The half cell reactions for the corrosion are, 2H++12O2+2e−→H2O;E0=1.23V and Fe2++2e−→Fe(s);E0=−0.44V. Find the ∆G0 (in kJ) for the overall reaction

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The half cell reactions for the corrosion are, $2 H^{+} + \frac{1}{2} O_{2} + 2 e^{-} \to H_{2} O; E^{0} = 1.23 V$ and ${Fe}^{2 +} + 2 e^{-} \to Fe (s); E^{0} = - 0.44 V$ . Find the $∆$ G⁰ (in kJ) for the overall reaction

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-76 K

-161 kJ

-152 kJ

-322 kJ

answer is B.

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Detailed Solution

$\begin{array}{l} E_{cell}^{0} = (E_{high SRP}^{0} - E_{low SRP}^{0}) \\ = [1.23 - (- 0.44)] = 1.67 V \\ ΔG = - 2 \times 96500 \times 1.67 = - 322.3 KJ \end{array}$