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The rusting of iron takes place as follo...

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

A

`-322kJ mol^(-10`

B

`-161kJmol^(-1)`

C

`-152kJ mol^(-1)`

D

`-76kJ mol^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
a
`E^(c-)._(cell)=1.23-(-0.44)=1.67V`
`impliesDeltaG^(c-)._(cell)=nFE^(c-)._(cell)=-2xx96500xx1.67 J mol^(-1)`
`=-332.3kJ mol ^(-1)`
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