The Gibbs energy for the decomposition of `Al_(2)O_(3)` at `500^(@)C` is as follows:
`(2)/(3)Al_(2)O_(3) rarr (4)/(3)Al + O_(2), Delta_(r)G = +966kJ mol^(-1)`
The potential difference needed for electrolytic reeduction of `Al_(2)O_(3)` at `500^(@)C` 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) Al_(2)O_(3) to (4)/(3) Al + O_(2) , Delta_(r) G = + 966k J mol^(-1)`
Thus , `(2)/(3) xx (3 O^(2-)) , i.e., 2 O^(2-) to O_(2) , n = 4 `
`Delta G =- n FE`
`therefore 966 xx 10^(3) = - 4 xx 96500 xx E `
or `E = -2.50 V`
Thus , minimum potential difference required = 2.50 V .