Question

Estimate the minimum potential difference needed to reduce $A{l}_2{O}_3$ at ${500}^o$C. The Gibbs energy change for the decomposition reaction :

$\frac{2}{3}A{l}_2{O}_3\to \frac{4}{3}Al+O_2$ is 960 kJ.

(Take F = 96500 C $mo{l}^{-1}$)

Answer 1

Given Reaction:

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

Cell Reaction:

{$AtCathode:\frac{4}{3}A{l}^{+3}\to \frac{4}{3}Al+4e^{-}$

$AtAnode:2O^{-2}+4e^{-}\to O_2$}

so, the no. of electrons involved in the cell reaction is $4$.

$\because \Delta G=-nF{E}^{\circ }$

$⟹$ $960\times 1000=-4\times 96500\times E^{\circ }$

$⟹$ $E^{\circ }=-2.49V$

So, the minimum potential difference required to reduce $A{l}_2{O}_{3}at{500}^{0}C$ is $-2.49V$.