Balancing Redox Reactions (Half-Reaction Method)

Core Concept

A redox (oxidation–reduction) reaction transfers electrons between species: the substance that loses electrons is oxidized (its oxidation number increases) and the substance that gains electrons is reduced (its oxidation number decreases). Because free electrons never appear in a balanced net equation, redox equations are balanced with the half-reaction method: split the reaction into an oxidation half-reaction and a reduction half-reaction, balance each for atoms and charge (using H⁺, H₂O in acid; then OH⁻ in base), equalize the electron counts, and add the halves together.

Key Ideas

• Oxidation = loss of e⁻ (OIL), reduction = gain of e⁻ (RIG). "OIL RIG" or "LEO says GER."
• Half-reactions separate the transfer. One species is oxidized (e⁻ on the product side), another is reduced (e⁻ on the reactant side).
• Balance O with H₂O, H with H⁺ (acid). These come from the solvent and are always available in aqueous acid.
• Basic solutions: neutralize H⁺ with OH⁻. Add OH⁻ to both sides to convert H⁺ → H₂O, then simplify.
• Electrons must cancel. Multiply each half-reaction so the electrons lost equal the electrons gained.

Equations and Variables

How It Works (acidic conditions)

1. Assign oxidation numbers to find what is oxidized and reduced.
2. Write the two unbalanced half-reactions.
3. Balance every atom except H and O.
4. Balance O by adding H₂O; balance H by adding H⁺.
5. Balance charge by adding electrons (e⁻).
6. Multiply each half-reaction so the electron counts match.
7. Add the halves and cancel species that appear on both sides.
8. (Basic) Add OH⁻ to both sides to cancel H⁺, forming H₂O, and simplify.

Worked Example

Balance MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺ in acidic solution.

Oxidation half-reaction: Fe²⁺ → Fe³⁺ + e⁻ Reduction half-reaction: MnO₄⁻ → Mn²⁺ (Mn goes +7 → +2, gains 5 e⁻)

• Balance O: MnO₄⁻ → Mn²⁺ + 4 H₂O
• Balance H: MnO₄⁻ + 8 H⁺ → Mn²⁺ + 4 H₂O
• Balance charge: MnO₄⁻ + 8 H⁺ + 5 e⁻ → Mn²⁺ + 4 H₂O (left charge +7 −5 = +2; right +2 ✓)

Multiply oxidation by 5: 5 Fe²⁺ → 5 Fe³⁺ + 5 e⁻ Add and cancel electrons:

MnO₄⁻ + 8 H⁺ + 5 Fe²⁺ → Mn²⁺ + 4 H₂O + 5 Fe³⁺

Check: atoms (1 Mn, 4 O, 8 H, 5 Fe on each side ✓) and charge (+7 left = +7 right ✓).

In basic solution, add 8 OH⁻ to both sides to neutralize the 8 H⁺: 8 H⁺ + 8 OH⁻ → 8 H₂O, giving MnO₄⁻ + 4 H₂O + 5 Fe²⁺ → Mn²⁺ + 5 Fe³⁺ + 8 OH⁻ (after canceling the 4 H₂O that appear on both sides).

Common confusions

• "Oxidation means adding oxygen." — Too narrow. Oxidation is loss of electrons (often accompanied by O, but not always).
• "Electrons appear in the final balanced equation." — Wrong. They must cancel between the half-reactions.
• "H⁺ can be added in basic solution freely." — No. In base, you neutralize H⁺ by adding OH⁻ to both sides.
• "Balancing charge is optional." — Wrong. Both atoms and charge must balance.
• "Reduction = loss of electrons." — Wrong. Reduction is the gain of electrons (remember "LEO the lion says GER").

Quick review

• Identify oxidation (e⁻ loss) and reduction (e⁻ gain).
• Split into half-reactions; balance atoms, then O (H₂O), H (H⁺), then charge (e⁻).
• Equalize electrons, add, and cancel.
• In base, add OH⁻ to remove H⁺.
• Verify atoms and charge balance.