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This lesson develops the procedural toolkit that all A-Level redox chemistry rests on: writing half-equations, combining them so that the electron count cancels, and using the resulting balanced ionic equations as the stoichiometric basis of redox titrations. The treatment is anchored in oxidation-number (ON) bookkeeping established in lesson 0, and it previews the quantitative use of standard electrode potentials in lesson 2 and the experimental practice of redox titration in lesson 6. The lesson covers the canonical A-Level redox systems — metal-acid displacement, halogen-halide displacement, and the manganate(VII) and dichromate(VI) oxidations of iron(II) — together with the iodine-thiosulfate system that anchors Required Practical 9 in the analytical course. We classify oxidising and reducing agents by their direction of electron flow, examine which couples can react and which cannot, and conclude with the practical-skills judgement required when choosing an oxidant or reductant for a given synthetic target.
Spec mapping (AQA 7405): This lesson maps to §3.1.7 (oxidation, reduction and redox equations) in its entirety. It builds directly on lesson 0 of this course (oxidation-number foundations) and is the procedural prerequisite for lesson 2 (standard electrode potentials, which quantify the direction of redox), lesson 6 (redox titrations, RP9 anchor), and §3.2.3 inorganic Group 7 redox chemistry (halogen-halide displacement). The half-equation balancing technique developed here is also a prerequisite for §3.1.11 (electrochemical cells) and §3.3.7 (transition-metal redox). Refer to the official AQA specification document for the exact wording of each section.
Assessment objectives: AO1 items include the definitions of oxidising agent (electron acceptor, species reduced) and reducing agent (electron donor, species oxidised) and the OIL RIG mnemonic linking electron transfer to oxidation-number change. AO2 dominates the practical exam-paper mark scheme: balancing and combining half-equations in acidic medium, applying the resulting stoichiometry to KMnO₄, K₂Cr₂O₇, and thiosulfate titrations, and computing concentrations or percentage purity from titration data. AO3 features in extended-response items that require students to rationalise oxidising-strength trends across Period 3 oxides, to predict which redox couples will react on the basis of qualitative ordering, and to identify which observable changes constitute evidence of electron transfer.
Redox is the only major topic in A-Level Chemistry where two parallel languages describe the same physical event. The electron-transfer language defines oxidation as the loss of electrons and reduction as the gain of electrons. The oxidation-number language defines oxidation as an increase in oxidation number (ON) and reduction as a decrease in oxidation number. These two definitions are equivalent for any reaction in which electrons are genuinely transferred, but the ON language extends to covalent reactions where no full electron transfer occurs. The textbook mnemonic OIL RIG (Oxidation Is Loss; Reduction Is Gain) captures the electron-transfer definition.
An oxidising agent (or oxidant) is the species that causes another to be oxidised: it accepts electrons and is itself reduced. Its ON decreases over the course of the reaction. A reducing agent (or reductant) is the species that causes another to be reduced: it donates electrons and is itself oxidised. Its ON increases. Identifying the oxidising and reducing agents in any redox equation is therefore a two-step procedure: assign ONs to every atom, locate the species whose ON has changed, and label the species whose ON has decreased as the oxidant and the species whose ON has increased as the reductant.
Worked classification: In MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O, manganese decreases from +7 (in MnO₄⁻) to +2 (in Mn²⁺), so MnO₄⁻ is the oxidising agent. Iron increases from +2 (in Fe²⁺) to +3 (in Fe³⁺), so Fe²⁺ is the reducing agent. Hydrogen (+1 throughout) and oxygen (−2 throughout) do not change ON and are spectators in the redox accounting, even though H⁺ is essential as a reagent to balance the equation.
The same species can act as oxidant in one reaction and reductant in another, depending on the partner couple. Hydrogen peroxide (H₂O₂) is the standard example: it oxidises Fe²⁺ to Fe³⁺ (acting as oxidant; O goes from −1 to −2) but reduces MnO₄⁻ to Mn²⁺ (acting as reductant; O goes from −1 to 0, producing O₂). Both behaviours are consistent because H₂O₂ contains oxygen in an intermediate ON (−1) between its most reduced form (−2 in H₂O) and its most oxidised form (0 in O₂).
