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 the agent does with electrons. In terms of electrons, the role of an oxidising agent is “an electron acceptor”; “chlorine accepts (gains) electrons” scores the same. Not “chlorine oxidises” on its own. For a fuller explanation, the frame “X oxidises Y by removing electrons from Y, and X is reduced” says what is oxidised as well.
- 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 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
“Gains electrons” or “accepts electrons” also scores. Do not accept “electron pair acceptor” (that is a Lewis acid) or “gain of electrons”, which is the definition of reduction rather than a description of the agent.
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 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 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.
In acidic solution, IO3− ions oxidise iodide ions to iodine: IO3− + 5I− + 6H+ → 3I2 + 3H2O. Give a half-equation for the oxidation of iodide ions to iodine. Deduce the half-equation to show the reduction process in this reaction.
Show answer
Oxidation: 2I− → I2 + 2e− 1 mark
Reduction: 2IO3− + 12H+ + 10e− → I2 + 6H2O 1 mark
Iodine is +5 in IO3− and 0 in I2, so each iodine gains 5 electrons, 10 for the two that build one I2. AQA allows multiples of the oxidation half-equation, and awards a mark if the two equations are written the other way round, but writing the two the wrong way round scores only 1 mark.
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 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 questions
Which equation does not show the reduction of a transition metal? Tick (✓) one box.
In which conversion is the metal reduced? Tick (✓) one box.
Which compound contains chlorine in an oxidation state of +1? Tick (✓) one box.
NO2− ions can be reduced in acidic solution to NO. How many electrons are gained when each NO2− ion is reduced? Tick (✓) one box.
Which compound contains a chlorine atom with an oxidation state of +4? Tick (✓) one box.
In which oxide is the named element in its highest oxidation state? 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.