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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

`-161 kJ mol^(-1)`

C

`-152 kJ mol^(-1)`

D

`-76 kJ mol^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
A
`Fe(s)rarrFe^(2+)+2e^(-), " "E^(@)=0.44 V`
`(2H^(+)+2e^(-)+1//2O_(2)rarrH_(2)O(l), " "E^(@)=+1.23 V)/(Fe(s) +2H^(+)+1//2O_(2)rarrFe^(2+)+H_(2)O,)`
`E_("cell")^(@)=0.44+1.23=1.67 V`
`thereforeDeltaG^(@)="-nF " E_("cell")^(@)=-2xx96500xx1.67=-322 kJ`.
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The half cell reaction for rusting of iron are: 2H^(+)+2e^(-)+(1)/(2)O_(2)rarrH_(2)O(l), E^(@)=+1.23V Fe^(2+)+2e^(-)rarrFe(s), E^(@)=-0.44V DeltaG^(@) (in KJ) for the reaction is

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