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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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Knowledge Check

  • The minimum voltage required to electrolyse alumina in the Half - Heroult process is Given, DeltaG_(f)^(@)(Al_(2)O_(3))=-1520 kJ mol^(-1) DeltaG_(f)^(@)(CO_(2))=-394 kJ mol^(-1)

    A
    1.575 V
    B
    1.60 V
    C
    1.312 V
    D
    `-2.62 V`

  • Based on given information DeltaG^(@)_(f)(CaO) =- 604.2kJ//mol and DeltaG^(@)_(f)(A1_(2)O_(3)) =- 1582kJ//mol , which of the following is feasible?

    A
    `2Ca +A1_(2)O_(3)rarr2A1+3CaO`
    B
    `3CaO +2A1 rarrA1_(2)O_(3) +3Ca`
    C
    Both `(a)` and `(b)`
    D
    None of the above

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    A
    `-100.5 kJ`
    B
    `-62.5 kJ`
    C
    `-80.3 kJ`
    D
    `-74.9 kJ`

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