Part 2: Half-equations and disproportionation

Balance atoms and charge independently, cancel the electron transfer and identify reactions where one element is both oxidised and reduced.

A half-equation displays one electron transfer

Write the species before and after the change, balance the atoms, then add electrons to balance charge. Oxidation has electrons on the product side; reduction has them on the reactant side. The charge sum, including each electron as −1, must be identical on both sides.

For Mg → Mg²⁺, atoms are already balanced. Add two electrons to the right so both sides have total charge zero. For Cl₂ → Cl⁻, first put 2 before Cl⁻ to conserve chlorine atoms, then add two electrons to the left. Starting with electrons before balancing the atoms can give the wrong factor.

Cancel equal electron numbers, not equal coefficients

Multiply each whole half-equation until the electron counts match. Then add, cancel electrons and cancel any identical species appearing on both sides. The final ionic equation contains the reacting species; ions that remain unchanged are spectators and are omitted.

Worked example: aluminium is oxidised to Al³⁺ and hydrogen ions are reduced to H₂. The oxidation half is Al → Al³⁺ + 3e⁻; the reduction half is 2H⁺ + 2e⁻ → H₂. The least common electron count is six, so multiply the first by two and the second by three.

The combined equation is 2Al(s) + 6H⁺(aq) → 2Al³⁺(aq) + 3H₂(g). Check atoms: two Al and six H on each side. Check charge: +6 on each side. No electrons remain because the amount lost equals the amount gained. A full equation with hydrochloric acid adds six chloride spectators to both sides, giving 2Al + 6HCl → 2AlCl₃ + 3H₂.

Balance oxygen and hydrogen using the stated medium

In acidic aqueous conditions, balance all atoms except O and H first. Add H₂O to balance oxygen, H⁺ to balance hydrogen, then electrons to balance charge. Finally recount atoms and charge. The order is an accounting method, not a proposed molecular mechanism.

Worked sulfur example: SO₄²⁻ → SO₂. Add 2H₂O to the right to balance four oxygens; add 4H⁺ to the left to balance four hydrogens. The left charge is initially +2, so add 2e⁻ to the left to make both sides neutral. Sulfur falls from +6 to +4, consistent with two electrons gained.

Combining that reduction with 2Br⁻ → Br₂ + 2e⁻ gives SO₄²⁻ + 4H⁺ + 2Br⁻ → SO₂ + 2H₂O + Br₂. Sulfate is reduced and bromide is oxidised. The charges sum to zero on both sides. This accounting helps explain the bromide/concentrated-sulfuric-acid chemistry in Topic 4.

In alkaline conditions, a final equation should not leave an unexplained supply of H⁺. An efficient method is to balance temporarily in acid, add equal OH⁻ to both sides to turn H⁺ into water, then cancel water. For Cl₂ + 2e⁻ → 2Cl⁻ no H or O occurs, so the half-equation is unchanged by the medium.

One starting oxidation number goes both up and down

Disproportionation is a reaction in which an element in a single starting species is simultaneously oxidised and reduced. Some atoms move to a higher oxidation number and others to a lower one. Ordinary redox between two different elements does not meet this definition.

In cold dilute alkali, chlorine forms chloride and chlorate(I): Cl₂ + 2OH⁻ → Cl⁻ + ClO⁻ + H₂O. Chlorine changes from 0 to −1 in chloride and from 0 to +1 in chlorate(I). One chlorine atom gains one electron while the other loses one, so the two product anions form in a 1:1 ratio.

Peroxide decomposition gives another example: 2H₂O₂ → 2H₂O + O₂. Oxygen begins at −1; some becomes −2 in water and some becomes 0 in oxygen gas. The same element in the same starting compound has both directions of change. Hydrogen remains +1 throughout.

Use oxidation-number changes to deduce a product ratio

Suppose a supplied reaction converts a halogen X₂ in hot alkali to X⁻ and XO₃⁻. X goes from 0 to −1 in X⁻ but from 0 to +5 in XO₃⁻. One atom oxidised loses five electrons; five atoms reduced each gain one. Therefore five halide ions form per one halate(V) ion, using six halogen atoms or three X₂ molecules.

Complete oxygen and hydrogen with the alkaline reactant: 3X₂ + 6OH⁻ → 5X⁻ + XO₃⁻ + 3H₂O. Check six X atoms, six O atoms and six H atoms on both sides, and charge −6 on each. The coefficient ratio comes from electron balance; OH⁻ and water complete atom and charge balance.

Pearson 8CH0/01 June 2023 Q6(b) used a supplied bromate(V) product and asked for justification of balancing. Its report warns against producing HBr in sodium hydroxide solution. In alkaline conditions a free hydrogen halide would react, so use the appropriate halide ion rather than mixing acid products into an alkaline equation.

A reliable final check has three parts

First check each element’s atom count. Second sum algebraic charge separately on each side. Third verify that the electron loss inferred from oxidation-number increases equals the gain inferred from decreases. Atom balance alone does not guarantee charge balance.

When a test-tube reaction is discussed, separate observation from inference. ‘The solution becomes orange’ is an observation; ‘bromide was oxidised to bromine’ is an interpretation supported by the products and electron accounting. Work on a microscale with eye protection under the prescribed risk assessment, especially for halogens and strongly acidic mixtures. Topic 4 develops the specific observations and controls.

Quick checks

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

Sources