Balancing (the Equation) of Redox Reactions

Balancing a redox equation by simple inspection quickly becomes impractical once several elements change oxidation state at once. Two systematic methods exist, both resting on the same underlying principle: in any redox reaction, the total number of electrons donated by the reducing agent must exactly equal the total number of electrons gained by the oxidising agent.

Method 1 -- the oxidation number method (worked in full as Example 1.12, for FeSO₄ + KMnO₄ + H₂SO₄ → Fe₂(SO₄)₃ + MnSO₄ + K₂SO₄ + H₂O):

1. Using oxidation numbers, identify which atoms are oxidised and which are reduced: Mn goes from +7 (in KMnO₄) to +2 (in MnSO₄) -- a gain of 5 electrons (reduction); Fe goes from +2 (in FeSO₄) to +3 (in Fe₂(SO₄)₃) -- a loss of 1 electron (oxidation).
2. Equalise electrons lost and gained by cross-multiplying with suitable integers. Since the product Fe₂(SO₄)₃ contains 2 iron atoms, the "1 electron lost" and "5 electrons gained" steps are both scaled by 2, giving 10 FeSO₄ and 2 KMnO₄.
3. Balance the oxidised/reduced species on the product side to match (5 Fe₂(SO₄)₃, 2 MnSO₄).
4. Balance every other element except H and O. Here, the sulphate (S) count on the product side (5×3 + 2 + 1 = 18 S atoms) exceeds the 10 S atoms so far on the reactant side (from 10 FeSO₄), so 8 more must come from H₂SO₄.
5. Finally balance H and O: 8 H₂SO₄ contributes 16 H atoms, which requires 8 H₂O on the product side; oxygen then balances automatically.

This gives the fully balanced equation: 10FeSO₄ + 2KMnO₄ + 8H₂SO₄ → 5Fe₂(SO₄)₃ + 2MnSO₄ + K₂SO₄ + 8H₂O.

Method 2 -- the ion-electron (half-reaction) method, used specifically for ionic redox reactions (worked in full as Example 1.13, for the same overall reaction written in ionic form, MnO₄⁻ + Fe²⁺ + H⁺ → Mn²⁺ + Fe³⁺ + H₂O):

1. Using oxidation numbers, identify the species oxidised and reduced, exactly as before.
2. Write two separate half-equations, one for oxidation and one for reduction: Fe²⁺ → Fe³⁺ + e⁻ (oxidation), and MnO₄⁻ + 5e⁻ → Mn²⁺ (reduction, unbalanced).
3. Balance the atoms and charges in each half-equation. The oxidation half needs no change. The reduction half has 4 oxygen atoms on the reactant side that must appear as 4H₂O on the product side, which in turn requires 8H⁺ added to the reactant side to balance the hydrogen: MnO₄⁻ + 5e⁻ + 8H⁺ → Mn²⁺ + 4H₂O. …