Part 1: Oxidation numbers and electron-transfer agents

Use a consistent electron-accounting system to distinguish oxidation, reduction and the role played by each reactant.

Oxidation number is an electron-accounting label

The oxidation number of an atom is the charge it would have if the bonding electrons were assigned to the more electronegative atom, with equal sharing for identical atoms. It helps track changes even when a reaction involves covalent molecules rather than a visible transfer between isolated ions.

For a monatomic ion, oxidation number equals ionic charge: Mg²⁺ is +2 and Cl⁻ is −1. In a covalent molecule, it is a formal bookkeeping value, not necessarily a real charge on the atom. The carbon in CO₂ has oxidation number +4, but CO₂ is not a mixture of isolated C⁴⁺ and O²⁻ ions.

The sum of all oxidation numbers, including the number of each atom, equals the total charge on the species. It is zero for a neutral compound and equals the charge for a polyatomic ion. Assign a value to one atom of the named element; two chlorine atoms each at −1 contribute −2 in total, rather than each being −2.

Use the common assignments and their exceptions

Use the formula to recognise an exception before solving. In Na₂O₂, two Na contribute +2, so the two oxygens together are −2 and each oxygen is −1: this is a peroxide. In CaH₂, calcium contributes +2 and each hydrogen is −1. In OF₂, two fluorines contribute −2, so oxygen must be +2.

A zero oxidation number belongs to the element in its uncombined state, regardless of whether it is monatomic or molecular. O₂ is zero, whereas the oxygen in H₂O₂ is −1; do not confuse the molecule O₂ with the peroxide group in a compound.

Worked calculations in compounds and ions

For sulfur in SO₃, let its oxidation number be x. The compound is neutral, so x + 3(−2) = 0 and x = +6. For sulfur in SO₃²⁻, the sum must instead be −2: x − 6 = −2, giving x = +4. The same atom ratio with a different overall charge gives a different answer.

For nitrogen in NH₄⁺, x + 4(+1) = +1, so x = −3. For chlorine in ClO₃⁻, x + 3(−2) = −1, so x = +5. Write the total charge on the right-hand side before rearranging; assuming every sum is zero is a common avoidable mistake.

For a formula containing more than one atom of the unknown element, include that coefficient. In S₂O₃²⁻, 2x − 6 = −2 gives an average sulfur oxidation number of +2. This average does not prove the two sulfur atoms are chemically equivalent; questions may distinguish individual atom environments when more structural information is supplied.

Roman numerals state an element’s oxidation number

Roman numerals in names distinguish compounds where an element has different oxidation numbers: sulfur(IV) oxide is SO₂ and sulfur(VI) oxide is SO₃. Sodium chlorate(I) contains ClO⁻, while sodium chlorate(V) contains ClO₃⁻. The numeral describes the named element’s oxidation number; it is not a subscript to copy into the formula.

Worked formula construction: phosphorus(V) oxide has P at +5 and O at −2. The simplest neutral ratio is two P to five O, giving empirical formula P₂O₅. If the question asks for the molecular formula of the common molecular form, that is P₄O₁₀; an oxidation number determines charge balance but does not by itself determine molecular size.

For aluminium oxide, Al is +3 and O is −2, so use the lowest total balanced charge, six: two aluminium atoms and three oxygens give Al₂O₃. Check 2(+3) + 3(−2) = 0. For sodium sulfate(VI), the sulfate ion is SO₄²⁻ and needs two Na⁺, giving Na₂SO₄.

Loss raises oxidation number; gain lowers it

Oxidation is loss of electrons and an increase in oxidation number. Reduction is gain of electrons and a decrease in oxidation number. Electrons lost by one species must be gained by another, so oxidation and reduction occur together in a redox reaction.

For Mg + Cl₂ → MgCl₂, magnesium changes from 0 to +2 and loses two electrons: it is oxidised. Each chlorine changes from 0 to −1 and gains one electron: chlorine is reduced. Metals generally form positive ions by electron loss; non-metals commonly form negative ions by electron gain, although non-metals can also be oxidised in other reactions.

Acid–base neutralisation is not automatically redox. In H⁺ + OH⁻ → H₂O, hydrogen remains +1 and oxygen remains −2. A reaction can transfer a proton without transferring electrons between oxidation states. Classify the reaction from the changes, not from whether an equation contains an acid.

An agent’s name describes what it does to the other reactant

An oxidising agent accepts electrons, causing another species to be oxidised; it is itself reduced. A reducing agent donates electrons, causing another species to be reduced; it is itself oxidised. In Br₂ + 2I⁻ → 2Br⁻ + I₂, Br₂ gains electrons and is the oxidising agent; I⁻ loses electrons and is the reducing agent.

Identify the reacting species, not its product. Iodide, I⁻, is the reducing agent in that reaction; the iodine produced is not the electron donor that started it. When referring to a compound reagent, be precise: ‘iodide ions in potassium iodide’ distinguishes the active ion from its spectator cation.

Pearson 8CH0/01 June 2023 Q6(a) specifically asked for electron transfer. The report notes that oxidation-number changes alone did not fully answer that request. If the command asks for electrons, explicitly state gain or loss and name the species; oxidation numbers can support that statement.

Quick checks

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

Sources