OxfordAQA A-Level Chemistry: Oxidation, reduction and redox equations (9620)

This study guide covers section 3.1.5, Oxidation, reduction and redox equations (subsection 3.1.5.1), of the OxfordAQA International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards. The specification labels the whole of 3.1.5 as International AS content, so it is also part of the full International A-level; nothing here is “International A-level only”. It sits in the content for Unit 1: Inorganic 1 and Physical 1, and Unit 5: Practical and synoptic can assess all content. The section lists no practical activity.

Use it with the revision notes and the practice questions. The OxfordAQA A-Level Chemistry hub and the printable checklist list every other section, and the free diagnostics show where your gaps are.

What this section covers

Spec section	What you must know or be able to do	Stage
3.1.5.1	Oxidation is the process of electron loss; oxidising agents are electron acceptors	International AS
3.1.5.1	Reduction is the process of electron gain; reducing agents are electron donors	International AS
3.1.5.1	The rules for assigning oxidation states	International AS
3.1.5.1	Work out the oxidation state of an element in a compound or ion from the formula	International AS
3.1.5.1	Write half-equations identifying the oxidation and reduction processes in redox reactions	International AS
3.1.5.1	Combine half-equations to give an overall redox equation	International AS

The specification’s overview says redox reactions involve a transfer of electrons from the reducing agent to the oxidising agent, and that the change in oxidation state of an element is used to identify which element has been oxidised or reduced.

Electron transfer: the four definitions

Learn these in the specification’s words:

• Oxidation is the process of electron loss.
• Reduction is the process of electron gain.
• An oxidising agent is an electron acceptor. Because it gains electrons, it is reduced.
• A reducing agent is an electron donor. Because it loses electrons, it is oxidised.

The two processes always happen together. Electrons lost by one species must be gained by another.

Worked example 1: calcium burning in oxygen

2Ca + O₂ → 2CaO

Calcium oxide is ionic, made of Ca²⁺ and O²⁻ ions.

1. Each calcium atom becomes Ca²⁺, so it loses two electrons: Ca → Ca²⁺ + 2e⁻. Calcium is oxidised.
2. Each O₂ molecule becomes two O²⁻ ions, so it gains four electrons: O₂ + 4e⁻ → 2O²⁻. Oxygen is reduced.
3. Two calcium atoms supply the four electrons one O₂ molecule needs. That is why the equation has 2Ca to 1O₂.
4. Calcium is the reducing agent (electron donor). Oxygen is the oxidising agent (electron acceptor).

“Gain of oxygen” happens to describe calcium here, but the specification defines oxidation by electrons. Use the electron definition in every answer.

Rules for assigning oxidation states

An oxidation state is the charge an atom would carry if all its bonds were ionic. Apply these rules. Where two rules clash, the one higher in the list wins.

Rule	Oxidation state
Atoms in an uncombined element (Ca, O₂, P₄)	0
Sum for a neutral compound	0
Sum for an ion	the charge on the ion
Fluorine in any compound	−1
Group 1 metals / Group 2 metals / aluminium in compounds	+1 / +2 / +3
Hydrogen in compounds	+1, but −1 when bonded to a metal (metal hydrides)
Oxygen in compounds	−2, but −1 in peroxides, and positive when bonded to fluorine
Chlorine in compounds	−1, but positive when bonded to oxygen or fluorine

Write an oxidation state sign first (+3). Write an ionic charge number first (3+). Roman numerals in names give the oxidation state: lead(IV) oxide has lead at +4; bromate(V) has bromine at +5.

Worked example 2: oxidation states from formulae

Use x for the unknown and make the total equal the overall charge.

(a) Nitrogen in NO₂⁻

x + 2(−2) = −1
x = +3

(b) Chlorine in ClO₄⁻

Chlorine is bonded to oxygen, so oxygen takes −2 and chlorine is positive:

x + 4(−2) = −1
x = +7

(c) Oxygen in barium peroxide, BaO₂

Barium is Group 2, so +2. This is a peroxide:

+2 + 2x = 0
x = −1

(d) Hydrogen in lithium hydride, LiH

Lithium, in Group 1, is +1 and the compound is neutral, so hydrogen is −1. Hydrogen bonded to a metal takes the negative value.

(e) Oxygen in O₂F₂

Fluorine is always −1, which ranks above the oxygen rule:

2x + 2(−1) = 0
x = +1

(f) Iron in Fe₃O₄

3x + 4(−2) = 0
x = +8/3 (about +2.67)

A fraction is an average over several atoms. Fe₃O₄ contains iron in two different oxidation states, and the formula only gives the mean. If asked, give +8/3.

Spotting what is oxidised and what is reduced

Write the oxidation state of every element on both sides. An increase means that element is oxidised. A decrease means it is reduced.

Worked example 3: two reactions with no obvious ions

(a) The thermite reaction

2Al + Fe₂O₃ → Al₂O₃ + 2Fe

Element	Left	Right	Change	Process
Al	0	+3	up 3	oxidised
Fe	+3	0	down 3	reduced
O	−2	−2	none	not involved

Aluminium is the reducing agent. Fe₂O₃ (specifically the Fe³⁺ in it) is the oxidising agent. Electron check: 2 Al atoms lose 3 each, 6 in total; 2 Fe atoms gain 3 each, 6 in total.

