6.1 Redox: electron transfer and oxidation number
- Syllabus
- 9701–2028–2029
- Topic
- 6.1
- Level
- AS
Learning objectives
6.1.1Oxidation numbers of elements in compounds• Calculate oxidation numbers of elements in compounds and ions6.1.2Changes in oxidation numbers to help balance• Use changes in oxidation numbers to help balance chemical equations6.1.3Redox Terminology• Explain and use the terms redox, oxidation, reduction and disproportionation in terms of electron transfer and changes in oxidation number6.1.4And use the terms oxidising agent and reducing• Explain and use the terms oxidising agent and reducing agent6.1.5Oxidation Numbers in Compound Names• Use a Roman numeral to indicate the magnitude of the oxidation number of an element
Calculate oxidation numbers by making their weighted sum match the charge
An oxidation number is an assigned electron-bookkeeping value for one atom in a species. It is not necessarily a real ionic charge, especially in a covalent molecule.
| Rule | Result | Important boundary |
|---|---|---|
| uncombined element | 0 | includes H₂, O₂ and Cl₂ |
| monatomic ion | ionic charge | Fe³⁺ is +3 |
| Group 1 / Group 2 in compounds | +1 / +2 | use before solving the unknown |
| fluorine in compounds | −1 | fluorine is the more electronegative element |
| oxygen in compounds | usually −2 | −1 in peroxides; +2 in OF₂ |
| hydrogen in compounds | usually +1 | −1 in metal hydrides |
| sum of all oxidation numbers | 0 for a neutral species; ion charge for a polyatomic ion | multiply each value by its atom count |
x+4(−2)=−2⇒x=+6 for S in SO42−
In H₂O₂, hydrogen contributes +2 in total and the molecule is neutral, so the two oxygen atoms contribute −2 in total: each oxygen is −1. Applying the usual −2 without checking for a peroxide would give the wrong result.
Solve from the overall charge, not from a memorised answer. Oxidation number belongs to each atom, so coefficients within a formula must be included before the terms are summed.
Balance redox equations by equalising total oxidation-number change
Assign oxidation numbers to the elements that change. Multiply each change by the number of affected atoms, choose coefficients so total increase equals total decrease, then balance O with H₂O and H with H⁺ in acidic solution. Finish by checking every atom and the total charge.
For MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺ in acid:
| Element | Oxidation-number change | Electron-equivalent change | Required ratio |
|---|---|---|---|
| Mn | +7 → +2 | gains 5 | 1 Mn |
| Fe | +2 → +3 | loses 1 | 5 Fe |
MnO4−+5Fe2++8H+→Mn2++5Fe3++4H2O
Atoms balance as Mn₁Fe₅O₄H₈ on both sides. Charge also balances: left = −1 + 10 + 8 = +17 and right = +2 + 15 = +17.
Equalising the numerical changes is only the redox core; it does not finish the equation. Never accept coefficients until both atom counts and total charge agree, and do not add H⁺ unless the stated medium is acidic.
Redox couples oxidation with reduction; disproportionation gives one species both roles
| Process | Electron transfer | Oxidation-number change |
|---|---|---|
| oxidation | loss of electrons | increase |
| reduction | gain of electrons | decrease |
| redox | oxidation and reduction occur together | at least one increase and one decrease |
| disproportionation | the same reactant species is both oxidised and reduced | one starting oxidation number produces a higher and a lower value |
In Zn + Cu²⁺ → Zn²⁺ + Cu, Zn loses two electrons and changes 0 → +2, while Cu²⁺ gains two electrons and changes +2 → 0. This is redox because the paired electron loss and gain occur in the same reaction.
In Cl₂ + 2OH⁻ → Cl⁻ + ClO⁻ + H₂O, chlorine starts at 0. Some chlorine atoms are reduced to −1 in Cl⁻ while others are oxidised to +1 in ClO⁻, so Cl₂ disproportionates.
A species merely appearing in two products is not enough for disproportionation: the same element in that one reactant must move to both a higher and a lower oxidation number. Oxidation and reduction are simultaneous electron-bookkeeping changes, not separate optional labels.
Oxidising agents accept electrons and reducing agents donate them
An oxidising agent causes another species to lose electrons and is reduced itself. A reducing agent causes another species to gain electrons and is oxidised itself.
Write the electron change for the agent, not just the reaction partner. Stronger agents are better electron acceptors or donors under the stated conditions.
In Cl₂ + 2Br⁻ → 2Cl⁻ + Br₂, chlorine accepts electrons and is the oxidising agent; bromide donates electrons and is the reducing agent.
The agent’s name describes what it does to the other species, not the change it undergoes. The oxidising agent does not undergo oxidation.
A Roman numeral states the magnitude of an element's oxidation number
In a systematic compound name, a Roman numeral in parentheses after an element states the magnitude of that element's oxidation number. The sign is not written inside the numeral, and the numeral is not an atom count.
| Formula | Oxidation-number calculation | Name |
|---|---|---|
| FeCl₃ | Fe + 3(−1) = 0, so Fe = +3 | iron(III) chloride |
| SO₂ | S + 2(−2) = 0, so S = +4 | sulfur(IV) oxide |
| NaClO₃ | +1 + Cl + 3(−2) = 0, so Cl = +5 | sodium chlorate(V) |
To move from name to formula, translate the numeral into the stated oxidation number and combine it with the known oxidation numbers or ion charges so the whole formula has the required charge. To move from formula to name, calculate first and then convert 1, 2, 3, 4, 5, 6 and 7 to I, II, III, IV, V, VI and VII.
The numeral can describe a metal or a central non-metal; do not restrict it to metal-ion charge. In iron(III) chloride, III means each Fe is +3, not that the formula contains three iron atoms.