At GCSE (Higher tier) you met OIL RIG — oxidation is loss, reduction is gain, of electrons — in displacement and electrolysis. A-Level keeps that definition and adds three tools: oxidation states to track electrons through any formula, the language of oxidising and reducing agents, and a routine for building half-equations and combining them into full redox equations — even for unfamiliar species.
Oxidation & reduction (electron transfer)
At A-Level, oxidation and reduction are defined by electrons — the oxygen definition from GCSE is left behind.
Oxidation is the loss of electrons. Reduction is the gain of electrons. An oxidising agent is an electron acceptor; a reducing agent is an electron donor.
The trap is the word “agent”. An oxidising agent does the oxidising to something else — and to do that, it takes the electrons, so it is itself reduced. The two things cross over, every time:
| This species… | …does its electrons… | …so it is… | …and it acts as the… |
|---|---|---|---|
| loses electrons | gives them away | oxidised | reducing agent (it reduces the other species) |
| gains electrons | takes them | reduced | oxidising agent (it oxidises the other species) |
- An oxidising agent is itself reduced. Examiner reports flag students confusing “is oxidised” with “is the oxidising agent” — they are opposites. The oxidising agent gains the electrons.
- Say what is oxidised, and what the agent does. Not “chlorine oxidises” on its own — state what it oxidises and that chlorine is itself reduced. Sentence frame: “X oxidises Y by removing electrons from Y, and X is reduced.”
- Mention electron transfer when the electron story is tested — an oxidation-state change on its own is not always the full answer.
GCSE groundwork: C4 — displacement & OIL RIG.
🧪 Exam-style questions
Which statement about this redox reaction is correct? 3Sn2+(aq) + Cr2O72−(aq) + 2H+(aq) → 2Cr3+(aq) + 3SnO2(s) + H2O(l) Tick (✓) one box.
State, in terms of electrons, the meaning of the term oxidising agent.
Show answer
An electron acceptor. 1 mark
Do not accept “electron pair acceptor” (that is a Lewis acid) or the vaguer “gains electrons” when the mark scheme wants the noun “electron acceptor”.
Source: AQA A-Level Chemistry past papers.
Oxidation states
An oxidation state (or oxidation number) is a bookkeeping figure for the electrons on an atom. Track how it changes and you can spot oxidation and reduction in any reaction — even one with unfamiliar species. A small set of rules assigns it.
Find the oxidation state of chromium in the dichromate ion, Cr2O72−.
Step 1 — put in the states you know: each O is −2, and there are 7 of them: 7 × (−2) = −14.
Step 2 — the states sum to the overall charge (−2):
2 × Cr + (−14) = −2 → 2 × Cr = +12 → Cr = +6
Divide by the number of atoms at the end — the oxidation state is per atom, so it is +6 for each chromium, not +12.
Calcium reacts with water: Ca + 2H2O → Ca(OH)2 + H2. What is oxidised and what is reduced?
Calcium: 0 → +2 — the state rises, so calcium is oxidised.
Hydrogen (the two that become H2): +1 → 0 — the state falls, so that hydrogen is reduced.
A rise in oxidation state is oxidation; a fall is reduction. Quote the numbers — “from 0 to +2” — not just “it goes up”.
Interactive — work out an oxidation state
Pick a formula
- The exceptions win. O is −2 except in peroxides (−1) and with fluorine (+2); H is +1 except in metal hydrides (−1); F is always −1.
- Per atom. When several atoms of an element share a total, divide at the end — the state is quoted for one atom.
- Rise = oxidation, fall = reduction — and quote the actual numbers when asked to explain.
🧪 Exam-style questions
Which of these oxidation states is correct? Tick (✓) one box.
In which substance is oxygen in the highest oxidation state? Tick (✓) one box.
In which species is chlorine in its highest oxidation state? Tick (✓) one box.
The reaction of calcium with water is a redox reaction: Ca + 2H2O → Ca(OH)2 + H2. Explain, in terms of oxidation states, why this reaction involves both oxidation and reduction.
Show answer
The oxidation state of calcium goes from 0 to +2, so calcium is oxidised. 1 mark
The oxidation state of hydrogen goes from +1 to 0, so hydrogen is reduced. 1 mark
Source: AQA A-Level Chemistry past papers.
Half-equations
A half-equation shows just the oxidation, or just the reduction, complete with the electrons that make it balance. You met these at the electrodes in GCSE electrolysis (Pb2+ + 2e− → Pb); A-Level extends them to reactions in acidic solution, where water and H+ join in.
There is a routine that works every time, in this order:
In acid, Cr2O72− oxidises SO32−, forming Cr3+ and SO42−.
Oxidation (SO32− → SO42−): S is balanced; add 1 H2O on the left for the extra O; add 2 H+ on the right for its H; add 2 e− on the right for the charge:
SO32− + H2O → SO42− + 2H+ + 2e−
Reduction (Cr2O72− → Cr3+): balance 2 Cr; add 7 H2O on the right for the O; add 14 H+ on the left for its H; add 6 e− on the left for the charge:
Cr2O72− + 14H+ + 6e− → 2Cr3+ + 7H2O
Check both: atoms balance, and the charge is the same on each side (−2 for the oxidation, +6 for the reduction). The electrons are the last thing in — if they are missing, you have not finished.
Interactive — build the half-equation, step by step
Pick a half-reaction
- Electrons on the correct side. Examiner reports flag electrons missing or on the wrong side — reduction gains them (electrons on the left), oxidation loses them (electrons on the right).
