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This lesson extends the GCSE and AS treatment of oxidation numbers to the depth required at A2. The basic rules — free elements are zero, monatomic ions take their ionic charge, and most compounds obey the familiar +1/−2 conventions for hydrogen and oxygen — are assumed as prior knowledge. What is new here is the careful handling of the awkward cases that AQA Paper 1 examiners use to separate the grades: peroxides where oxygen sits at −1, superoxides where the same atom is assigned a half-integer value, metal hydrides where hydrogen is anomalously −1, and mixed-state organic species. We also formalise fractional oxidation numbers (the iron in magnetite, Fe₃O₄, is assigned +8/3 as a population average), the systematic balancing of half-equations in both acidic and alkaline media, and the concept of disproportionation, where a single species is simultaneously oxidised and reduced. Mastery of these techniques is essential for §3.1.7 itself and for every subsequent redox-driven topic on the A-Level syllabus.
Spec mapping (AQA 7405): This lesson anchors §3.1.7 (oxidation, reduction and redox equations) and is a direct extension of lesson 1 (redox-reactions-equations) of this course. It feeds forward into lesson 2 (standard electrode potentials and the electrochemical series), §3.2.5 (transition metals and variable oxidation states), §3.2.6 (reactions of transition-metal ions in aqueous solution, including disproportionation of copper(I)), and the organic synthesis routes in §3.3 (alcohol oxidation, carbonyl oxidation, side-chain oxidation of arenes). Refer to the official AQA specification document for the exact wording of each subsection.
Assessment objectives: Recall of the six priority rules for assigning oxidation numbers is AO1. Assigning oxidation numbers to atoms in complex species — peroxides, superoxides, polythionates, mixed-valence oxides — and balancing half-equations in acidic and alkaline media are AO2. Rationalising the disproportionation of halogens in alkali, identifying oxidising and reducing agents from oxidation-number changes, and selecting the appropriate convention when more than one assignment is mathematically possible are AO3.
Oxidation numbers (ON) are assigned by following a hierarchy. When two rules conflict, the rule higher in the list wins. The hierarchy below is the standard A-Level convention; it is identical to the IUPAC recommended procedure for compounds of the main-group and first-row transition elements.
Free elements have ON = 0. This applies regardless of how the element occurs structurally. Sodium metal (Na), chlorine gas (Cl₂), the cyclic S₈ ring, the tetrahedral P₄ molecule, and graphite are all assigned ON = 0. A common slip is to assume the molecular formula changes the answer — it does not. Every atom in S₈ is at ON = 0, even though S₈ is a covalent ring.
Monatomic ions take ON equal to the ionic charge. Na⁺ is at +1, Cl⁻ is at −1, Fe²⁺ is at +2, Fe³⁺ is at +3, Al³⁺ is at +3, S²⁻ is at −2. Polyatomic ions are not covered by this rule; the ON of individual atoms inside a polyatomic ion is found using the rules below and the total-charge condition (rule 7).
Group 1 metals in compounds are at +1; Group 2 metals are at +2. This rule covers Li, Na, K, Rb, Cs in their compounds and Be, Mg, Ca, Sr, Ba. There are no exceptions for A-Level purposes (a handful of low-temperature alkali-metal compounds with anomalous oxidation states exist in advanced inorganic chemistry but lie far outside the AQA specification).
Fluorine in compounds is always at −1. Fluorine is the most electronegative element, and no compound contains fluorine at any oxidation number other than −1. This is the single rule with no exceptions at all.
Hydrogen in compounds is at +1, except in metal hydrides where it is at −1. In water, in hydrocarbons, in ammonia, in mineral acids, and in nearly every other compound that a student will meet, hydrogen is at +1. The exceptions are the s-block and some d-block hydrides — LiH, NaH, KH, CaH₂, MgH₂ — where hydrogen is more electronegative than the metal partner and is therefore assigned −1. NaBH₄ and LiAlH₄, the standard A-Level reducing agents, also contain hydrogen at −1, formally bonded to boron or aluminium as the hydride ligand.
Oxygen in compounds is at −2, except: in peroxides (the −O−O− linkage) where oxygen is at −1 (H₂O₂, Na₂O₂, BaO₂); in superoxides (the O₂⁻ ion) where oxygen is assigned −½ (KO₂, NaO₂, RbO₂); in oxygen difluoride (OF₂) where oxygen is at +2 because fluorine is more electronegative and takes priority by rule 4. Ozone (O₃) is itself a free element and is therefore at 0 by rule 1, even though it is structurally unsymmetric.
