Topic 1 of 7

Redox and half-equations

Balance atoms, charge and electrons before using the data.

A-Level 9476 (2026-2027)

Track electrons before calculating a voltage

Oxidation loses electrons and raises oxidation number; reduction does the reverse.

In a redox reaction, oxidation and reduction occur together. Oxidation is electron loss or an increase in oxidation number. Reduction is electron gain or a decrease in oxidation number. The oxidising agent accepts electrons and is itself reduced; the reducing agent supplies electrons and is itself oxidised.

For Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), zinc changes from 0 to +2 and releases two electrons. Copper changes from +2 to 0 and accepts them. Separating the two half-reactions can make those electrons travel through an external wire, delivering electrical energy.

Balance a half-equation in acidic solution

Balance the changing element
Write its reactant and product species with the correct charge.
Balance oxygen and hydrogen
Add H₂O to balance O, then H⁺ to balance H.
Balance charge
Add electrons to the more positive side; check both atoms and charge.
Combine
Multiply half-equations to cancel electrons, add them, and cancel common species.

Worked example

Combine an oxyanion reduction with peroxide oxidation

Construct the reaction of acidified MnO₄^- with H₂O₂, forming Mn²⁺ and O₂.

Reduction: MnO₄^- + 8H⁺ + 5e^- → Mn²⁺ + 4H₂O.
Oxidation: H₂O₂ → O₂ + 2H⁺ + 2e^-. Oxygen rises from -1 to 0.
Multiply the first by 2 and the second by 5. Cancel ten electrons and ten of the sixteen H⁺ ions.

Answer

2MnO₄^- + 6H⁺ + 5H₂O₂ → 2Mn²⁺ + 8H₂O + 5O₂. All species except water and oxygen are aqueous; water is liquid and oxygen is gas. Both sides have net charge +4.

For a half-equation required in alkaline solution, an acid-balanced equation can be converted by adding enough OH^- to both sides to consume every H⁺, then cancelling water. Use the actual stated product: permanganate, for example, does not necessarily form Mn²⁺ outside acidic conditions.