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. `-76kJ`
B. `-322kJ`
C. `-161J`
D. `-152kJ`

Answer

B

Explanation

`Fe(s) to Fe^(2+)+2e^(-) " "triangleG^(1)^(@)`
`2H^(+)+2e^(-)+1/2O_(2) to H_(2)O(l), " "triangleG_(2)^(0)`
`=Fe(s)+2H^(+)+1/2O_(2) to Fe^(2+) +H_(2)O" "triangleG_(3)^(@)`
Applying `triangleG_(1)^(@)+triangleG_(2)^@=triangleG_(3)^@`
`triangleG_(3)^@=(-2F xx 0.44)+(-2F xx 1.23)`
`=(-2 xx 96500xx 0.44)+(-2 xx 96500xx 1.23)`
=322310J=-322 kJ