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Balancing Redox by Half-Reactions

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Why split a redox reaction in half?

In a redox reaction electrons move from the species being oxidized to the species being reduced. Trying to balance the whole thing at once is error-prone, so we separate the electron donor and the electron acceptor into two half-reactions — one oxidation, one reduction — balance each independently, then recombine them so the electrons lost exactly equal the electrons gained. First assign oxidation states to spot who changes: the atom whose number rises is oxidized, the atom whose number falls is reduced.

The half-reaction algorithm (acidic solution)

For each half-reaction, work in a fixed order: (1) balance every atom except O and H; (2) balance O by adding H₂O to the deficient side; (3) balance H by adding H⁺; (4) balance charge by adding electrons (e⁻) to the more-positive side. Then scale the two half-reactions by whole numbers so the electron counts match, add them, and cancel anything common to both sides (electrons, and often some H⁺ or H₂O). The final equation must balance for both mass and charge.

Adapting to basic solution

H⁺ cannot exist in appreciable amount in a basic solution, so balance the reaction as if it were acidic first, then neutralize. Count the H⁺ in the finished equation and add that same number of OH⁻ to both sides. On the side with H⁺, each H⁺ + OH⁻ combines into H₂O; on the other side the OH⁻ stays free. Finally cancel any H₂O that now appears on both sides. The result contains OH⁻ instead of H⁺ — exactly what a basic medium requires.

Half-reaction balancing checklist
atoms (≠ O,H) → O with H₂O → H with H⁺ → charge with e⁻ → scale & add → (basic: + OH⁻ both sides, combine to H₂O, cancel)
Electrons always land on the more-positive side of a half-reaction: the reactant side for reduction, the product side for oxidation.
Worked example

Balance the reduction half-reaction MnO₄⁻ → Mn²⁺ in acidic solution.

  1. 1.Balance atoms other than O and H: Mn is already 1 on each side.
  2. 2.Balance O by adding H₂O: the left has 4 O, so add 4 H₂O to the right → MnO₄⁻ → Mn²⁺ + 4 H₂O.
  3. 3.Balance H by adding H⁺: the right now has 8 H, so add 8 H⁺ to the left → MnO₄⁻ + 8 H⁺ → Mn²⁺ + 4 H₂O.
  4. 4.Balance charge with e⁻: left charge = (−1) + 8(+1) = +7; right charge = +2. Add 5 e⁻ to the left to drop +7 down to +2 → MnO₄⁻ + 8 H⁺ + 5 e⁻ → Mn²⁺ + 4 H₂O.
  5. 5.Verify: atoms Mn 1=1, O 4=4, H 8=8; charge left = +7 − 5 = +2, right = +2. Balanced for mass and charge.
Answer: MnO₄⁻ + 8 H⁺ + 5 e⁻ → Mn²⁺ + 4 H₂O
Worked example

Combine that reduction with the oxidation Fe²⁺ → Fe³⁺ to give the full balanced equation in acid.

  1. 1.Oxidation half-reaction, balanced for charge: Fe²⁺ → Fe³⁺ + e⁻ (loses 1 electron).
  2. 2.Match electrons: reduction uses 5 e⁻, oxidation releases 1 e⁻, so multiply the oxidation by 5 → 5 Fe²⁺ → 5 Fe³⁺ + 5 e⁻.
  3. 3.Add the two half-reactions and cancel the 5 e⁻ common to both sides → MnO₄⁻ + 8 H⁺ + 5 Fe²⁺ → Mn²⁺ + 4 H₂O + 5 Fe³⁺.
  4. 4.Check charge: left = (−1) + 8 + (5)(+2) = +17; right = (+2) + 0 + (5)(+3) = +17. Atoms and charge both balance.
Answer: MnO₄⁻ + 8 H⁺ + 5 Fe²⁺ → Mn²⁺ + 4 H₂O + 5 Fe³⁺
Tip

The electron count in a reduction half-reaction equals the total drop in oxidation number. Mn goes from +7 in MnO₄⁻ to +2, a drop of 5 — so exactly 5 e⁻ appear. Use this as an independent check on step 4.

Checkpoint

The dichromate half-reaction in acid is Cr₂O₇²⁻ + 14 H⁺ + ? e⁻ → 2 Cr³⁺ + 7 H₂O. How many electrons are needed, and on which side?

Checkpoint

The half-reaction ClO⁻ + 2 H⁺ + 2 e⁻ → Cl⁻ + H₂O is balanced in acid. What is the correct form in basic solution?

On the exam

On the AP exam a redox equation earns full credit only when it is balanced for mass and charge. Always end by summing the charges on each side — if they are not equal, you mis-scaled the electrons or lost an H⁺/H₂O when combining the halves.

Answer the 2 checkpoints as you read.

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