The sum of oxidation numbers equals zero in a neutral compound and equals the ionic charge in a polyatomic ion. This is the closing condition that fixes the oxidation number of any remaining atom.
Key Point: The rules are applied in strict priority order. When fluorine is present alongside oxygen, rule 4 overrides rule 6 — this is why OF₂ has oxygen at +2 rather than −2. When hydrogen is present alongside a Group 1 metal in a hydride, the metal is fixed at +1 by rule 3 and hydrogen must therefore be at −1 by rule 7. Always work from the top of the list downward.
Chlorine spans a wide range of oxidation numbers across its oxoacids. Assigning each is straightforward provided the priority order is followed.
| Species | H | O | Cl | Working |
|---|---|---|---|---|
| HCl | +1 | — | −1 | +1 + Cl = 0 |
| HClO | +1 | −2 | +1 | +1 + Cl + (−2) = 0 |
| HClO₂ | +1 | −2 | +3 | +1 + Cl + 2(−2) = 0 |
| HClO₃ | +1 | −2 | +5 | +1 + Cl + 3(−2) = 0 |
| HClO₄ | +1 | −2 | +7 | +1 + Cl + 4(−2) = 0 |
Chlorine therefore spans −1 (in the chloride ion) all the way to +7 (in the perchlorate ion ClO₄⁻ and perchloric acid HClO₄), a total range of eight oxidation numbers. The acid strengths correlate with the oxidation number — perchloric acid is one of the strongest mineral acids, and hypochlorous acid (HClO) is comparatively weak — because the conjugate base ClO₄⁻ delocalises negative charge over four equivalent oxygens, whereas ClO⁻ has only one oxygen to spread charge over.
Sulfur is the canonical example of an element with multiple stable oxidation numbers, and AQA examiners use it routinely.
| Species | H | O | S | Working |
|---|---|---|---|---|
| H₂S | +1 | — | −2 | 2(+1) + S = 0 |
| S₈ | — | — | 0 | free element |
| SO₂ | — | −2 | +4 | S + 2(−2) = 0 |
| SO₃ | — | −2 | +6 | S + 3(−2) = 0 |
| H₂SO₄ | +1 | −2 | +6 | 2(+1) + S + 4(−2) = 0 |
| S₂O₃²⁻ (thiosulfate) | — | −2 | +2 (average) | 2S + 3(−2) = −2 |
The thiosulfate ion S₂O₃²⁻ deserves comment. The mathematical average oxidation number is +2, but structurally the ion contains a central sulfur formally at +5 bonded to a terminal sulfur at −1; the two are not equivalent. For the purposes of AQA mark schemes the average value +2 is accepted, and reaction stoichiometry is unaffected because the total number of electrons transferred when thiosulfate is oxidised to tetrathionate (2S₂O₃²⁻ → S₄O₆²⁻ + 2e⁻) depends on the sum of oxidation numbers, not on their distribution between the two sulfur sites.
Manganese and chromium are the two transition metals most often examined in redox contexts.
| Species | K | O | Cl | Metal | Working |
|---|---|---|---|---|---|
| MnO₂ | — | −2 | — | +4 | Mn + 2(−2) = 0 |
| MnO₄⁻ | — | −2 | — | +7 | Mn + 4(−2) = −1 |
| K₂Cr₂O₇ | +1 | −2 | — | +6 (each Cr) | 2(+1) + 2Cr + 7(−2) = 0 |
| CrO₄²⁻ | — | −2 | — | +6 | Cr + 4(−2) = −2 |
The change Mn(+7) → Mn(+2) in acidic solution is a five-electron reduction; the change Cr(+6) → Cr(+3) in acidic solution is a three-electron reduction per chromium atom, or six electrons per dichromate ion. These two reductions underlie almost all permanganate and dichromate titrations on the A-Level syllabus.
Mixed-valence compounds — those containing the same element in two different oxidation states — produce non-integer averages when treated by the standard rules. The textbook example is magnetite, Fe₃O₄. Applying rules 6 and 7:
3 × Fe + 4 × (−2) = 0 3 × Fe = +8 Fe = +8/3
The fractional value is not a physical absurdity. It is a population average: structurally, magnetite contains one Fe²⁺ ion and two Fe³⁺ ions per formula unit, and the average of 2 + 3 + 3 divided by 3 is indeed 8/3. The same approach applies to lead red (Pb₃O₄, average ON = +8/3, structurally 2 Pb²⁺ + 1 Pb⁴⁺) and to several mineral oxides outside the syllabus. For thiosulfate, the average ON +2 is the population average of one S(+5) and one S(−1); for the tetrathionate ion S₄O₆²⁻ produced when thiosulfate is oxidised by iodine, the average ON of sulfur is +5/2.
