Given the half-cell reactions:2H^+ + frac{1}{ | vedclass

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(A) The overall corrosion reaction is: $Fe(s) + 2H^+ + \frac{1}{2}O_2 	o Fe^{2+} + H_2O(l)$.The standard cell potential is calculated as: $E^o_{cell}

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$-322$

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(A) The overall corrosion reaction is: $Fe(s) + 2H^+ + \frac{1}{2}O_2 	o Fe^{2+} + H_2O(l)$.The standard cell potential is calculated as: $E^o_{cell} = E^o_{cathode} - E^o_{anode}$.$E^o_{cell} = 1.23 \ V - (-0.44 \ V) = 1.67 \ V$.The Gibbs free energy change is given by: $\Delta G^o = -nFE^o_{cell}$.Here,$n = 2$ (number of electrons transferred),$F = 96500 \ C/mol$.$\Delta G^o = -2 	imes 96500 	imes 1.67 \ J/mol$.$\Delta G^o = -322310 \ J/mol = -322.31 \ kJ/mol$.Rounding to the nearest integer,the value is $-322 \ kJ$.

Given the half-cell reactions:
$2H^+ + \frac{1}{2}O_2 + 2e^- \to H_2O(l), E^o = 1.23 \ V$
$Fe^{2+} + 2e^- \to Fe(s), E^o = -0.44 \ V$
Calculate the value of $\Delta G^o$ in $kJ$ for the overall corrosion reaction.

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Given the half-cell reactions:$2H^+ + \frac{1}{2}O_2 + 2e^- \to H_2O(l), E^o = 1.23 \ V$$Fe^{2+} + 2e^- \to Fe(s), E^o = -0.44 \ V$Calculate the value of $\Delta G^o$ in $kJ$ for the overall corrosion reaction.