The rusting of iron takes place as follows2H+ + 2e– + $\dfrac{1}{2}$O2 → H2O(l); E° = +1.23 VFe2+ + 2e– → Fe(s); E° =

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The rusting of iron takes place as follows

2H⁺ + 2e^– + $\dfrac{1}{2}$O₂ → H₂O(l); E° = +1.23 V

Fe²⁺ + 2e^– → Fe(s); E° = –0.44 V

Calculate ΔG° for the net process

–322 kJ mol^–1

–161 kJ mol^–1

–152 kJ mol^–1

–76 kJ mol^–1

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The reduction potential of hydrogen half–cell will be negative if:

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