A half-equation isolates the oxidation or the reduction half of a redox process and shows electrons explicitly. The standard A-Level procedure for balancing a half-equation in acidic aqueous solution follows five steps.
The electrons appear on the left for a reduction (the species is gaining electrons) and on the right for an oxidation (the species is losing electrons). Always check at the end that both atom counts and net charge balance.
Worked half-equation 1 — reduction of permanganate. MnO₄⁻ → Mn²⁺ in acidic solution. Mn is already balanced. Add 4 H₂O on the right to balance the four oxygens: MnO₄⁻ → Mn²⁺ + 4 H₂O. Add 8 H⁺ on the left to balance the eight hydrogens now on the right: MnO₄⁻ + 8 H⁺ → Mn²⁺ + 4 H₂O. Charge on the left is (−1) + 8 = +7; charge on the right is +2. Add 5 electrons on the left to give net +2 on both sides:
MnO₄⁻ + 8 H⁺ + 5 e⁻ → Mn²⁺ + 4 H₂O (ΔON for Mn = −5; reduction)
Worked half-equation 2 — reduction of dichromate. Cr₂O₇²⁻ → 2 Cr³⁺. Balance Cr first (factor of 2 on the right). Balance O with 7 H₂O on the right. Balance H with 14 H⁺ on the left. Balance charge: left = −2 + 14 = +12; right = +6. Add 6 electrons on the left:
Cr₂O₇²⁻ + 14 H⁺ + 6 e⁻ → 2 Cr³⁺ + 7 H₂O (ΔON for each Cr = −3; reduction)
Worked half-equation 3 — oxidation of iron(II). Fe²⁺ → Fe³⁺. Atom count already balanced. Charge: left +2, right +3. Add 1 electron on the right:
Fe²⁺ → Fe³⁺ + e⁻ (ΔON for Fe = +1; oxidation)
Worked half-equation 4 — oxidation of thiosulfate. Two thiosulfate ions combine to form tetrathionate during iodine titration: 2 S₂O₃²⁻ → S₄O₆²⁻. Sulfur balances (4 on each side). No O or H to balance. Charge: left −4, right −2. Add 2 electrons on the right:
2 S₂O₃²⁻ → S₄O₆²⁻ + 2 e⁻ (mean ON of S changes from +2 to +2.5; oxidation)
Worked half-equation 5 — reduction of iodine. I₂ → 2 I⁻. Iodine balances. Charge: left 0, right −2. Add 2 electrons on the left:
I₂ + 2 e⁻ → 2 I⁻ (ΔON for each I = −1; reduction)
These five half-equations recur throughout A-Level redox chemistry; committing them to memory accelerates every titration calculation.
Common Pitfall: Students sometimes add electrons to balance atoms rather than charge. Electrons only ever balance the net charge difference once atom balancing is complete. If you find yourself adding electrons before atoms balance, you have made an error earlier in the procedure.
To obtain the overall ionic equation, the two half-equations are scaled so that the number of electrons released by the oxidation equals the number consumed by the reduction. The two scaled equations are then added; the electrons cancel; and any species appearing on both sides (such as H⁺ or H₂O) are simplified.
Worked combination 1 — manganate(VII)/iron(II). Reduction: MnO₄⁻ + 8 H⁺ + 5 e⁻ → Mn²⁺ + 4 H₂O (consumes 5 e⁻). Oxidation: Fe²⁺ → Fe³⁺ + e⁻ (releases 1 e⁻). Multiply the oxidation by 5 so that both half-equations involve 5 e⁻:
5 Fe²⁺ → 5 Fe³⁺ + 5 e⁻
Add the half-equations and cancel electrons:
MnO₄⁻ + 8 H⁺ + 5 Fe²⁺ → Mn²⁺ + 4 H₂O + 5 Fe³⁺
This is the foundation equation of the KMnO₄/Fe²⁺ titration. The 1 : 5 mole ratio of MnO₄⁻ : Fe²⁺ is the single fact most often demanded in A-Level titration calculations.
