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

These revision notes cover section 3.1.5, Oxidation, reduction and redox equations (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. All of 3.1.5 is International AS content and is also part of the full International A-level. It belongs to Unit 1: Inorganic 1 and Physical 1, and Unit 5: Practical and synoptic can assess all content.

For full explanations and longer worked examples, read the study guide. To test yourself on exam-style questions, use the practice questions. The OxfordAQA A-Level Chemistry hub and printable checklist cover the rest of the course, and the free diagnostics help you decide what to revise next.

3.1.5.1 Definitions

Term	Specification wording
Oxidation	the process of electron loss
Reduction	the process of electron gain
Oxidising agent	electron acceptor (so it is reduced)
Reducing agent	electron donor (so it is oxidised)
Redox reaction	electrons transfer from the reducing agent to the oxidising agent

Using oxidation states:

• oxidation state rises → that element is oxidised
• oxidation state falls → that element is reduced

Oxidation state rules

Higher rules beat lower ones when they clash.

Rule	Value
Uncombined element	0
Neutral compound (sum)	0
Ion (sum)	charge on the ion
F in compounds	−1 always
Group 1 / Group 2 / Al in compounds	+1 / +2 / +3
H	+1; −1 in metal hydrides
O	−2; −1 in peroxides; positive with F
Cl	−1; positive with O or F

Convention: oxidation state +3; ionic charge 3+. Roman numerals in names give oxidation states: iodate(VII), lead(IV) oxide, copper(I) oxide.

Oxidation states to recognise quickly

Each value follows from the rules. In an exam, still show a line of working if the question says “deduce”.

Element	Species and oxidation state
Nitrogen	N₂O₅ +5, NO₂⁻ +3, N₂ 0, Li₃N −3
Chlorine	ClO₄⁻ +7, ClO₂ +4, Cl₂O +1, Cl⁻ −1
Iodine	IO₄⁻ +7, IO₃⁻ +5, ICl₃ +3, I₂ 0
Lead	PbO₂ +4, Pb²⁺ +2, Pb 0
Copper	CuO and Cu²⁺ +2, Cu₂O and Cu⁺ +1
Carbon	CO₂ +4, HCOOH +2, HCHO 0, CH₃OH −2
Hydrogen	H₂O +1, H₂ 0, LiH −1
Oxygen	most oxides −2, BaO₂ −1, O₂ 0, O₂F₂ +1

Half-equations without H⁺ or H₂O

Simple atoms and monatomic ions only need electrons to balance the charge. Balance the atoms first, then add electrons.

Change	Half-equation	Process
nickel atom to nickel ion	Ni → Ni²⁺ + 2e⁻	oxidation
lead ion to lead atom	Pb²⁺ + 2e⁻ → Pb	reduction
oxide ions to oxygen	2O²⁻ → O₂ + 4e⁻	oxidation
cobalt(III) ion to cobalt(II) ion	Co³⁺ + e⁻ → Co²⁺	reduction

The number in front of e⁻ equals the total change in oxidation state. In 2O²⁻ → O₂ + 4e⁻, two oxygen atoms each rise by 2, so four electrons are lost.

Method in steps

Oxidation state from a formula

1. Let x be the unknown element’s oxidation state.
2. Fill in known values from the rules.
3. Set the total equal to 0 (compound) or the ion charge.
4. Solve for x per atom. A fractional answer is an average and stands as it is.

Reminder: N in N₂O₅ → 2x + 5(−2) = 0 → x = +5.

Deciding what is oxidised and reduced

1. Label every element on both sides with its oxidation state.
2. Ignore elements that do not change.
3. Up = oxidised (the species containing it is the reducing agent).
4. Down = reduced (the species containing it is the oxidising agent).

Reminder: WO₃ + 3H₂ → W + 3H₂O. W falls from +6 to 0, so WO₃ is reduced and is the oxidising agent. H rises from 0 to +1, so H₂ is oxidised and is the reducing agent. Six electrons move per W atom, supplied by three H₂ molecules.

Half-equation in acidic solution

1. Balance the element that changes.
2. Balance O with H₂O.
3. Balance H with H⁺.
4. Balance charge with e⁻.
5. Check: e⁻ = change in oxidation state × atoms changing.

Reminder (iodate(VII) to iodate(V)):

IO₄⁻ + 2H⁺ + 2e⁻ → IO₃⁻ + H₂O

I falls from +7 to +5: two electrons gained, so this is a reduction.

Combining half-equations

1. Scale so that electrons lost = electrons gained.
2. Add the two half-equations.
3. Cancel e⁻, then any H⁺ and H₂O on both sides.
4. Check atoms and total charge.