Fractional oxidation numbers should not be rejected as errors. They are correct outputs of the rules and indicate either genuine fractional charge distribution (rare) or a mixture of integer oxidation states within a single formula unit (the usual case).
Exam Tip: If a calculation gives a fractional oxidation number, the student should report it as a fraction or the equivalent decimal and, if asked, comment briefly that the species is mixed-valence. AQA mark schemes accept fractional answers for Fe₃O₄ and similar species without penalty.
The modern definition of oxidation and reduction is framed entirely in oxidation-number language:
The agent that brings about each change is named for the partner process:
This nomenclature is the source of one of the most common A-Level errors. When asked to identify the oxidising agent in a reaction, students sometimes name the species that gains oxygen or loses hydrogen — a hangover from the pre-electron definition of oxidation. The correct procedure is to assign oxidation numbers on both sides of the equation, identify the atom whose oxidation number has decreased, and name the species containing that atom as the oxidising agent.
Consider the reaction in acidic solution:
MnO₄⁻ + 5 Fe²⁺ + 8 H⁺ → Mn²⁺ + 5 Fe³⁺ + 4 H₂O
Oxidation-number changes:
The manganate(VII) ion MnO₄⁻ is therefore the oxidising agent (it was reduced). The iron(II) ion Fe²⁺ is the reducing agent (it was oxidised). The factor of 5 in the equation arises from electron balance: each Mn gains 5 electrons, each Fe loses 1, and the totals must match.
Combining half-equations is the central computational technique of A-Level redox. The five-step recipe below applies in any acidic aqueous medium. The same five steps appear in every AQA mark scheme.
Reduction half-equation:
Step 1: MnO₄⁻ → Mn²⁺. Mn already balanced. Step 2: Add 4 H₂O to the right to balance the 4 oxygens. MnO₄⁻ → Mn²⁺ + 4 H₂O. Step 3: Add 8 H⁺ to the left to balance the 8 hydrogens now on the right. MnO₄⁻ + 8 H⁺ → Mn²⁺ + 4 H₂O. Step 4: Left charge = −1 + 8 = +7. Right charge = +2. Add 5 e⁻ to the left to drop its charge from +7 to +2. MnO₄⁻ + 8 H⁺ + 5 e⁻ → Mn²⁺ + 4 H₂O.
Oxidation half-equation (iron(II) to iron(III)):
Fe²⁺ → Fe³⁺ + e⁻ (already balanced for atoms and charge).
Combining:
Multiply the iron half-equation by 5 to match electrons:
5 Fe²⁺ → 5 Fe³⁺ + 5 e⁻
Add to the manganate half-equation; the 5 e⁻ cancel:
MnO₄⁻ + 8 H⁺ + 5 Fe²⁺ → Mn²⁺ + 4 H₂O + 5 Fe³⁺
Atom check: 1 Mn, 4 O, 8 H, 5 Fe each side. Charge check: left = −1 + 8 + 10 = +17; right = +2 + 0 + 15 = +17. Balanced.
A similar treatment yields the standard dichromate reduction:
Cr₂O₇²⁻ + 14 H⁺ + 6 e⁻ → 2 Cr³⁺ + 7 H₂O
Six electrons are transferred per dichromate ion, reflecting the two chromium atoms each undergoing a (+6 → +3) three-electron reduction.
In alkaline solution, the proton concentration is vanishingly small and writing H⁺ into the equation is misleading. The procedure is to balance first as if the medium were acidic, then convert the protons to hydroxide ions by adding an equal number of OH⁻ to both sides. Protons combine with hydroxide to form water, and any water on the same side as added water can be cancelled.
1–4. Identical to the acidic procedure. 5. Add OH⁻ to both sides in a number equal to the H⁺ present. On the H⁺ side, the protons combine with hydroxide to form water. 6. Cancel any water that now appears on both sides.