Worked combination 2 — dichromate(VI)/iron(II). Reduction: Cr₂O₇²⁻ + 14 H⁺ + 6 e⁻ → 2 Cr³⁺ + 7 H₂O. Oxidation: Fe²⁺ → Fe³⁺ + e⁻. Multiply the oxidation by 6:
6 Fe²⁺ → 6 Fe³⁺ + 6 e⁻
Add and cancel:
Cr₂O₇²⁻ + 14 H⁺ + 6 Fe²⁺ → 2 Cr³⁺ + 7 H₂O + 6 Fe³⁺
Mole ratio Cr₂O₇²⁻ : Fe²⁺ = 1 : 6.
Worked combination 3 — iodine/thiosulfate. Reduction: I₂ + 2 e⁻ → 2 I⁻. Oxidation: 2 S₂O₃²⁻ → S₄O₆²⁻ + 2 e⁻. The electron counts already match; add directly:
I₂ + 2 S₂O₃²⁻ → 2 I⁻ + S₄O₆²⁻
Mole ratio I₂ : S₂O₃²⁻ = 1 : 2. This equation anchors RP9 in the analytical course and is used in iodometric determinations of oxidising agents such as ClO⁻, Cu²⁺ (via the Cu²⁺/I⁻ reaction), and IO₃⁻.
Worked combination 4 — zinc/copper(II). Reduction: Cu²⁺ + 2 e⁻ → Cu. Oxidation: Zn → Zn²⁺ + 2 e⁻. Electrons already match:
Zn + Cu²⁺ → Zn²⁺ + Cu
The classical school-laboratory demonstration: a zinc rod dipped in copper(II) sulfate solution acquires a copper coating while the solution colour fades from blue.
Worked combination 5 — magnesium/hydrochloric acid. Reduction: 2 H⁺ + 2 e⁻ → H₂. Oxidation: Mg → Mg²⁺ + 2 e⁻. Electrons match:
Mg + 2 H⁺ → Mg²⁺ + H₂
In molecular form (recognising 2 Cl⁻ as spectator ions):
Mg + 2 HCl → MgCl₂ + H₂
Several redox couples appear repeatedly in A-Level questions, and a working knowledge of which direction each runs in is a quick route to AO3 marks.
Halogen-halide displacement. A halogen higher in Group 7 oxidises the halide ion of any halogen below it. Chlorine therefore oxidises Br⁻ to Br₂ and oxidises I⁻ to I₂. Bromine oxidises I⁻ to I₂ but cannot oxidise Cl⁻ (chlorine sits above bromine). Iodine oxidises neither Cl⁻ nor Br⁻. The underlying trend is the decrease in oxidising power down Group 7 as atomic radius increases and electron-gain enthalpy becomes less exothermic. Sample equation: Cl₂ + 2 KBr → 2 KCl + Br₂ (Br⁻ oxidised from −1 to 0; Cl₂ reduced from 0 to −1).
Metal-cation displacement. A metal higher in the electrochemical series reduces the cation of any metal below it. Zinc reduces Cu²⁺ to Cu (Zn + Cu²⁺ → Zn²⁺ + Cu) because Zn lies above Cu in the activity series. Copper does not reduce Zn²⁺; the reverse reaction is not feasible. The qualitative ordering mirrors the more positive E° of the Cu²⁺/Cu couple (+0.34 V) relative to the Zn²⁺/Zn couple (−0.76 V), a quantitative criterion developed in lesson 2.
KMnO₄ in acidic solution. Manganate(VII) is one of the strongest oxidising agents routinely used in A-Level practical work, with E°(MnO₄⁻/Mn²⁺) = +1.51 V in acidic conditions. It oxidises Fe²⁺ to Fe³⁺, oxalate to CO₂, ethanedioate (C₂O₄²⁻), I⁻ to I₂, and many alcohols to aldehydes or carboxylic acids depending on conditions. The titration is self-indicating because the purple MnO₄⁻ ion is reduced to the near-colourless Mn²⁺ ion. The endpoint is recognised as the first persistent pale pink colour, indicating a slight excess of MnO₄⁻.
Important caution on titration medium. KMnO₄ titrations must be carried out in dilute sulfuric acid rather than hydrochloric acid. The reason is that Cl⁻ (E°(Cl₂/Cl⁻) = +1.36 V) is itself oxidised by MnO₄⁻ (E° = +1.51 V), so HCl would be partially oxidised to Cl₂ and the titration end-point would over-read the titre. Nitric acid is also unsuitable because NO₃⁻ is a competing oxidant. Dichromate(VI) titrations, by contrast, may be performed in HCl because Cr₂O₇²⁻ (E° = +1.33 V) is not strong enough to oxidise Cl⁻ under standard titration conditions.