Reminder (copper(I) ions oxidised by oxygen in acid):

O₂ + 4H⁺ + 4e⁻ → 2H₂O
Cu⁺ → Cu²⁺ + e⁻          (× 4)
4Cu⁺ + O₂ + 4H⁺ → 4Cu²⁺ + 2H₂O

Charge: left +4 + 4 = +8; right +8.

Is it a redox reaction at all?

1. Work out the oxidation state of each element on both sides.
2. If at least one element rises and another falls, the reaction is redox.
3. If nothing changes, it is not redox, however many ions are involved.

Reminder: SO₃ + H₂O → H₂SO₄. Sulfur is +6 on both sides, hydrogen stays +1 and oxygen stays −2, so this is not a redox reaction. Acid–base reactions and precipitation reactions are usually in this group, but check the oxidation states rather than relying on the reaction type.

Must-know distinctions

Pair	How to tell them apart
Oxidising agent / substance oxidised	The oxidising agent takes electrons, so its own oxidation state falls
+2 / 2+	Sign first for an oxidation state; number first for the charge on an ion
Half-equation / full equation	Only a half-equation shows e⁻
Reduction / oxidation half-equation	Look at where e⁻ sits: with the reactants for reduction, with the products for oxidation
H at +1 / H at −1	−1 only when hydrogen is bonded to a metal (LiH, MgH₂)
O at −2 / O at −1 or above	Peroxides give −1; bonding to fluorine makes O positive
Per-atom value / element total	Divide by the number of atoms of that element
All atoms change / some atoms change	Track each atom: in PbO₂ + 4HCl, two Cl atoms stay at −1

Final checks on any redox equation

• Count each element on the left and on the right.
• Add up the charges on each side; the two totals must agree.
• The electron count from the oxidation must match the count from the reduction, and the final equation shows none.
• After cancelling, no H⁺ or H₂O is left on both sides.
• At least one oxidation state rises and another falls; if not, it is not redox.

Quick self-test

Cover the answers and work through each one.

1. State the oxidation state of manganese in Mn₂O₃.
2. State the oxidation state of chlorine in Cl₂O.
3. State the oxidation state of phosphorus in P₄O₁₀.
4. State the oxidation state of iodine in ICl₃.
5. State the oxidation state of sulfur in Na₂S₂O₅.
6. State the oxidation state of carbon in methanol, CH₃OH.
7. State the oxidation state of copper in Cu₂O.
8. Write a half-equation for the oxidation of methanol, CH₃OH, to methanal, HCHO, in acidic solution.
9. Combine Cl₂ + 2e⁻ → 2Cl⁻ with Ti³⁺ + H₂O → TiO²⁺ + 2H⁺ + e⁻ to give the overall equation.
10. Sodium reacts with water: 2Na + 2H₂O → 2NaOH + H₂. Identify the oxidising agent.
11. Is 2NO₂ → N₂O₄ a redox reaction? Explain.

Answers

1. +3. 2x + 3(−2) = 0.
2. +1. 2x + (−2) = 0. Chlorine is bonded to oxygen, so it is positive.
3. +5. 4x + 10(−2) = 0, so 4x = +20.
4. +3. Chlorine is bonded to iodine, not O or F, so it is −1: x + 3(−1) = 0.
5. +4. 2(+1) + 2x + 5(−2) = 0, so 2x = +8.
6. −2. x + 4(+1) + (−2) = 0.
7. +1. 2x + (−2) = 0.
8. CH₃OH → HCHO + 2H⁺ + 2e⁻. Carbon rises from −2 to 0, so two electrons are lost.
9. Multiply the titanium half-equation by 2, then add: Cl₂ + 2Ti³⁺ + 2H₂O → 2Cl⁻ + 2TiO²⁺ + 4H⁺. Charge: left +6, right −2 + 4 + 4 = +6.
10. H₂O. Hydrogen falls from +1 in H₂O to 0 in H₂, so water accepts electrons from sodium.
11. No. Nitrogen is +4 in NO₂ and +4 in N₂O₄, and oxygen stays −2. No oxidation state changes.

Where marks are usually lost

• Writing “oxidation is gain of oxygen” when the question asks for a definition. Use electron loss.
• Saying the oxidising agent is oxidised.
• Writing an oxidation state without a sign, or as a charge (2+ instead of +2).
• Using O = −2 in a compound with fluorine, or in a peroxide.
• Giving H as +1 in a metal hydride.
• Rounding an average such as +8/3 to a whole number.
• Giving the total for all the atoms instead of the value per atom (for example +20 for P in P₄O₁₀).
• Balancing a half-equation’s charge with H⁺ instead of electrons.
• Leaving electrons, or H⁺ and H₂O on both sides, in the final equation.
• Naming an element (“chlorine”) when the question asks for a species such as Cl₂ or Cl⁻.

Related pages

• Study guide for this section
• Practice questions for this section
• Atomic structure: electron arrangements and ions
• Amount of substance: using reacting ratios
• 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.