Balancing redox equations

Oxidation number method: write the skeletal equation, find the atoms whose oxidation numbers change, and multiply so that the total increase equals the total decrease.

Then balance the charge (with H⁺ in acid or OH⁻ in base), then hydrogen with water, and finally check oxygen.

Example: dichromate and sulphite in acid give Cr₂O₇²⁻ + 3SO₃²⁻ + 8H⁺ → 2Cr³⁺ + 3SO₄²⁻ + 4H₂O. Each Cr falls by 3 (6 for the pair) and each S rises by 2, so three sulphites are needed.

Half-reaction method: split the reaction into oxidation and reduction halves, balance the atoms other than O and H, then O with H₂O and H with H⁺ (in acid). Add electrons to balance charge, scale the halves to equal electron counts and add them.

Example: Fe²⁺ with dichromate in acid gives 6Fe²⁺ + Cr₂O₇²⁻ + 14H⁺ → 6Fe³⁺ + 2Cr³⁺ + 7H₂O, since each Fe²⁺ gives one electron and one Cr₂O₇²⁻ takes six.

In basic medium, balance as if acidic and then add as many OH⁻ to both sides as there are H⁺, combining H⁺ + OH⁻ into water. Permanganate and iodide in base give 2MnO₄⁻ + 6I⁻ + 4H₂O → 2MnO₂ + 3I₂ + 8OH⁻.

A balanced redox equation must balance atoms of every element and the total charge on both sides.