OCR A Level Chemistry A: Redox (H432)

This study guide teaches section 2.1.5, Redox, of the OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017. The section sits in Module 2, Foundations in chemistry, and has six lettered learning outcomes, 2.1.5(a) to (f), all covered below. H432 is the full A level. The specification states that Module 2 content is assessed in Periodic table, elements and physical chemistry (01), Synthesis and analytical techniques (02) and Unified chemistry (03). No PAG is listed against this section.

Every H432 topic is linked from the OCR chemistry hub; track outcomes on the printable checklist.

Outcomes in 2.1.5

Outcome	Skill you need	Additional guidance in the specification
2.1.5(a)	Apply the rules for assigning and calculating oxidation numbers for atoms in elements, compounds and ions	Know the oxidation number of O in peroxides and H in metal hydrides
2.1.5(b)	Write formulae using oxidation numbers	HSW8: appropriate use of oxidation numbers in written communication
2.1.5(c)	Use a Roman numeral to show the oxidation number when an element has more than one	Iron(II) and iron(III), and others; formulae from names such as chlorate(I) and chlorate(III) and back; “nitrate” and “sulfate” with no number mean NO₃⁻ and SO₄²⁻; HSW8
2.1.5(d)	Describe oxidation and reduction as (i) electron transfer and (ii) changes in oxidation number	Examples from s-, p- and d-block elements
2.1.5(e)	Write full equations for redox reactions of metals with acids to form salts	Metals from s-, p- and d-blocks, e.g. Mg, Al, Fe, Zn; ionic equations not required; only reactions giving a salt and hydrogen
2.1.5(f)	Interpret the equations in (e), and unfamiliar redox reactions, to make predictions using oxidation numbers and electron loss or gain	M0.2 (ratios, fractions and percentages); nitric acid and concentrated sulfuric acid reactions could appear here

The terms oxidising agent and reducing agent, and building equations from half-equations, belong to 5.2.3 later in the course. This guide does not rely on them.

2.1.5(a): Rules for oxidation numbers

An oxidation number shows how many electrons an atom has lost (positive) or gained (negative) compared with the uncombined element, counting every bond as if it were ionic. Apply these rules in order; when two clash, the higher one wins.

1. An atom in an uncombined element has oxidation number 0. This applies to Fe, Cu, N₂, O₂ and S₈ alike.
2. In a compound, the oxidation numbers add up to 0.
3. In an ion, they add up to the charge on the ion. For a simple ion such as Br⁻ or Ca²⁺, the oxidation number is just the charge.
4. Fluorine in a compound is always −1.
5. Group 1 metals are +1, Group 2 metals are +2 and aluminium is +3 in their compounds.
6. Hydrogen is +1, except in metal hydrides, where it is −1.
7. Oxygen is −2, except in peroxides, where it is −1.
8. Chlorine is −1, except when it is bonded to oxygen or fluorine.

Learn the two named exceptions as a pair. In a metal hydride such as KH, hydrogen takes the negative value. In a peroxide, the two oxygen atoms are bonded to each other in an O–O unit, so each carries only −1.

Write oxidation numbers with the sign first: +3, −2. Write ionic charges with the number first: 3+, 2−. This is part of the HSW8 point about clear communication.

Worked example 1: oxidation numbers from formulae

Let the unknown oxidation number be x. Add up the known contributions and set the total equal to the overall charge.

Phosphorus in H₂PO₄⁻

2(+1) + x + 4(−2) = −1
x − 6 = −1
x = +5

Carbon in the ethanedioate ion, C₂O₄²⁻

2x + 4(−2) = −2
2x = +6
x = +3

Manganese in Mn₂O₇

2x + 7(−2) = 0
x = +7

Oxygen in lithium peroxide, Li₂O₂

Lithium is in Group 1, so it is +1. The compound is neutral.

2(+1) + 2x = 0
x = −1

Hydrogen in potassium hydride, KH

K is a Group 1 metal at +1. KH has no overall charge, which leaves hydrogen at −1.

2.1.5(b) and (c): Formulae and Roman numerals

When an element, often a d-block metal, has more than one oxidation number, the name shows which one with a Roman numeral in brackets: iron(II), iron(III), tin(IV). For an anion containing oxygen, the numeral refers to the central element, so chlorate(I) means chlorine is +1.

