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The rusting of iron takes place as follo...

The rusting of iron takes place as follows:
`2H^(+) + 2e + 1//2O(2) rarr H_(2)O_((l)), E^(@) = +1.23 V`
`Fe^(2+) + 2e rarr Fe_((s)), E^(@) = -0.44 V`
The `DeltaG^(@)` for the net process is :

A

`-322 kJ mol^(-1)`

B

`161 kJ mol^(-1)`

C

`-152 kJ mol^(-1)`

D

`-76 kJ mol^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
A
`E_(cell)^(@) = E_(OPFe)^(@) + E_(RPH_(2)O)^(@)`
`:. DeltaG^(@) = -nE^(@)F`
`= -2 xx 1.67 xx 96500 J`
`= -322.31 kJ mol^(-1)`
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