(b) Lead(IV) oxide with concentrated hydrochloric acid

PbO₂ + 4HCl → PbCl₂ + Cl₂ + 2H₂O

Lead goes from +4 to +2, so PbO₂ is reduced and is the oxidising agent. Chlorine starts at −1 in all four HCl. Two of those chlorine atoms end at −1 in PbCl₂ and do not change. The other two end at 0 in Cl₂, so they are oxidised. Only half the HCl acts as the reducing agent. The rest supplies chloride ions and hydrogen. Electron check: Pb gains 2; two Cl atoms lose 1 each.

Writing half-equations

A half-equation shows either the oxidation or the reduction on its own, with electrons included. Electrons on the left mean reduction. Electrons on the right mean oxidation.

For species containing oxygen in acidic solution:

1. Balance the element whose oxidation state changes.
2. Balance oxygen with H₂O.
3. Balance hydrogen with H⁺.
4. Balance charge with electrons.
5. Check: electrons = change in oxidation state × number of atoms that change.

Worked example 4: bromate(V) ions reduced to bromide ions

BrO₃⁻ → Br⁻

1. One Br on each side.
2. Three O on the left: add 3H₂O on the right.
3. Six H on the right: add 6H⁺ on the left.
4. Charge on left = −1 + 6 = +5. Charge on right = −1. Add 6e⁻ to the left.

BrO₃⁻ + 6H⁺ + 6e⁻ → Br⁻ + 3H₂O

5. Check: Br goes from +5 to −1, a fall of 6, and 6 electrons are gained. This is a reduction.

Worked example 5: nitrite ions oxidised to nitrate ions

NO₂⁻ → NO₃⁻

1. One N on each side.
2. The right has one extra O: add H₂O on the left.
3. Two H on the left: add 2H⁺ on the right.
4. Charge on left = −1. Charge on right = −1 + 2 = +1. Add 2e⁻ to the right.

NO₂⁻ + H₂O → NO₃⁻ + 2H⁺ + 2e⁻

5. Check: N goes from +3 to +5, a rise of 2, and 2 electrons are lost. This is an oxidation.

Combining half-equations

1. Write both half-equations.
2. Multiply so electrons lost equal electrons gained.
3. Add them and cancel the electrons.
4. Cancel any H⁺ and H₂O that appear on both sides.
5. Check atoms and total charge.

Worked example 6: bromate(V) and nitrite ions

From worked examples 4 and 5, the reduction takes 6e⁻ and the oxidation releases 2e⁻. Multiply the oxidation by 3:

BrO₃⁻ + 6H⁺ + 6e⁻ → Br⁻ + 3H₂O
3NO₂⁻ + 3H₂O → 3NO₃⁻ + 6H⁺ + 6e⁻

Add. The 6e⁻, 6H⁺ and 3H₂O all appear on both sides and cancel completely:

BrO₃⁻ + 3NO₂⁻ → Br⁻ + 3NO₃⁻

Charge check: left = −1 − 3 = −4; right = −1 − 3 = −4. An overall equation can end up with no H⁺ or H₂O even though both half-equations needed them.

Worked example 7: lead(IV) oxide and manganese(II) ions

Lead(IV) oxide oxidises Mn²⁺ ions to purple manganate(VII) ions in acid.

Reduction:  PbO₂ + 4H⁺ + 2e⁻ → Pb²⁺ + 2H₂O
Oxidation:  Mn²⁺ + 4H₂O → MnO₄⁻ + 8H⁺ + 5e⁻

The lowest common multiple of 2 and 5 is 10. Multiply the reduction by 5 and the oxidation by 2:

5PbO₂ + 20H⁺ + 10e⁻ → 5Pb²⁺ + 10H₂O
2Mn²⁺ + 8H₂O → 2MnO₄⁻ + 16H⁺ + 10e⁻

Add and cancel. 20H⁺ on the left minus 16H⁺ on the right leaves 4H⁺ on the left. 10H₂O on the right minus 8H₂O on the left leaves 2H₂O on the right:

5PbO₂ + 4H⁺ + 2Mn²⁺ → 5Pb²⁺ + 2MnO₄⁻ + 2H₂O

Charge check: left = +4 + 4 = +8; right = +10 − 2 = +8. Balanced. The reacting ratio PbO₂ : Mn²⁺ is 5 : 2, which you would use in any mole calculation (see amount of substance).

Common errors

• Calling the oxidising agent “the substance oxidised”. It accepts electrons, so it is reduced.
• Applying the oxygen rule before the fluorine rule, which gives the wrong value in O₂F₂.
• Treating hydrogen as +1 in a metal hydride, or oxygen as −2 in a peroxide.
• Giving the total for all atoms of an element (for example +8 for iron in Fe₃O₄) instead of the value per atom.
• Balancing charge with extra H⁺ rather than with electrons.
• Putting electrons on the right of a reduction half-equation.
• Forgetting to cancel H⁺ or H₂O that end up on both sides, as in worked example 7.
• Assuming every atom of an element changes. In worked example 3(b), only half the chlorine atoms are oxidised.

Where to go next

• Condensed summary and a quick self-test: revision notes
• Original questions with mark-by-mark answers: practice questions
• Ion charges from electron arrangements: atomic structure
• Ionic bonding and ions: bonding and structure
• Mole ratios from balanced equations: amount of substance
• Find your weak areas: free diagnostics

Official syllabus

OxfordAQA International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards, section 3.1.5 Oxidation, reduction and redox equations (3.1.5.1). Published by OxfordAQA.