Permanganate oxidises bromide to bromate(V) in concentrated alkaline solution. In a strongly alkaline reaction mixture, manganese is reduced only as far as manganese(IV) oxide (MnO₂) rather than the manganese(II) ion. The unbalanced skeleton is:
MnO₄⁻ + Br⁻ → MnO₂ + BrO₃⁻
Reduction half-equation:
MnO₄⁻ + 4 H⁺ + 3 e⁻ → MnO₂ + 2 H₂O (acidic-style first)
Convert by adding 4 OH⁻ to both sides:
MnO₄⁻ + 4 H₂O + 3 e⁻ → MnO₂ + 2 H₂O + 4 OH⁻
Cancel 2 H₂O from each side:
MnO₄⁻ + 2 H₂O + 3 e⁻ → MnO₂ + 4 OH⁻
Oxidation half-equation:
Br⁻ + 6 OH⁻ → BrO₃⁻ + 3 H₂O + 6 e⁻ (already in alkaline form, derived by the same procedure)
Combining (ratio 2:1 to balance electrons):
2 MnO₄⁻ + Br⁻ + H₂O → 2 MnO₂ + BrO₃⁻ + 2 OH⁻
Atom and charge checks confirm balance. The presence of OH⁻ (rather than H⁺) on the product side correctly signals the alkaline medium.
Key Point: Never leave H⁺ on the product side of an alkaline-medium equation. Examiners deduct a mark for a thermodynamically nonsensical free proton in a basic solution, even when the atoms balance numerically.
Disproportionation is a redox reaction in which a single species is simultaneously oxidised and reduced. The same element appears in three different oxidation numbers across the products: the starting value, a higher value (oxidation product), and a lower value (reduction product).
Cold dilute alkali produces a mixture of chloride and chlorate(I):
Cl₂ + 2 OH⁻ → Cl⁻ + ClO⁻ + H₂O
Chlorine in Cl₂ is at 0. In the products it is at −1 (chloride) and +1 (chlorate(I), ClO⁻). One chlorine atom has been reduced, the other oxidised.
Hot concentrated alkali drives the oxidation further, producing chlorate(V):
3 Cl₂ + 6 OH⁻ → 5 Cl⁻ + ClO₃⁻ + 3 H₂O
Here chlorine spans 0 → −1 (five atoms reduced) and 0 → +5 (one atom oxidised). The 5:1 stoichiometry reflects electron balance: each oxidation transfers 5 electrons (one atom 0 → +5), and each reduction transfers 1 electron (one atom 0 → −1), so five reductions are required per oxidation.
Hydrogen peroxide decomposes in the presence of a catalyst (e.g. MnO₂) by disproportionation of oxygen:
2 H₂O₂ → 2 H₂O + O₂
Oxygen in H₂O₂ is at −1 (peroxide). In H₂O it is at −2 (reduced); in O₂ it is at 0 (oxidised). Two oxygen atoms move from −1 to −2 (gain one electron each, total 2 e⁻ gained), and two move from −1 to 0 (lose one electron each, total 2 e⁻ lost). Electron balance is automatic.
The copper(I) ion is unstable in aqueous solution and disproportionates to copper(II) and copper metal:
2 Cu⁺(aq) → Cu²⁺(aq) + Cu(s)
Copper in Cu⁺ is at +1. In Cu²⁺ it is at +2 (oxidised); in metallic Cu it is at 0 (reduced). This reaction explains why copper(I) compounds are stable as solids (CuCl, Cu₂O) but rapidly disproportionate when dissolved, unless stabilised by a complexing ligand such as ammonia or chloride at high concentration.
Comproportionation is the reverse process: two species at different oxidation states of the same element combine to give a single product at an intermediate oxidation state. The classic A-Level example is the reaction of chloride and chlorate(V) in concentrated hydrochloric acid:
Cl⁻ + ClO₃⁻ + 6 H⁺ → 3 Cl₂ + 3 H₂O
Wait — the stoichiometry as written here is unbalanced for chlorine. The correctly balanced form is:
5 Cl⁻ + ClO₃⁻ + 6 H⁺ → 3 Cl₂ + 3 H₂O
Chlorine in Cl⁻ is at −1 (oxidised to 0 in Cl₂); chlorine in ClO₃⁻ is at +5 (reduced to 0 in Cl₂). Five chloride ions and one chlorate(V) ion combine to give three chlorine molecules. The reaction is the reverse of the chlorate(V) disproportionation in hot alkali and is used in the laboratory preparation of chlorine gas.
Common Misconception: Disproportionation is sometimes confused with simple redox. The defining feature is that a single starting species ends up at two different oxidation states. If two different starting elements change oxidation state, the reaction is ordinary redox, not disproportionation.
For any equation, the most efficient diagnostic is to assign oxidation numbers to every atom and look for changes.
