Balancing redox equations
Redox equations do not balance by inspection, and guessing wastes time you do not have. The half-reaction method is a fixed procedure: split, balance each half completely, scale until the electrons cancel, add. Here it is, including the awkward cases.
Why redox equations need a procedure
Most equations balance by inspection: count atoms, adjust coefficients, done. Redox equations often will not, because the charges have to balance too, and in solution the missing pieces — water and hydrogen ions — are not written in the question. Guessing coefficients can take ten minutes and still fail.
The half-reaction method works because it separates the two things that are happening. One substance loses electrons and another gains them; balancing each half on its own, then scaling until the electrons cancel, turns a puzzle into a procedure that always terminates.
It is also the same skeleton as the rest of the unit. A balanced half-reaction is what a cell diagram, a cell potential and a Faraday calculation are all built from — get the half-reaction right and the rest of the question usually follows.
Where it turns up
- Batteries — every cell, from a phone battery to a car battery, is two half-reactions and the electrons moving between them.
- Corrosion — rust is iron being oxidized, and protecting a bridge or a pipeline means arranging for something else to be oxidized instead.
- Water treatment and bleaching — chlorine and permanganate work by oxidizing what they are cleaning — dosing them means balancing the reaction.
- Breathalyzers — the classic design uses an orange dichromate solution that turns green as alcohol reduces it — a colour change that is literally a half-reaction.
- Why it matters
- 1. The procedure
- 2. Worked in acidic solution
- 3. Finishing in basic solution
- 4. When the half-reaction is not in the table
The procedure
The idea: balance atoms first, charge last, and never mix the two halves until the electrons match.
- Assign oxidation numbers and identify what was oxidized and what was reduced.
- Write two half-reactions, one for each.
- Balance the main element in each half.
- Balance oxygen with H₂O — one water per missing oxygen.
- Balance hydrogen with H⁺.
- Balance charge with electrons, added to the more positive side.
- Scale both halves so electrons lost equal electrons gained.
- Add and cancel anything that appears identically on both sides.
- In basic solution, neutralize every H⁺ with OH⁻ on both sides, then simplify the water.
Check: atoms balance, charges balance, and no electrons remain in the final equation. All three must hold.
Worked in acidic solution
The idea: one full example, every step shown, using the permanganate reaction that turns up constantly.
MnO₄⁻(aq) + Fe²⁺(aq) → Mn²⁺(aq) + Fe³⁺(aq)
Oxidation numbers: Mn goes +7 → +2 (reduced); Fe goes +2 → +3 (oxidized).
Reduction half:
- MnO₄⁻ → Mn²⁺
- Oxygen: MnO₄⁻ → Mn²⁺ + 4H₂O
- Hydrogen: MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O
- Charge: left is +7, right is +2, so add 5 electrons to the left
- MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
Oxidation half: Fe²⁺ → Fe³⁺ + e⁻
Scale: multiply the iron half by 5 so both halves move 5 electrons.
Add: MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺
Check: charge is +7 + 10 = +17 on the left and +2 + 15 = +17 on the right. ✓
Finishing in basic solution
The idea: balance as though it were acidic, then convert. Do not try to balance with OH⁻ from the start.
Converting the last step
Suppose a balanced acidic half-reaction contains 8H⁺ on the left.
- Add 8 OH⁻ to both sides. The left now has 8H⁺ + 8OH⁻, which is 8H₂O.
- Simplify: if the right already had 4H₂O, cancel four from each side, leaving 4H₂O on the left and 8OH⁻ on the right.
Nothing else changes — the electrons and the main species are already correct, which is why converting last is so much less work than starting over.
When the half-reaction is not in the table
The idea: Alberta reports that students manage when both half-reactions appear in the data booklet and struggle when one does not. That case is the one worth practising.
- Build it yourself with the nine steps above. The table is a convenience, not a requirement.
- Work from oxidation numbers: the change per atom, multiplied by the number of atoms, is the number of electrons.
- Skeleton equations — where H⁺ and H₂O are not shown — are the exam's way of testing exactly this. Add them yourself in steps 4 and 5.
- Disproportionation: when one substance is both oxidized and reduced, write two half-reactions for the same species and proceed normally.
Worked: an unlisted half-reaction
C₂O₄²⁻ → CO₂
Carbon is +3 in oxalate and +4 in carbon dioxide, and there are two carbons, so two electrons are lost in total:
C₂O₄²⁻ → 2CO₂ + 2e⁻
Charge: −2 on the left, 0 + (−2) on the right. ✓ No table needed.