HookThe roadside test that ran on stolen electrons
On 8 October 1967, Barbara Castle's Road Safety Act put a new object in every patrol car: the breathalyser. The original device was a glass tube of orange crystals — potassium dichromate(VI) on silica, acidified — and a bag. If a motorist's breath carried enough ethanol, the crystals turned green in front of the officer. Road deaths fell by roughly 1,100 in the first year. The colour change that did it is an oxidation-state change: chromium in \(\text{Cr}_2\text{O}_7^{2-}\) sits at +6 (orange) and is dragged down to +3 (green \(\text{Cr}^{3+}\)) as it strips electrons from ethanol, which is oxidised towards ethanoic acid. A drink-driving conviction, decided by electron transfer.
Redox is the accountancy of that transfer. Oxidation states are the ledger — a number pinned to every atom so you can see exactly where electrons went. Half-equations are the itemised receipts, showing each loss and gain separately, and the full redox equation is the settled invoice, electrons cancelled to nothing. Master this small section properly and half the course opens up: halide displacement in 3.2.3, manganate(VII) titrations in 3.2.5.5 and the whole of electrode potentials in 3.1.11 are this machinery, reused.
ModelOxidation states — the electron ledger
An oxidation state is the charge an atom would carry if every bond in the species were treated as fully ionic. It is a bookkeeping fiction — most bonds are nowhere near fully ionic — but the fiction makes electron transfer countable. The rules, in priority order: any uncombined element is 0 (that covers \(\text{Na}\), \(\text{O}_2\) and \(\text{Cl}_2\) alike); fluorine is always −1; Group 1 metals are +1 and Group 2 are +2 in their compounds; hydrogen is +1, except in metal hydrides such as \(\text{NaH}\), where it is −1; oxygen is −2, except in peroxides (−1) and in \(\text{OF}_2\) (+2, because fluorine outranks it).
The final rule does the actual work: the oxidation states in a species must sum to its total charge — zero for a neutral compound, the ion's charge otherwise. So you treat the unknown element as \(x\) and solve a one-line equation. Two conventions matter to the mark scheme: the state belongs to one atom, not the whole formula unit, and the sign is written first — +6, never 6+ or a bare 6.
Three assignments you should manage in under a minute, then the sneaky fourth. Chromium in \(\text{Cr}_2\text{O}_7^{2-}\): seven oxygens give \(7 \times (-2) = -14\), and the whole ion is −2, so \(2x - 14 = -2\), which gives \(x = +6\) per chromium. Sulfur in \(\text{SO}_4^{2-}\): \(x + 4(-2) = -2\), so \(x = +6\). Nitrogen in \(\text{NH}_4^{+}\): hydrogen counts +1 here, so \(x + 4(+1) = +1\) and \(x = -3\). Now chlorine in household bleach, \(\text{NaClO}\): sodium +1 and oxygen −2 force \(+1 + x - 2 = 0\), so chlorine sits at +1 — which is exactly why the ion is named chlorate(I). The Roman numeral in a compound's name is an oxidation state, so you can often read states straight off the name before touching any algebra.
ModelOxidation is loss — and the agents work in mirrors
At A-level, retire the GCSE idea that oxidation means gaining oxygen. Oxidation is loss of electrons, shown by a rise in oxidation state; reduction is gain of electrons, shown by a fall. OIL RIG still holds — but the mark scheme pays for electrons and oxidation states, not oxygen. The proof that oxygen is optional: \(\text{Cl}_2 + 2\text{Br}^- \rightarrow 2\text{Cl}^- + \text{Br}_2\) is a complete redox reaction with not one oxygen atom in it. Bromine climbs from −1 to 0 (oxidised); chlorine falls from 0 to −1 (reduced).
The agents are mirror twins, and students swap them constantly. An oxidising agent accepts electrons — it causes oxidation in something else and is itself reduced. A reducing agent donates electrons and is itself oxidised. Every redox reaction has exactly one of each, and naming one means quoting the whole species: \(\text{Cr}_2\text{O}_7^{2-}\) is the oxidising agent in the breathalyser, not "chromium". In that reaction the average oxidation state of ethanol's carbons climbs from −2 (in \(\text{C}_2\text{H}_5\text{OH}\)) to 0 (in \(\text{CH}_3\text{COOH}\)) — a four-electron surrender per molecule — so ethanol is the reducing agent. The same logic runs the railways: Network Rail joins track by aluminothermic (thermite) welding, \(\text{Fe}_2\text{O}_3 + 2\text{Al} \rightarrow \text{Al}_2\text{O}_3 + 2\text{Fe}\), which pours out iron at around 2,500 °C. Aluminium (0 to +3) is the reducing agent; iron(III) oxide, whose iron drops from +3 to 0, is the oxidising agent.
MechanismHalf-equations — the four-step build
A half-equation shows one half of the electron transfer — the oxidation alone, or the reduction alone — with the electrons written explicitly. AQA works in acidic solution, which hands you exactly three balancing tools: \(\text{H}_2\text{O}\), \(\text{H}^+\) and \(\text{e}^-\). The build order never changes. Step 1: balance the central element. Step 2: balance oxygen by adding \(\text{H}_2\text{O}\). Step 3: balance hydrogen by adding \(\text{H}^+\). Step 4: balance the charge by adding electrons to the more positive side.
