The half-cell reaction for the corrosion, `{:(2H^(+)+,(1)/(2)O_(2)+2e^(-) rarr H_(2)O,,,E_(@) = 123 V),(,Fe^(2+)+2e^(-) rarrFe(s),,,E^(@) = -0.44 V):}`
Find the `Delya G^(@)` (in kJ) for the overall reaction :

Options

A. `-76`
B. `-322`
C. `-161`
D. `152`

Answer

B

Explanation

Applying `DeltaG^(@)=-nFE^(@)`
`Fe(s) to Fe^(2+) + 2e^(-) , DeltaG_(1)^(@)`
`2H^(+) + 2e^(-) + 1/2 O_(2) to H_(2)O_((l)) , DeltaG_(2)^(@)`
`Fe_((s)) + 2H^(+) + 1/2 O_(2) to Fe^(2+) + H_(2)O , DeltaG_(3)^(@)`
Applying `DeltaG_(1)^(@) + DeltaG_(2)^(@) = DeltaG_(3)^(@)`
`DeltaG_(3)^(@) =(-2Fxx0.44)+(-2Fxx1.23)`
`=-(2xx96500xx0.44+2 xx 96500 xx1.23)`
`=-322310 J`
`therefore DeltaG_(3)^(@)=-322 KJ`