The Gibbs energy for the decomposition of `Al_(2)O_(3)` at `500^(@)C` is as follow :
`(2)/(3)Al_(2)O_(3) rarr (4)/(3)Al+O_(2), Delta_(r)G= +960 kJ mol^(-1)`
The potential difference needed for the electrolytic reduction of aluminium oxide `(Al_(2)O_(3))` at `500^(@)C` is

Options

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

Answer

C

Explanation

`Delta G^(@) =- nRE^@`
`F = 96500, Delta G^@ = + 960 xx 10^3 J // mol `
`2/3 Al_2 O_3 rarr 4/3 Al + O_2`
Total number of Al atoms in `Al_2O_3 rarr 3/3 xx 2 = 4/3`
As `3e^(-)` change occur for each Al atom
`therefore` Total `n= 4/3 xx 3 =4`
`E^@ =- (Delta G^@)/(nF)= -(960 xx 1000)/(4 xx96500) rArr E^@=-2.48 V =-2.5 V`