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The rusting of iron takes place as follo...

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

A

`-76`

B

`-322`

C

`-122`

D

`-176`

Text Solution

Verified by Experts

The correct Answer is:
B
`DeltaG^(@) =- nFE^(@)`
`Fe^(+2) +2e^(Theta) harr Fe(n = 2)`
`E_(cell)^(@) = E_(H^(+)//H_(2)O)^(@) - E_(Fe^(2+)//Fe)^(@)`
`= 1.23 - (-0.44) = 1.67V`
`DeltaG^(@) =- 2 xx (96500 xx 1.67)/(1000) =- 322.3 kJ`
`DeltaG^(@) =- 322 kJ//mol`
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