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9476 / 2027
Electrochemistry

Lesson 1 of 7

Redox and half-equations

Balance atoms, charge and electrons before using the data.

A-Level 9476 (2026-2027)

  • I can explain redox using electrons and oxidation numbers.Syllabus

    Syllabus 9476, 12(a). Explain redox using electrons and oxidation numbers.

  • I can construct overall redox equations from half-equations.Syllabus

    Syllabus 9476, 12(h). Construct overall redox equations from half-equations.

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
  1. Balance the changing element

    Write its reactant and product species with the correct charge.

  2. Balance oxygen and hydrogen

    Add H2O to balance O, then H+ to balance H.

  3. Balance charge

    Add electrons to the more positive side; check both atoms and charge.

  4. Combine

    Multiply half-equations to cancel electrons, add them, and cancel common species.

Worked exampleCombine an oxyanion reduction with peroxide oxidation

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

Try it first. Then show the worked stepsHide the worked steps
  1. Reduction: MnO4- + 8H+ + 5e- → Mn2+ + 4H2O.
  2. Oxidation: H2O2 → O2 + 2H+ + 2e-. Oxygen rises from -1 to 0.
  3. 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.

Check your understanding

In Zn(s) + Cu2+(aq) -> Zn2+(aq) + Cu(s), which species is the reducing agent?

  1. Cu, because it is the metal formed at the end
  2. Cu2+, because it is reduced to copper metal
  3. Zn, because it gives electrons to Cu2+ and is oxidised
Show the answer

Zn, because it gives electrons to Cu2+ and is oxidised

Zinc goes from 0 to +2, losing two electrons. The reducing agent supplies electrons and is itself oxidised.