If decomposition of `Al_(2)O_(3)` at 500 `""^(@)C` temperature gives following reaction and gives Gibb's free energy :
`(2)/(3)Al_(2)O_(3) to (4)/(3) Al+O_(2) . . . DeltaG=966kJ" "mol^(-1)`
then what is the difference of minimum required electrical energy for the reduction of `Al_(2)O_(3)` by electrolysis ?

Options

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

Answer

C

Explanation

`(2)/(3)Al_(2)O_(3)" "therefore DeltaG=966kJ`
`therefore Al_(2)O_(3)" "DeltaG=(966xx3)/(2)=1449kJ`
`2Al^(3+)+6e^(-) to 2Al`
`DeltaG=nFE_(cell)`, `therefore E_(cell)=(DeltaG)/(nF)|"Where, "n=6," "DeltaG=1449xx10^(3)J," "E=96500`
`=(1449xx1000)/(6xx96500)`
`=2.5V`