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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^(c-)=+1.23V`
`Fe^(2+)+2e^(-) rarr Fe(s)," "E^(c-)=-0.44V`
Calculae `DeltaG^(c-)` for the net process.

A

`-322 kJ mol^(-1)`

B

`-152 kJ mol^(-1)`

C

`-76kJ mol^(-1)`

D

`-161kJ mol^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
a

Cathode `:2H^(o+)(aq)+2e^(-) +(1)/(2) O_(2)(g)rarr H_(2)O(l)`
`DeltaG^(c-)._(1)=-2xxFE^(c-)=-2xxF(1.23)`
Anode`:Fe(s) rarr Fe^(2+)(aq)+2e^(-)`
`DeltaG^(c-)._(2)=-2xxFE^(c-)=-2xxF(0.44)`
`ulbar(2H^(o+)(aq)+(1)/(2)O_(2)(g)+Fe(s) rarr H_(2)O(l) +Fe^(2+)(aq))`
`DeltaG^(c-)._(n et)=[-2F(1.23)]+[-2F(0.44)]=-322.3 k J mol^(-1)`
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