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Learn9701Paper 2 · AS Level Structured QuestionsElectrochemistryRedox and oxidation numbers
Electrochemistry · Subtopic 1 of 1

Redox and Oxidation Numbers

Oxidation number rules: elements 0; monatomic ions = charge; O usually −2 (−1 in peroxides); H +1 (−1 in metal hydrides); F always −1; sum = overall charge. Oxidation = increase in oxidation number (loss of electrons); reduction = decrease. Half-equations in acid: balance atoms other than O and H, add H₂O for O, H⁺ for H, then electrons for charge. Combine by matching electrons. Disproportionation: one element is both oxidised and reduced. Redox titrations: MnO4−\mathrm{MnO_4^-} (self-indicating, purple → colourless) and iodine–thiosulfate (starch indicator).

Cambridge 9701 Paper 1Cambridge 9701 Paper 2Edexcel YCH11 WCH12AQA 7405 Paper 1≈ 45 minRevision 1 · updated 2026-10-30
Section 1 of 14

Learning objectives

  • Assign oxidation numbers.
  • Write half-equations and combine them into full redox equations.
  • Identify disproportionation.
  • Carry out redox titration calculations.
Pause and think

In Cl2+2NaOH→NaCl+NaClO+H2O\mathrm{Cl_2+2NaOH\rightarrow NaCl+NaClO+H_2O}, chlorine starts at 0. What happens to it?

Section 2 of 14

Core idea

Start hereCounting electrons

Balancing complex redox equations by trial and error is painful. Half-equations turn it into simple bookkeeping.

The big idea

Oxidation numbers track electron loss and gain. Balance each half-equation separately, then match electrons to combine them.

1. Building a half-equation

Half-equation recipe (acidic conditions)

Work through in order.

Worked exampleManganate(VII) and ethanedioate

MnO4−→Mn2+\mathrm{MnO_4^-\rightarrow Mn^{2+}} and C2O42−→CO2\mathrm{C_2O_4^{2-}\rightarrow CO_2}.

Goal
Overall equation.
Strategy
Half-equations; match electrons.
  1. MnO4−+8H++5e−→Mn2++4H2O\mathrm{MnO_4^-+8H^++5e^-\rightarrow Mn^{2+}+4H_2O}
    Reduction (×2).
  2. C2O42−→2CO2+2e−\mathrm{C_2O_4^{2-}\rightarrow2CO_2+2e^-}
    Oxidation (×5).
  3. 2MnO4−+16H++5C2O42−→2Mn2++8H2O+10CO2\mathrm{2MnO_4^-+16H^++5C_2O_4^{2-}\rightarrow2Mn^{2+}+8H_2O+10CO_2}
    10 electrons cancel.
AnswerRatio MnO₄⁻ : C₂O₄²⁻ = 2 : 5.
✓ CheckCharge: −2+16−10=+4-2+16-10=+4 on both sides ✓.
Key takeawayDisproportionation The same species is both oxidised and reduced, e.g. 2H2O2→2H2O+O2\mathrm{2H_2O_2\rightarrow2H_2O+O_2} (O: −1 → −2 and 0).
Pause and think

Why can't hydrochloric acid be used to acidify manganate(VII) titrations?

Summary

  • Oxidation number rules.
  • H₂O, H⁺, e⁻.
  • Match electrons to combine.
Key words
Oxidation number
Charge an atom would have if all bonds were ionic.
Half-equation
Equation showing either oxidation or reduction with electrons.
Disproportionation
Simultaneous oxidation and reduction of the same element.
HistoryRoman numerals. The Stock system of writing oxidation numbers in names, like iron(III), was proposed by Alfred Stock in 1919.
Fun factBreathalyser chemistry. Early breathalysers used orange dichromate(VI), reduced to green Cr³⁺ by ethanol in breath.
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