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ExamsJEE MainChemistry

During the rusting of iron the relevant half reactions are: 2H+ + (1/2)O2 + 2e⁻ -> H2O, E° = 1.23 V and Fe²+ + 2e⁻ -> Fe, E° = -0.44 V. Calculate the standard Gibbs free energy change (in kJ) for the overall reaction.

  1. -76
  2. -322
  3. -122
  4. -176

Correct answer: -322

Solution

Fe is oxidised (anode) and O2 is reduced (cathode), so the cell emf is 1.23 - (-0.44) = 1.67 V for a 2-electron process. deltaG° = -nFE° = -2 * 96500 * 1.67 ~ -3.22 x 10⁵ J = -322 kJ.

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