The Gibbs energy for the decomposition of `Al_(2)O_(3)` at `500^(@)C` is as follows
`(2)/(3)A1_(2)O_(3)to (4)/(3) Al+O_(2), Delta_(r) G = + 966KJmol^(-1)`
The potential difference needed for electrolytic reduction of `Al_(2)O_(3)` at `500^(@)` is at least :

Options

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

Answer

A

Explanation

`2/3 overset(+3) (A) l_(2) O_(3) to (4)/(3) overset(0)(A) l + O_(2)`
No. of electrons involved = `2/3 ( 3 xx 2) = 4`
`Delta G^(@) = - n FE^(@)`
`966 xx 10^(3) = -4 xx 96500 xx E^(@)`
`therefore " " E^(@) = (966 xx 10^(3))/(4 xx 96500) = -2.5 V `
Thus , minimum potential difference required = 2.50 V