The chlorate family shows how the numeral fixes the formula:

Name	Oxidation number of Cl	Formula of ion
chlorate(I)	+1	ClO⁻
chlorate(III)	+3	ClO₂⁻
chlorate(V)	+5	ClO₃⁻
chlorate(VII)	+7	ClO₄⁻

Each ion is 1−, so the oxidation number fixes the number of oxygens. Check chlorate(III): x + 2(−2) = −1 gives x = +3.

When a name says only “nitrate” or “sulfate”, with no numeral, the specification treats it as NO₃⁻ or SO₄²⁻. Other oxidation numbers must be shown, for example nitrate(III) for NO₂⁻ and sulfate(IV) for SO₃²⁻.

Worked example 2: from name to formula

Treat the oxidation number as the ion’s charge, then balance charges to zero.

• Chromium(III) sulfate. Cr is +3, sulfate is SO₄²⁻. The lowest total charge both fit into is 6, so 2Cr³⁺ and 3SO₄²⁻: Cr₂(SO₄)₃.
• Calcium chlorate(I). Ca²⁺ needs two ClO⁻ ions: Ca(ClO)₂.
• Potassium nitrate(III). Nitrogen +3 with 1− overall gives NO₂⁻, so KNO₂.

Worked example 3: from formula to name

Find the varying element’s oxidation number, then add the numeral.

• NaBrO₃. Na +1, three O give −6, so Br is +5: sodium bromate(V).
• K₂SO₃. 2(+1) + x + 3(−2) = 0, so S is +4: potassium sulfate(IV).
• Co(NO₃)₂. Two nitrate ions give −2, so Co is +2: cobalt(II) nitrate.

Do not add a numeral where an element has only one oxidation number in its compounds: not “sodium(I) chloride”.

2.1.5(d): Oxidation and reduction

	(i) Electron transfer	(ii) Oxidation number
Oxidation	loss of electrons	increase in oxidation number
Reduction	gain of electrons	decrease in oxidation number

Both happen at once, and total electrons lost equal total gained. The oxidation number view still works when a reaction has no obvious ions.

Worked example 4: one example from each block

s-block: lithium burning in nitrogen

6Li + N₂ → 2Li₃N

Each Li goes from 0 to +1, losing one electron: Li → Li⁺ + e⁻. Lithium is oxidised. Each N goes from 0 to −3, gaining three electrons. Nitrogen is reduced. Electron check: 6 Li atoms lose 6 electrons; 2 N atoms gain 2 × 3 = 6.

p-block: aluminium with bromine

2Al + 3Br₂ → 2AlBr₃

Al: 0 to +3, oxidised, three electrons lost per atom, 6 in total. Br: 0 to −1, reduced, one electron gained per atom; 6 Br atoms gain 6.

d-block: iron(II) chloride with chlorine

2FeCl₂ + Cl₂ → 2FeCl₃

Fe: +2 to +3, oxidised, one electron lost per ion. Cl in Cl₂: 0 to −1, reduced. The chloride already in FeCl₂ stays at −1, so it is neither oxidised nor reduced. Electron check: 2 Fe lose 2; 2 Cl atoms from Cl₂ gain 2.

So say which atoms change: one population of an element can change while another does not.

2.1.5(e): Metals with acids

A reactive metal and a dilute acid give a salt and hydrogen. You need full equations with correct salt formulae. The specification states that ionic equations are not required here, and limits (e) to reactions that produce a salt and hydrogen.

Four examples, one per metal named in the guidance:

Mg(s) + 2CH₃COOH(aq) → (CH₃COO)₂Mg(aq) + H₂(g)
2Al(s) + 3H₂SO₄(aq) → Al₂(SO₄)₃(aq) + 3H₂(g)
Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g)
Fe(s) + H₂SO₄(aq) → FeSO₄(aq) + H₂(g)

Iron gives the iron(II) salt with dilute acids: FeSO₄, not Fe₂(SO₄)₃. The salt formula follows from the metal’s oxidation number, so (e) builds on (b) and (c).

