Q1(e). Spontaneity of Fe²⁺ oxidation

Answer in brief

Yes, under standard conditions with H₂O₂ as oxidant: E°cell = +1.00 V and ΔG° = −193 kJ mol⁻¹ of reaction.

Key points

• Equation (i) is written as an oxidation, so its supplied potential is −0.77 V. Do not reverse its sign a second time.
• Add the half-reactions: 2Fe²⁺ + H₂O₂ + 2H⁺ → 2Fe³⁺ + 2H₂O.
• E°cell = E°oxidation + E°reduction = −0.77 + 1.77 = +1.00 V. Electrode potentials are not multiplied by stoichiometric coefficients.
• ΔG° = −nFE° = −2 × 96485 × 1.00 = −192970 J mol⁻¹ ≈ −193 kJ mol⁻¹.
• A positive cell potential and negative ΔG° establish standard-state spontaneity; they do not specify the reaction rate.