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The half cell reaction for rusting of ir...

The half cell reaction for rusting of iron are:
`2H^(+) + 1/2 O_(2) + 2e^(-) to H_(2)O, E^(@) = 1.23 V`
`Fe^(2+) + 2e^(-) to Fe, E^(@) = -0.44 V`
`DeltaG^(@)` for the reaction.
`4H^(+) + O_(2) + 2Fe to 2Fe^(2+) + 2H_(2)O` is:

A

`- 76` kJ

B

`- 644` kJ

C

`-122` kJ

D

`-176` kJ

Text Solution

Verified by Experts

The correct Answer is:
B
`4H^(+) + O_(2) + 4e^(-) to 2H_(2)O`
`DeltaG^(@) = -nFE^(@) = -4 xx 96500 xx 1.67`
`= -644 kJ`
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