1. Oxidation and reduction
| Process | Electron transfer | Oxidation number |
|---|---|---|
| Oxidation | Loss of electrons | Increases |
| Reduction | Gain of electrons | Decreases |
Remember OIL RIG: oxidation is loss, reduction is gain. In a redox reaction, one species loses electrons while another gains them. Electron loss and gain occur together; electrons are transferred, not created or destroyed.
An oxidation number (oxidation state) is a bookkeeping value assigned to an atom. For a monatomic ion it equals its charge. In a covalent substance it need not be the actual charge on the atom. Use the sign: an increase from −2 to 0 is oxidation, even though the magnitude becomes smaller.
2. Oxidation-number rules
| Species or element | Usual oxidation number | Important qualification |
|---|---|---|
| Element in its uncombined form | 0 | Includes O₂, H₂, Cl₂, S₈ and metals |
| Monatomic ion | Its ionic charge | Fe³⁺ is +3; Cl⁻ is −1 |
| Group 1 / Group 2 metals | +1 / +2 | In their compounds |
| Aluminium | +3 | In its usual compounds |
| Fluorine | −1 | In compounds; F₂ itself is 0 |
| Hydrogen | +1 | −1 in metal hydrides such as NaH |
| Oxygen | −2 | −1 in peroxides; positive in compounds with fluorine, e.g. +2 in OF₂ |
| Other halogens | Usually −1 | Can be positive with oxygen or a more electronegative halogen |
The sum of all oxidation numbers, counting every atom, is zero in a neutral compound and equals the overall charge in a polyatomic ion. An oxidation number is assigned per atom. In CaCl₂ each Cl is −1; the two chlorine atoms contribute −2 in total.
Roman numerals can specify an oxidation state: iron(II) contains Fe²⁺, iron(III) contains Fe³⁺. In sulfate(VI), sulfur is +6; that does not mean the sulfate ion has charge +6. SO₄²⁻ has overall charge −2.
3. Worked oxidation numbers
| Species | Calculation | Unknown oxidation number |
|---|---|---|
| H₂SO₄ | 2(+1) + x + 4(−2) = 0 | S: +6 |
| Cr₂O₇²⁻ | 2x + 7(−2) = −2; 2x = 12 | Each Cr: +6 |
| NH₄⁺ | x + 4(+1) = +1 | N: −3 |
| H₂O₂ | 2(+1) + 2x = 0 | Each O: −1 (peroxide) |
| NaH | (+1) + x = 0 | H: −1 (hydride) |
| OF₂ | x + 2(−1) = 0 | O: +2 |
4. Oxidising and reducing agents
| Agent | What it does | What happens to the agent |
|---|---|---|
| Oxidising agent | Accepts electrons from another species | It is reduced; its relevant oxidation number falls |
| Reducing agent | Donates electrons to another species | It is oxidised; its relevant oxidation number rises |
Zn goes from 0 to +2, losing two electrons. Zinc is oxidised and is the reducing agent. Cu²⁺ goes from +2 to 0, accepting those two electrons. Cu²⁺ is reduced and is the oxidising agent.
Identify the actual reacting species, including its charge: for example Cu²⁺, Fe²⁺ or MnO₄⁻. Giving only an element name or an oxidation number may not identify the agent. Read whether the question asks for the atom whose oxidation state changes, the species, or the reagent.
5. Disproportionation and non-redox reactions
Disproportionation occurs when the same element in the same initial oxidation state is both oxidised and reduced. In chlorine reacting with cold, dilute alkali:
Chlorine starts at 0. In Cl⁻ it becomes −1 (reduction); in ClO⁻ it becomes +1 (oxidation), because x − 2 = −1. Show both changes to justify disproportionation.
Not every ionic reaction is redox. In H⁺ + OH⁻ → H₂O, hydrogen stays +1 and oxygen stays −2. In Ag⁺ + Cl⁻ → AgCl, silver stays +1 and chlorine stays −1. Neutralisation and precipitation in these examples involve no oxidation-number change.
Quick checks
Original Finesse questions with indicative worked solutions, not official AQA mark allocations.
Q1. Find the oxidation number of Mn in MnO₄⁻ and N in NO₃⁻.Show answer
Mn: x + 4(−2) = −1, so x = +7. N: x + 3(−2) = −1, so x = +5. Use the ion charge, not zero, for the sum.
Q2. State the oxidation number of oxygen in O₂, H₂O₂ and OF₂.Show answer
O₂: 0, an element. H₂O₂: −1, a peroxide. OF₂: +2, because fluorine is −1 and the total is zero.
Q3. For Cl₂ + 2Br⁻ → 2Cl⁻ + Br₂, identify both agents and justify your answer.Show answer
Cl₂ accepts electrons, chlorine 0 → −1; Cl₂ is the oxidising agent. Br⁻ loses electrons, bromine −1 → 0; Br⁻ is the reducing agent.
Q4. Fe²⁺ becomes Fe³⁺. Is this oxidation or reduction, and how many electrons are transferred per ion?Show answer
Oxidation number increases from +2 to +3, so this is oxidation. One electron is lost per Fe²⁺ ion: Fe²⁺ → Fe³⁺ + e⁻.
Q5. Explain why Cl₂ + 2OH⁻ → Cl⁻ + ClO⁻ + H₂O is disproportionation.Show answer
Chlorine starts at 0 and becomes both −1 in Cl⁻ and +1 in ClO⁻. The same element is reduced and oxidised from the same starting state.
Sources
Sources and examiner guidance (reviewed 1 October 2026)
- Chemrevise — AQA 1.7 Redox revision guide (N. Goalby) — Oxidation states, electron transfer and agents; identify full ionic species rather than excluding ions.
- AQA 7405 specification — 3.1.7 Oxidation, reduction and redox equations — 3.1.7: oxidation, reduction and oxidation states.
- AQA 7404/1 mark scheme, June 2022 — Q06.1 and Q19: electron acceptor and OF₂.
- AQA 7404/1 examiner report, June 2022 — Q06: definitions and identifying the requested species.
- AQA 7404/1 mark scheme, 2021 (November archive; June header) — Q03.1: oxidising-agent definition.
Finesse Tuition is not endorsed by AQA or Chemrevise. All explanations and examples here are our own.