Look at the aluminium reaction with oxidation numbers:

Element	Before	After	Change
Al	0	+3 in Al₂(SO₄)₃	up 3, oxidised
H	+1 in H₂SO₄	0 in H₂	down 1, reduced
S	+6	+6	none
O	−2	−2	none

Every metal-with-dilute-acid reaction follows this pattern: the metal is oxidised, hydrogen is reduced, and the acid’s anion passes through unchanged.

Worked example 5: amounts and electrons

0.486 g of magnesium ribbon reacts completely with excess ethanoic acid. Use Ar(Mg) = 24.3; one mole of gas occupies 24.0 dm³ at room temperature and pressure. Find the volume of hydrogen and the amount of electrons transferred.

1. Amount of Mg = 0.486 ÷ 24.3 = 0.0200 mol.
2. The equation gives 1 mol H₂ per 1 mol Mg, so 0.0200 mol H₂.
3. Volume = 0.0200 × 24.0 = 0.480 dm³ (480 cm³).
4. Each Mg atom loses two electrons (0 to +2), so electrons transferred = 2 × 0.0200 = 0.0400 mol.
5. Check: forming 0.0200 mol H₂ takes up the same 0.0400 mol electrons.

For mole and gas-volume methods, see Amount of substance. For the acid formulae and neutralisation, see Acids.

2.1.5(f): Interpreting unfamiliar redox reactions

Outcome (f) asks you to read any redox equation and make predictions from oxidation numbers and electron loss or gain; nitric acid and concentrated sulfuric acid reactions may appear. A routine:

1. Write the oxidation number of every atom on both sides.
2. Mark which atoms go up (oxidised) and which go down (reduced). Ignore the ones that stay the same.
3. Multiply the change per atom by the number of atoms that change.
4. Check that total electrons lost equal total gained. Use this to predict ratios or missing products.

Worked example 6: copper with concentrated nitric acid

Copper does not release hydrogen from dilute hydrochloric or sulfuric acid, yet it dissolves in concentrated nitric acid:

Cu(s) + 4HNO₃(aq) → Cu(NO₃)₂(aq) + 2NO₂(g) + 2H₂O(l)

• Cu: 0 to +2, oxidised, 2 electrons lost.
• N in HNO₃ is +5. In NO₂ it is +4. Two N atoms go down by 1 each, gaining 2 electrons in total.
• The other two N atoms end up in Cu(NO₃)₂ as nitrate, still +5. They are not reduced.
• H stays +1 and O stays −2.

Electrons lost (2) equal electrons gained (2 × 1). Nitrogen, not hydrogen, is reduced, so the gas is NO₂, not H₂.

Worked example 7: predicting a reacting ratio

Tin(II) chloride solution reacts with iron(III) chloride solution. Tin ends up as tin(IV) and iron as iron(II). Predict the mole ratio and write the equation.

1. Sn: +2 to +4, so each Sn loses 2 electrons.
2. Fe: +3 to +2, so each Fe gains 1 electron.
3. To balance electrons, 1 Sn needs 2 Fe. The ratio SnCl₂ : FeCl₃ is 1 : 2.
4. Equation: SnCl₂ + 2FeCl₃ → SnCl₄ + 2FeCl₂. Check chlorine: 2 + 6 = 8 on the left; 4 + 4 = 8 on the right.

Common errors

• Giving a peroxide oxygen −2 or a hydride hydrogen +1. Check for these two exceptions before anything else.
• Writing 2+ when you mean the oxidation number +2, or leaving the sign off.
• Reporting the total for a group of atoms rather than the value per atom (writing +6 for carbon in C₂O₄²⁻ instead of +3).
• Writing Fe₂(SO₄)₃ or FeCl₃ as the product of iron and a dilute acid.
• Wrong salt formulae, such as AlSO₄ or MgCH₃COO.
• Saying “nitrogen is reduced” in the copper reaction without noting that only the nitrogen ending up in NO₂ changes.

Next steps

• Condensed tables and a quick self-test: Redox revision notes
• Find your gaps: free diagnostics
• Original questions with mark-by-mark answers: Redox practice questions
• Formulae of ions and equation writing: Compounds, formulae and equations

Official syllabus

OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017, section 2.1.5 Redox, learning outcomes (a) to (f). Published by OCR.