1. Oxidation and Reduction
A redox reaction is a reaction in which oxidation and reduction happen at the same time. Electrons move from one species to another.
Oxidation is the process of electron loss.
Reduction is the process of electron gain.
A useful memory aid is OIL RIG: Oxidation Is Loss, Reduction Is Gain (of electrons).
Oxidising and reducing agents
- An oxidising agent is an electron acceptor. It takes electrons from another species, so it is itself reduced.
- A reducing agent is an electron donor. It gives electrons to another species, so it is itself oxidised.
- Each Cl atom in Cl₂ gains an electron to form Cl⁻. Chlorine is reduced.
- Each Br⁻ ion loses an electron to form Br. Bromide is oxidised.
- Cl₂ is the oxidising agent. It accepts electrons.
- Br⁻ is the reducing agent. It donates electrons.
You will meet this displacement reaction again on the Group 7 page.
2. Rules for Oxidation States
An oxidation state shows how many electrons an atom has lost or gained, compared with the uncombined element. An increase in oxidation state is oxidation. A decrease is reduction.
| Rule | Examples |
|---|---|
| An uncombined element has an oxidation state of 0. | Na, Cl₂, O₂, S₈ are all 0 |
| The oxidation states in a neutral compound add up to 0. In an ion they add up to the charge on the ion. | H₂O: 2(+1) + (-2) = 0 SO₄²⁻: (+6) + 4(-2) = -2 |
| Group 1 metals are +1, Group 2 metals are +2 and aluminium is +3. | Na in NaCl is +1; Mg in MgO is +2; Al in Al₂O₃ is +3 |
| Fluorine is always -1. | F in HF, OF₂ and CaF₂ is -1 |
| Hydrogen is +1, except in metal hydrides, where it is -1. | H in HCl is +1; H in NaH and CaH₂ is -1 |
| Oxygen is -2, except in peroxides (-1) and when combined with fluorine. | O in H₂O₂ is -1; O in OF₂ is +2 |
| Chlorine is -1, except when it is combined with oxygen or fluorine. | Cl in NaCl is -1; Cl in NaClO is +1 |
Write the sign first, then the number: +6, not 6+. Charges on ions are written the other way round (2+).
Roman numerals in a name give the oxidation state: iron(III) means Fe is +3, chlorate(I) means Cl is +1 and manganate(VII) means Mn is +7.
3. Working Out Oxidation States
Call the unknown oxidation state x. Use the rules to fill in the other atoms. Then make the total equal to the overall charge.
Step 1: O is -2. There are four O atoms.
Step 2: The total must equal the charge on the ion, -2.
Step 3: x + 4(-2) = -2, so x - 8 = -2.
Answer: x = +6.
| Species | Working | Answer |
|---|---|---|
| S in SO₄²⁻ | x + 4(-2) = -2 | +6 |
| Cl in ClO₃⁻ | x + 3(-2) = -1 | +5 |
| Cr in Cr₂O₇²⁻ | 2x + 7(-2) = -2 | +6 |
| N in NH₄⁺ | x + 4(+1) = +1 | -3 |
| O in H₂O₂ | 2(+1) + 2x = 0 | -1 |
| H in NaH | +1 + x = 0 | -1 |
| Mn in MnO₄⁻ | x + 4(-2) = -1 | +7 |
| S in S₂O₃²⁻ | 2x + 3(-2) = -2 | +2 |
| Cl in NaClO | +1 + x + (-2) = 0 | +1 |
In Cr₂O₇²⁻ and S₂O₃²⁻ there are two atoms of the element. Write 2x, not x. The answer is the oxidation state of one atom.
4. Writing Half-Equations
A half-equation shows either the oxidation or the reduction on its own. It includes the electrons. Electrons on the left mean reduction (gain). Electrons on the right mean oxidation (loss).
Method for half-equations in acid
- Balance the element being oxidised or reduced.
- Balance O by adding H₂O.
- Balance H by adding H⁺.
- Balance charge by adding electrons.
Step 1 (Mn): one Mn on each side. MnO₄⁻ → Mn²⁺
Step 2 (O): four O on the left, so add 4H₂O on the right. MnO₄⁻ → Mn²⁺ + 4H₂O
Step 3 (H): eight H on the right, so add 8H⁺ on the left. MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O
Step 4 (charge): the left is -1 + 8 = +7 and the right is +2. Add 5e⁻ to the left.
MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
Check: charge on the left -1 + 8 - 5 = +2, on the right +2. Mn goes from +7 to +2, which is a gain of 5 electrons.
Half-equations to know how to write
| Half-equation | Process |
|---|---|
| Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O | Reduction |
| SO₄²⁻ + 4H⁺ + 2e⁻ → SO₂ + 2H₂O | Reduction |
| SO₄²⁻ + 8H⁺ + 6e⁻ → S + 4H₂O | Reduction |
| SO₄²⁻ + 10H⁺ + 8e⁻ → H₂S + 4H₂O | Reduction |
| 2Br⁻ → Br₂ + 2e⁻ | Oxidation |
| 2I⁻ → I₂ + 2e⁻ | Oxidation |
| Fe²⁺ → Fe³⁺ + e⁻ | Oxidation |
| H₂O₂ → O₂ + 2H⁺ + 2e⁻ | Oxidation |
The three sulfate half-equations show sulfur being reduced from +6 to +4 (SO₂), 0 (S) and -2 (H₂S). The number of electrons matches each change in oxidation state.
5. Combining Half-Equations
Method
- Multiply so the electrons lost equal the electrons gained.
- Add the two half-equations.
- Cancel the electrons, and any H⁺ or H₂O that appears on both sides.
- Check that the atoms and the charge balance.
Reduction: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O (gains 5e⁻)
Oxidation: Fe²⁺ → Fe³⁺ + e⁻ (loses 1e⁻), so multiply by 5: 5Fe²⁺ → 5Fe³⁺ + 5e⁻
Add and cancel 5e⁻:
MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O
Check: charge on the left -1 + 8 + 10 = +17, on the right +2 + 15 = +17.
This reaction is used in manganate(VII) titrations. See Transition Metals: Variable Oxidation States and Catalysts (Unit 3).
Reduction: SO₄²⁻ + 4H⁺ + 2e⁻ → SO₂ + 2H₂O (gains 2e⁻)
Oxidation: 2Br⁻ → Br₂ + 2e⁻ (loses 2e⁻). The electrons already match.
2Br⁻ + SO₄²⁻ + 4H⁺ → Br₂ + SO₂ + 2H₂O
Check: charge on the left -2 - 2 + 4 = 0, on the right 0.
See the reactions of solid halides with concentrated sulfuric acid on the Group 7 page.
Reduction: SO₄²⁻ + 10H⁺ + 8e⁻ → H₂S + 4H₂O (gains 8e⁻)
Oxidation: 2I⁻ → I₂ + 2e⁻ (loses 2e⁻), so multiply by 4: 8I⁻ → 4I₂ + 8e⁻
8I⁻ + SO₄²⁻ + 10H⁺ → 4I₂ + H₂S + 4H₂O
Check: charge -8 - 2 + 10 = 0 on both sides; H 10 = 2 + 8; O 4 = 4.
Go deeper
These UK AQA pages cover the same ideas in more detail. They may include content that is not in 9620.
6. Practice Questions
Show Answer & Mark Scheme
- N in NO₃⁻: x + 3(-2) = -1, so +5 (1 mark)
- V in VO₂⁺: x + 2(-2) = +1, so +5 (1 mark)
- V in VO²⁺: x + (-2) = +2, so +4 (1 mark)
Show Answer & Mark Scheme
- Fe³⁺ (1 mark)
- It accepts an electron and is reduced to Fe²⁺ (1 mark)
Show Answer & Mark Scheme
H₂O₂ → O₂ + 2H⁺ + 2e⁻ (1 mark)
Show Answer & Mark Scheme
Multiply the MnO₄⁻ half-equation by 2 and the H₂O₂ half-equation by 5, so 10 electrons are gained and 10 are lost (1 mark). Then cancel 10H⁺ from both sides.
2MnO₄⁻ + 5H₂O₂ + 6H⁺ → 2Mn²⁺ + 5O₂ + 8H₂O (1 mark)
Check: e⁻ 10 = 10; charge -2 + 6 = +4 on both sides; H 10 + 6 = 16; O 8 + 10 = 18 = 10 + 8.
Show Answer & Mark Scheme
- Cl₂ 0, NaCl -1, NaClO₃ +5 (1 mark)
- Chlorine is both oxidised and reduced in the same reaction (1 mark)
Beyond the OxfordAQA specification: the reaction of chlorine with hot alkali is not in 9620. It is used here only to practise oxidation states.