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The half-cell reaction for the corrosion...

The half-cell reaction for the corrosion.
`2H^(+)+1/2O_(2)rarr H_(2)Ol,E^(@)=1.23 V`
`Fe ^(2+)+2e ^(−) →Fe (s);E^(@)` =−0.44V Find the `triangle G^(@)` (in kJ) for the overall reaction:

A

`-76kJ`

B

`-322kJ`

C

`-161kJ`

D

`-152kJ`

Text Solution

Verified by Experts

The correct Answer is:
B
E.P of (cathode - anode) `=E_("cell")= 1.67`
`Delta G^(@)- nFE_(n)= -2xx 96500 xx 1.67 = -322310J= -322kJ`
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