Where the electrons land is itself a check: electrons on the left means reduction; electrons on the right means oxidation — no memorising required, the charge audit forces it. Simple oxidations need only step 4: \(\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + \text{e}^-\). And the method scales to organic species without modification — the ethanol half of the breathalyser reaction is \(\text{CH}_3\text{CH}_2\text{OH} + \text{H}_2\text{O} \rightarrow \text{CH}_3\text{COOH} + 4\text{H}^+ + 4\text{e}^-\), built with the same four steps.
Build the dichromate(VI) reduction that turned the breathalyser green. Skeleton: \(\text{Cr}_2\text{O}_7^{2-} \rightarrow \text{Cr}^{3+}\). Step 1 — two chromiums: \(\text{Cr}_2\text{O}_7^{2-} \rightarrow 2\text{Cr}^{3+}\). Step 2 — seven oxygens on the left need seven waters on the right. Step 3 — those waters carry fourteen hydrogens, so add \(14\text{H}^+\) to the left. Step 4 — audit the charge: left is \(-2 + 14 = +12\); right is \(2 \times (+3) = +6\); adding six electrons to the left brings +12 down to +6. \[ \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{e}^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \] Electrons on the left confirm a reduction, as expected for an oxidising agent. The identical recipe produces the other workhorse of the course: \(\text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}\). Learn the method, not the equations — AQA can ask for half-equations you have never seen.
MechanismCombining half-equations — make the electrons vanish
Electrons never appear in a full redox equation, because every electron donated is accepted — the two half-equations must be scaled until their electron counts match, then added. The recipe: find the lowest common multiple of the electrons, multiply each half-equation accordingly, add left sides to left sides and right to right, then cancel any species that appears on both sides (usually \(\text{H}_2\text{O}\) or \(\text{H}^+\)).
Then run the double audit: atoms balance and total charge balances. If electrons survive into your final answer, the scaling is wrong; if the charges disagree, something slipped in the adding. This is not exam paranoia — it is the exact chemistry of the manganate(VII) titration you will meet at 3.2.5.5, where deep purple \(\text{MnO}_4^-\) is decolourised by iron(II) and the redox equation is the first two marks of a six-mark calculation.
Combine \(\text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}\) with \(\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + \text{e}^-\). Manganate takes five electrons; each iron gives one — so multiply the iron half-equation by five and add: \[ \text{MnO}_4^- + 8\text{H}^+ + 5\text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} + 5\text{Fe}^{3+} \] Audit. Atoms: one Mn, four O, eight H and five Fe on each side. Charge: left \(-1 + 8 + 10 = +17\); right \(+2 + 15 = +17\). Balanced both ways. That one-line charge check has rescued more redox marks than any other habit in the course — write it even when the question does not ask.
CaseOne species, both directions — and the redox that isn't
The examiners' favourite twist is disproportionation: a single element in a single species simultaneously oxidised and reduced. Bubble chlorine into cold dilute sodium hydroxide — this is how household bleach is manufactured — and you get \(\text{Cl}_2 + 2\text{NaOH} \rightarrow \text{NaCl} + \text{NaClO} + \text{H}_2\text{O}\). Chlorine starts at 0 and finishes at both −1 (in \(\text{NaCl}\)) and +1 (in \(\text{NaClO}\)): it has oxidised itself and reduced itself in one move. The reaction is formally examined in 3.2.3.2, but proving disproportionation is pure 3.1.7 — assign states to every atom of the element and show one went up while another went down.
The opposite trap is declaring redox where there is none. Precipitation and acid–base reactions shuffle whole ions without any electron transfer: in \(\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}\), every element keeps its oxidation state, so nothing is oxidised or reduced — it is not redox, however dramatic the temperature change. Before you use the word, run the ledger. If no oxidation state changes, the answer to "explain, in terms of oxidation states…" is that there is nothing to explain — and AQA has asked exactly that.
VocabularyKey terms the mark scheme pays for
TrapsMisconceptions that cost marks
ExamWhat examiners want
Definitions are AO1 gifts, but only in the right currency: define oxidation and reduction in terms of electrons or oxidation states, never oxygen. Write oxidation states sign-first (+2, never 2+ or a bare 2) — AQA mark schemes state the convention explicitly. When a question says "in terms of oxidation states, explain what has been oxidised", the mark needs numbers on both ends: "chromium falls from +6 to +3, so dichromate(VI) is reduced" — two states plus a direction word, for every species you mention.
Half-equation marks follow the four-step build: correct species, oxygen balanced with water, hydrogen with \(\text{H}^+\), then electrons on the side that makes charge balance. For combined equations the two silent killers are electrons surviving into the final answer (your scaling was wrong) and unbalanced charge (your adding was). Finish every redox equation with the one-line charge audit — examiner reports repeatedly note that candidates who verify charge outscore those who reason verbally.
When naming agents (AO2), quote the entire species — "\(\text{MnO}_4^-\) is the oxidising agent", not "manganese" — and expect disproportionation proofs, where you must show the same element going up and down from a single starting state. This section is bread-and-butter on Paper 1 and Paper 2, and it compounds: the halide reactions of 3.2.3, the redox titrations of 3.2.5.5 and all of electrode potentials in 3.1.11 assume you can produce and combine half-equations on demand.