When more than one valid assignment is mathematically possible (e.g. in a polyatomic ion with several heteroatoms), apply the priority order strictly. The hierarchy is designed so that exactly one assignment is consistent with all higher-priority rules.
Question 1. [13 marks total]
(a) Assign the oxidation number of the named atom in each of the following species: [3 marks]
(i) Sulfur in S₂O₃²⁻. (ii) Oxygen in KO₂. (iii) Iron in Fe₃O₄.
(b) Balance the following half-equation in acidic medium, showing all five steps of the procedure: [4 marks]
VO₂⁺ → V²⁺
(c) In the reaction below, identify the oxidising agent and the reducing agent, and state the oxidation-number change of each: [3 marks]
2 MnO₄⁻ + 5 H₂C₂O₄ + 6 H⁺ → 2 Mn²⁺ + 10 CO₂ + 8 H₂O
(d) Explain why chlorine undergoes disproportionation when added to cold, dilute aqueous sodium hydroxide solution. Include the balanced equation and assignment of oxidation numbers to chlorine in each species. [3 marks]
(a) Oxidation-number assignment [3 marks, AO1 + AO2]
(b) Half-equation balancing [4 marks, AO2]
Working from the recipe:
(c) Oxidising and reducing agents [3 marks, AO3]
(d) Disproportionation of chlorine in alkali [3 marks, AO2 + AO3]
Three responses are shown, covering the meaningful A-Level range: Grade C (the borderline-pass floor), Grade B (solid mark-scheme coverage), and Grade A* (top-band synthesis). Grade D and E responses are omitted because no A-Level student is targeting those bands and modelling failure adds nothing pedagogically. Editorial commentary follows each response.
(a)(i) In thiosulfate S₂O₃²⁻, oxygen is at −2, so 2 S + 3(−2) = −2, giving S = +2. (a)(ii) In KO₂, potassium is at +1 by Group 1 rule, so 2 O = −1, and oxygen is at −½. (a)(iii) In Fe₃O₄, oxygen is at −2, so 3 Fe = +8 and each Fe is at +8/3.
(b) Balancing VO₂⁺ → V²⁺ in acid: balance oxygen with H₂O to give VO₂⁺ → V²⁺ + 2 H₂O. Balance hydrogen with H⁺ to give VO₂⁺ + 4 H⁺ → V²⁺ + 2 H₂O. Left charge = +1 + 4 = +5, right charge = +2, so add 3 e⁻ to the left. Final equation: VO₂⁺ + 4 H⁺ + 3 e⁻ → V²⁺ + 2 H₂O.
(c) In MnO₄⁻, manganese is at +7; in Mn²⁺ it is at +2. Mn is reduced (gain of 5 electrons), so MnO₄⁻ is the oxidising agent. In H₂C₂O₄, each carbon is at +3; in CO₂ it is at +4. Carbon is oxidised (loss of 1 electron per C), so H₂C₂O₄ is the reducing agent.
(d) Chlorine reacts with cold dilute sodium hydroxide as:
Cl₂ + 2 OH⁻ → Cl⁻ + ClO⁻ + H₂O
Chlorine in Cl₂ is at 0. In Cl⁻ it is at −1 (reduced); in ClO⁻ it is at +1 (oxidised). Because chlorine is simultaneously oxidised and reduced in the same reaction, this is disproportionation.
Editorial commentary (Grade C): Mostly correct. The oxidation-number assignments are right, the half-equation is balanced, and the disproportionation argument is logically clean. To reach B, the student should comment on the structural meaning of the +2 average in thiosulfate (mixed +5 / −1) and explicitly link "oxidising agent reduced" / "reducing agent oxidised".
(a)(i) S in S₂O₃²⁻: applying rules 6 and 7, oxygen is at −2 (no peroxide linkage), so 2 S + 3(−2) = −2 gives S = +2 (average). Structurally, thiosulfate contains one central sulfur at +5 and one terminal sulfur at −1; the average is +2. (a)(ii) O in KO₂: KO₂ is a superoxide, the O₂⁻ ion paired with K⁺. K is at +1 by Group 1 priority, so 2 O = −1 and each oxygen is at −½. (a)(iii) Fe in Fe₃O₄: 3 Fe + 4(−2) = 0 gives Fe = +8/3. Magnetite is mixed-valence (1 Fe²⁺ + 2 Fe³⁺ per formula unit), and the fractional value is the population average.
