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

Answer

A

Explanation

`2//3Al_(2)O_(3)rarr4//3Al+O_(2)`.
`DeltaG= +966kJ "mol"^(-1)`
`4//3Al^(3+)+4e^(-)rarr4//3Al`
`DeltaG=-nFE`
` therefore 966 xx10^(3)=-4xx96500xxE`.
`therefore E=2.50V`
`therefore` the minimum potential difference required = 2.50V.