- No electrons? Not finished. A half-equation without electrons is almost always incomplete.
- Keep the order: main atom, then O with H2O, then H with H+, then charge with e−. Doing charge before hydrogen is where it goes wrong.
🧪 Exam-style questions
Which incomplete half-equation is balanced by adding two H+ ions and one electron to the left-hand side? Tick (✓) one box.
In acid, Cr2O72− oxidises SO32− to SO42−, forming Cr3+. Deduce the half-equation for the oxidation of SO32−, the half-equation for the reduction of Cr2O72−, and the overall equation.
Show answer
Oxidation: SO32− + H2O → SO42− + 2H+ + 2e− 1 mark
Reduction: Cr2O72− + 14H+ + 6e− → 2Cr3+ + 7H2O 1 mark
Overall: Cr2O72− + 3SO32− + 8H+ → 2Cr3+ + 3SO42− + 4H2O 1 mark
Scale the oxidation by 3 so its 6 electrons match the reduction’s 6, add, then cancel the electrons and tidy the H+/H2O.
Source: AQA A-Level Chemistry past papers.
Combining half-equations
Two half-equations combine into the full redox equation once the electrons match. The electrons lost by one species must exactly equal the electrons gained by the other — so scale each half-equation until they do, then add and tidy up.
Combine the oxidation of copper with the reduction of nitrate ions.
Step 1 — the two half-equations:
Cu → Cu2+ + 2e− (loses 2 electrons)
NO3− + 2H+ + e− → NO2 + H2O (gains 1 electron)
Step 2 — scale so the electrons match (×2 the reduction):
2NO3− + 4H+ + 2e− → 2NO2 + 2H2O
Step 3 — add and cancel the electrons:
Cu + 2NO3− + 4H+ → Cu2+ + 2NO2 + 2H2O
Two electrons on each side cancel exactly. Finish by checking the charge: left = 0 (Cu) + 2(−1) (nitrate) + 4(+1) (H+) = +2; right = +2 (Cu2+) + 0 + 0 = +2. Balanced for atoms and charge.
Interactive — combine the half-equations
Set each multiplier so the electrons cancel, then combine.
- Match the electrons first. Multiply each half-equation up to the lowest common number of electrons so they cancel exactly — this is where most overall equations go wrong.
- Cancel duplicated species. The electrons must disappear; then tidy any H+ or H2O that appears on both sides.
- Check atoms and charge on the final equation — examiner reports flag uncancelled species and wrong ionic forms.
🧪 Exam-style questions
Give a half-equation for the oxidation of copper to copper(II) ions, a half-equation for the reduction of NO3− ions in acidic solution to NO2, and hence deduce an overall equation for the reduction of NO3− by copper.
Show answer
Oxidation: Cu → Cu2+ + 2e− 1 mark
Reduction: NO3− + 2H+ + e− → NO2 + H2O 1 mark
Overall: Cu + 2NO3− + 4H+ → Cu2+ + 2NO2 + 2H2O 1 mark
Ammonia is oxidised as shown: w NH3 + x O2 → y H2O + z NO. Which whole-number values for w, x, y and z balance the equation? Tick (✓) one box.
Source: AQA A-Level Chemistry past papers.
Capstone: analysing a redox reaction
Put the whole topic together on one reaction. The classic redox-titration reaction — acidified manganate(VII) with iron(II) — runs through every skill: assign oxidation states, name what is oxidised and reduced and both agents, split into half-equations, and combine.
MnO4− + 8H+ + 5Fe2+ → Mn2+ + 4H2O + 5Fe3+
| Species | Oxidation state change | What happens | Its role |
|---|---|---|---|
| Mn (in MnO4−) | +7 → +2 (falls) | gains electrons — reduced | the oxidising agent |
| Fe (as Fe2+) | +2 → +3 (rises) | loses electrons — oxidised | the reducing agent |
The half-equations behind it: reduction MnO4− + 8H+ + 5e− → Mn2+ + 4H2O, and oxidation Fe2+ → Fe3+ + e−. Scaling the iron by 5 matches the five electrons manganese needs, and adding them gives the overall equation above. That is the entire topic in one worked analysis: states tell you who changed, electrons tell you the equation, and the agent is always the species that changed the other way.
Interactive — who does what to whom?
🧪 Exam-style questions
Which equation does not show the reduction of a transition metal? Tick (✓) one box.
Source: AQA A-Level Chemistry past papers.
- Oxidation = loss of electrons; reduction = gain of electrons (OIL RIG).
- Oxidising agent = electron acceptor (itself reduced); reducing agent = electron donor (itself oxidised). The agent and what happens to it are opposites.
- Oxidation states: element = 0; simple ion = its charge; the states sum to the overall charge. Group 1 = +1, Group 2 = +2, Al = +3.
- Exceptions: H = +1 (but −1 in metal hydrides); O = −2 (but −1 in peroxides, +2 in OF2/F2O); F = −1 always. A rise in oxidation state is oxidation; a fall is reduction.
- Half-equation routine: (1) balance the atom; (2) balance O with H2O; (3) balance H with H+; (4) balance the charge with e−. If there are no electrons, it is not finished.
- Reduction gains electrons (on the left); oxidation loses them (on the right).
- Combining: multiply the half-equations so the electrons cancel, add them, then cancel any species (e−, spare H+/H2O) on both sides; check atoms and charge balance.