(b) Five-step balance of VO₂⁺ → V²⁺ in acid:
(c) MnO₄⁻ contains Mn at +7; Mn²⁺ is at +2. Manganese is reduced by 5 units, so MnO₄⁻ is the oxidising agent (it is itself reduced). In ethanedioic acid H₂C₂O₄, each carbon is at +3 (rule 7); in CO₂ each carbon is at +4. Carbon is oxidised by 1 unit per atom, so H₂C₂O₄ is the reducing agent (it is itself oxidised). Stoichiometry confirms electron balance: 2 × 5 = 10 electrons gained by Mn; 10 × 1 = 10 electrons lost by C.
(d) Cl₂ + 2 OH⁻ → Cl⁻ + ClO⁻ + H₂O. Cl in Cl₂ = 0; in Cl⁻ = −1 (reduced); in ClO⁻ = +1 (oxidised). The same element is simultaneously oxidised and reduced — the definition of disproportionation.
Editorial commentary (Grade B): A* in everything except synoptic depth. To progress, the student should note the structural origin of the +2 in thiosulfate, the link to standard electrode potentials, or comment on why mixed-valence assignments are not physical absurdities.
(a)(i) S in S₂O₃²⁻ has an average oxidation number of +2, derived from 2 S + 3(−2) = −2. The average masks the structural reality: thiosulfate is a tetrahedral [SSO₃]²⁻ ion in which the central sulfur is formally at +5 (bonded to three terminal oxygens) and the terminal sulfur is at −1 (formally a sulfide). The population average is (5 + (−1))/2 = +2, consistent with the algebraic result. The mixed-valence character explains why thiosulfate is oxidised to tetrathionate S₄O₆²⁻ in iodometric titrations rather than to sulfate — the terminal sulfide-like sulfur is the kinetically accessible site. (a)(ii) O in KO₂ = −½. Potassium superoxide contains the O₂⁻ radical anion; one electron is shared over two oxygens, giving the fractional oxidation number. This is a genuine non-integer value in the molecular-orbital sense — the unpaired electron occupies a π* orbital, and the bond order is 1.5. (a)(iii) Fe in Fe₃O₄ = +8/3. Magnetite has the inverse-spinel structure with one Fe²⁺ in octahedral sites and two Fe³⁺ split equally between tetrahedral and octahedral sites per formula unit. The room-temperature ferrimagnetism of magnetite arises from antiparallel alignment of the tetrahedral and octahedral sublattice moments.
(b) Five-step balance: V balanced; add 2 H₂O to right; add 4 H⁺ to left; add 3 e⁻ to left to bring charge from +5 to +2. VO₂⁺ + 4 H⁺ + 3 e⁻ → V²⁺ + 2 H₂O. Vanadium descends through the (+5, +4, +3, +2) series in three sequential reductions; this half-equation telescopes the full series into a single three-electron step.
(c) MnO₄⁻ is the oxidising agent (Mn: +7 → +2, Δ = −5); H₂C₂O₄ is the reducing agent (C: +3 → +4, Δ = +1 per C). Stoichiometric electron count: 2 × 5 = 10 = 10 × 1. Synoptically, this is the permanganate–ethanedioate titration of §3.1.7 / Required Practical (volumetric analysis), with autocatalysis by Mn²⁺ giving the characteristic induction period.
(d) Cl₂ + 2 OH⁻ → Cl⁻ + ClO⁻ + H₂O. Cl: 0 → −1 (reduced) and 0 → +1 (oxidised) — disproportionation. The reaction is thermodynamically allowed because E°(Cl₂/Cl⁻) = +1.36 V exceeds E°(ClO⁻/Cl₂) = +0.42 V (alkaline), so Cl₂ is unstable with respect to disproportionation in alkali (cell EMF = +0.94 V). Hot concentrated alkali drives the oxidation further to chlorate(V): 3 Cl₂ + 6 OH⁻ → 5 Cl⁻ + ClO₃⁻ + 3 H₂O.
Editorial commentary (Grade A):* Genuinely A*: structural rationalisation of the +2 in thiosulfate, MO-level interpretation of the superoxide −½, magnetite's inverse-spinel structure and ferrimagnetism, an E° argument for the thermodynamic feasibility of chlorine disproportionation, and the link forward to chlorate(V) under hot conditions. This response demonstrates synoptic command of the whole §3.1.7 → §3.2.5 → §3.2.6 corridor.
This lesson completes the AQA 7405 §3.1.7 oxidation-number toolkit and prepares the student for lesson 2 (standard electrode potentials), where every electrode reaction is itself a half-equation balanced by the procedures developed here.