Iodine-thiosulfate. This is the standard titration system for determining oxidising agents in solution. The procedure is iodometric: a known excess of KI is added to a sample of oxidising agent, which liberates iodine in stoichiometric proportion. The liberated iodine is then titrated against standardised sodium thiosulfate solution. Starch is added near the end-point as an indicator: starch forms a deep blue-black complex with I₂ in the presence of I⁻, and the end-point is the disappearance of this colour as the last I₂ is consumed. The titration is best performed at room temperature because the starch-iodine complex destabilises on warming and adsorbs less reliably.
The balanced ionic equation supplies the mole ratio between titrant and analyte. The titration calculation then follows the standard three-step pattern used for acid-base titrations.
Worked KMnO₄ titration. A 25.0 cm³ aliquot of iron(II) sulfate solution required 24.20 cm³ of 0.0200 mol dm⁻³ KMnO₄ for full oxidation in acidic conditions. Calculate the concentration of Fe²⁺ in the original solution.
Step 1 — moles of MnO₄⁻ delivered: n(MnO₄⁻) = c × V = 0.0200 × (24.20 / 1000) = 4.840 × 10⁻⁴ mol.
Step 2 — apply the 1 : 5 mole ratio from the balanced equation MnO₄⁻ + 5 Fe²⁺ + 8 H⁺ → Mn²⁺ + 5 Fe³⁺ + 4 H₂O: n(Fe²⁺) = 5 × 4.840 × 10⁻⁴ = 2.420 × 10⁻³ mol.
Step 3 — concentration of Fe²⁺ in the aliquot: c(Fe²⁺) = n / V = 2.420 × 10⁻³ / 0.0250 = 0.0968 mol dm⁻³ (3 s.f.).
Worked thiosulfate titration. A 25.0 cm³ portion of a solution of iodine in KI was titrated with 0.100 mol dm⁻³ sodium thiosulfate, requiring 22.50 cm³ to reach the starch-disappearance end-point. Calculate the concentration of I₂.
n(S₂O₃²⁻) = 0.100 × 0.02250 = 2.250 × 10⁻³ mol. From I₂ + 2 S₂O₃²⁻ → 2 I⁻ + S₄O₆²⁻, mole ratio I₂ : S₂O₃²⁻ = 1 : 2, so n(I₂) = 2.250 × 10⁻³ / 2 = 1.125 × 10⁻³ mol. c(I₂) = 1.125 × 10⁻³ / 0.0250 = 0.0450 mol dm⁻³.
These titration calculations and the experimental procedures behind them are developed in detail in lesson 6 (Redox Titrations), which anchors Required Practical 9.
A disproportionation reaction is a redox reaction in which a single species is simultaneously oxidised and reduced. The Group 7 chemistry of chlorine supplies two textbook examples.
Chlorine in cold dilute alkali: Cl₂ + 2 NaOH → NaCl + NaClO + H₂O. Chlorine has ON 0 in Cl₂; in NaCl it is −1 (reduction) and in NaClO it is +1 (oxidation). Both products arise from the same Cl₂ starting material.
Chlorine in hot concentrated alkali: 3 Cl₂ + 6 NaOH → 5 NaCl + NaClO₃ + 3 H₂O. Here one chlorine is oxidised from 0 to +5 (in ClO₃⁻) while five are reduced from 0 to −1 (in Cl⁻). Both reactions are essential to the AQA inorganic specification §3.2.3.
A comproportionation reaction is the reverse: two species of the same element in different ONs combine to give a single product of intermediate ON. The acidification of a mixture of ClO⁻ and Cl⁻ regenerates Cl₂ (ClO⁻ + Cl⁻ + 2 H⁺ → Cl₂ + H₂O), recovering the +1 chlorine and the −1 chlorine as the 0 element. Disproportionation and comproportionation are formally inverse processes.
When planning a synthetic procedure, the choice of oxidising or reducing agent depends on three considerations: the required degree of oxidation, the compatibility of the reagent with the rest of the molecule, and the practical convenience of the work-up.
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