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):}

Question

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):}

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

Monjur Alahi

5 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Correct Answer - B
`{:(" "Fe(s) rarr Fe^(2+)+2e^(-),,Delta G_(1)^(@)),(2H^(+)+2e^(-)+(1)/(2)O_(2) rarr H_(2)O(l), ,Delta G_(2)^(@)):}/(ulbar(Fe(s)+2H^(+)+(1)/(2)O_(2) rarrFe^(2+)+H_(2)O,Delta G_(3)^(@)))`
Applying `Delta G_(1)^(@)+Delta G_(2)^(@) = Delta G_(3)^(@)`
`Delta G_(3)^(@) = (-2F xx 0.44)+(-2F xx 1.23)`
`= (-2 xx 96500 xx 0.44) + (-2 xx 96500 xx 1.23)`
`= - 322310 J = -322 kJ`