Part 1: Oxidation numbers and electron transfer

Track electron loss and gain, use oxidation states in names and formulae, and recognise redox in familiar and unfamiliar reactions.

Assign oxidation numbers to individual atoms

An uncombined element has oxidation number zero; a monatomic ion has its ionic charge. The sum for a neutral compound is zero; for a polyatomic ion it equals the ion charge. Group 1 metals are +1, Group 2 +2 and fluorine −1 in their compounds. Oxygen is normally −2 but −1 in peroxides; hydrogen normally +1 but −1 in metal hydrides.

For SO₄²⁻: x + 4(−2) = −2, so S is +6. For Cr₂O₇²⁻: 2x − 14 = −2, so each Cr is +6, not +12. An oxidation number is formal electron bookkeeping; +6 does not mean a free S⁶⁺ ion exists in sulfate.

Solve the total first, then the value per atom

For NO₂⁻, let nitrogen be x. Oxygen normally contributes two lots of −2, and the whole ion has charge −1: x − 4 = −1, so x = +3. For NH₄⁺, x + 4(+1) = +1 gives nitrogen −3. An element need not have the same oxidation state in all its compounds.

For S₂O₈²⁻, knowing only the formula is insufficient to apply “every oxygen is −2” blindly: this ion contains a peroxide linkage. For unfamiliar compounds use supplied structural information and stated exceptions. At AS, the safe routine is to apply the familiar rules, but check whether a question supplies a reason to depart from them.

Oxygen in OF₂ is +2 because fluorine is −1. Oxygen in H₂O₂ is −1 because hydrogen is +1 and there are two oxygen atoms. These are oxidation numbers, not claims that isolated O⁺ or O²⁺ ions are present in the molecules.

Use Roman numerals to remove ambiguity

Iron(III) oxide is Fe₂O₃: two +3 iron centres balance three −2 oxygens. Sodium chlorate(I) is NaClO, while sodium chlorate(V) is NaClO₃. If an oxyanion’s oxidation state is specified, use total charge to derive its oxygen count. Ordinary nitrate and sulfate refer to NO₃⁻ and SO₄²⁻ in this course unless indicated otherwise.

Oxidation and reduction happen together

Oxidation is loss of electrons and an increase in oxidation number. Reduction is gain of electrons and a decrease. An oxidising agent accepts electrons and is reduced; a reducing agent donates electrons and is oxidised.

In Zn + Cu²⁺ → Zn²⁺ + Cu, zinc changes 0 → +2 and copper +2 → 0. Zinc is the reducing agent. The following half-equations make the electron balance visible; add them and cancel electrons.

Name the agent by what it does to its partner

An oxidising agent causes another species to lose electrons by accepting those electrons itself. It is therefore reduced. A reducing agent gives electrons to another species and is itself oxidised. The names describe the effect on the partner, which is why the agent’s own change sounds opposite.

For Cl₂ + 2Br⁻ → 2Cl⁻ + Br₂, Br⁻ loses electrons and is the reducing agent. Cl₂ gains them and is the oxidising agent. Name the reactant species, including its charge: bromine and bromide are not interchangeable answers.

For Mg + 2H⁺ → Mg²⁺ + H₂, Mg loses two electrons and two hydrogen ions gain two electrons altogether. Chloride ions in a hydrochloric-acid version remain −1 throughout; they are spectators, not the species being reduced.

Combine half-equations with unequal electron counts

Take aluminium reducing copper(II) ions. Aluminium oxidation is Al → Al³⁺ + 3e⁻. Copper reduction is Cu²⁺ + 2e⁻ → Cu. The smallest common electron count is six: multiply the aluminium equation by two and the copper equation by three.

Adding gives 2Al + 3Cu²⁺ + 6e⁻ → 2Al³⁺ + 6e⁻ + 3Cu. Cancel six electrons on each side. The final equation is 2Al + 3Cu²⁺ → 2Al³⁺ + 3Cu. Total charge is +6 on both sides; there are two Al and three Cu atoms on each side.

Electrons are transferred, not created or destroyed overall. If they remain in a proposed complete redox equation, the half-equations have not been matched. This simple electron accounting supports AS redox; balancing complex acidic manganate/dichromate titration equations is developed later in the course.

Metals with dilute non-oxidising acids

Mg, Al, Fe and Zn can form a salt and hydrogen with dilute hydrochloric or sulfuric acid. An oxide coating can initially slow aluminium. Iron forms Fe²⁺ with these dilute acids. Nitric acid is oxidising and should not be assumed to give the same hydrogen reaction.

Observe the metal disappearing and effervescence; hydrogen gives a squeaky pop with a lighted splint on a small collected sample under supervised conditions. In Mg + 2HCl, magnesium is oxidised 0 → +2 and hydrogen is reduced +1 → 0.

Distinguish redox from acid–base change

Neutralisation H⁺ + OH⁻ → H₂O leaves H at +1 and O at −2: no redox. In Cl₂ + 2OH⁻ → Cl⁻ + ClO⁻ + H₂O, chlorine changes from 0 to −1 and +1: the same element is reduced and oxidised. This is disproportionation.

For an unfamiliar equation, assign oxidation numbers on both sides before naming the oxidised or reduced species. Multiply the per-atom change by the number of those atoms when checking total electrons.

Audit disproportionation atom by atom

For 2H₂O₂ → 2H₂O + O₂, peroxide oxygen starts at −1. Some oxygen becomes −2 in water, while some becomes 0 in oxygen gas. Oxygen is both reduced and oxidised, so the reaction is disproportionation. Hydrogen remains +1 and does not cause that classification.

For a metal reacting with acid, different elements undergo oxidation and reduction; that is redox but not disproportionation. For acid–base neutralisation no oxidation numbers change. Being a reaction, having ions or producing a gas does not alone make a reaction redox.

In an unfamiliar equation, mark only the atoms whose oxidation state changes and count how many of those atoms change. A +2 to +3 change loses one electron per atom; a coefficient of three means three electrons overall. This prevents confusing an oxidation number with the total electron transfer.

Quick checks

Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official OCR A mark allocations.

Sources