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The minimum voltage required to electrol...

The minimum voltage required to electrolyse alumina in the Hall-Heroul process is [Given, `DeltaG^(@)_(f)(A1_(2)O_(3)) =- 1520kJ//mol` and `DeltaG^(@)_(f)(CO_(2)) = 394kJ//mol]`

A

`1.60V`

B

`1.575V`

C

`1.312V`

D

`-2.62V`

Text Solution

Verified by Experts

The correct Answer is:
A
In hall-Heroult process the following reactions occur
`3C+2A1_(2)O_(3)rarr4A1+3CO_(2)`
`DeltaG^(@) = 3Delta_(f)G^(@)(CO_(2))-2Delta_(f)G^(@)(A1_(2)O_(2))`
`=3 (-394)-2(-1520) = 1858kJ`
`DeltaG^(@) = nFE^(@)=` or `=-E^(@)`
`=(DeltaG^(@))/(nF)= = (1858xx1000)/(12xx96500) = 1.60V`
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