Cathode

$A{l}^{3+}\to Al+3e^{-}]\times 4$

Net reaction

$4Al+6O^{2-}\to 3O_2+4Al$

or

$\frac{4}{3}Al+2O^{2-}\to O_2+\frac{4}{3}Al$

$n=\frac{12}{3}=4$

$\Delta G^o=-nF{E}^o$

$\Delta G^o=+960kJ/mol=960\times 1000J/mol$

$F=96500Coulomb$

Substituting values

$960\times 1000=-4\times 96500\times E^o$

$E^o=-\frac{960000}{4\times 96500}$

$E^o=-2.48V$

Hence, potential difference $\approx 2.5V$