What is a redox reaction in the context of Cambridge IGCSE Chemistry?

In Cambridge IGCSE Chemistry, a redox reaction is a type of chemical reaction where oxidation and reduction occur simultaneously. Oxidation involves the loss of electrons, while reduction involves the gain of electrons. These reactions are crucial for understanding processes such as combustion, respiration, and corrosion.

How can you identify a redox reaction?

A redox reaction can be identified by changes in oxidation states of the elements involved. If an element's oxidation state increases, it is oxidized, and if it decreases, it is reduced. Additionally, the presence of electron transfer between reactants is a key indicator of a redox reaction.

What are some common examples of redox reactions?

Common examples of redox reactions include the rusting of iron, where iron is oxidized and oxygen is reduced, and the reaction between hydrogen and oxygen to form water, where hydrogen is oxidized and oxygen is reduced. These examples illustrate the practical importance of redox reactions in everyday life.

Why are redox reactions important in chemical processes?

Redox reactions are fundamental to many chemical processes because they involve energy transfer through electron movement. They are essential in biological systems for energy production, such as in cellular respiration, and in industrial processes like metal extraction and electroplating.

What role do oxidizing and reducing agents play in redox reactions?

In redox reactions, an oxidizing agent is a substance that gains electrons and is reduced, while a reducing agent loses electrons and is oxidized. These agents facilitate the transfer of electrons, driving the chemical reaction forward. Understanding their roles is crucial for predicting the direction and outcome of redox reactions.

Why is "Inverting OIL RIG (saying oxidation is gain of electrons)." a common mistake in Redox?

Why it happens: Memory slip — the oxygen and electron definitions feel opposite. How to avoid it: OIL = Oxidation Is Loss (of electrons). RIG = Reduction Is Gain. With oxygen it is the reverse: oxidation is gain of oxygen. Source: 0620/42 — recurring

Why is "Calling the species that GETS oxidised the oxidising agent." a common mistake in Redox?

Why it happens: The naming feels backwards. How to avoid it: An oxidising AGENT does the oxidising — so it itself gets REDUCED. The reducing agent is the one that gets oxidised.

Why is "Mixing up the manganate(VII) and iodide colour changes." a common mistake in Redox?

Why it happens: Both are diagnostic tests, so they blur together. How to avoid it: Acidified KMnO₄ (oxidising agent): purple → colourless. Iodide being oxidised: colourless → brown (I⁻ → I₂). Source: 0620 Examiner Reports

Why is "Putting electrons on the wrong side of a half-equation or not balancing charge." a common mistake in Redox?

Why it happens: Rushing without checking the charges. How to avoid it: Oxidation: electrons on the RIGHT (lost). Reduction: electrons on the LEFT (gained). Always check the total charge balances both sides.

Cambridge IGCSE Chemistry (0620)

Redox

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Short Notes - Redox

Detailed Study Notes

Detailed notes on Chemical Reactions for Cambridge IGCSE Chemistry, covering key concepts, explanations, examples, and exam-focused revision points.

Redox Reactions — Cambridge IGCSE 0620 Chemistry Extended (2026)

Reduction and oxidation happen together. Three lenses: oxygen, hydrogen, and (Extended) electrons / oxidation numbers. Plus the redox tests.

At a glance

OIL RIG: Oxidation Is Loss; Reduction Is Gain (of electrons).
Three definitions, all consistent: gain/loss of oxygen, hydrogen, OR electrons.
Oxidising agent ITSELF is reduced (it gains electrons from the substrate).
Reducing agent ITSELF is oxidised (it gives electrons to the substrate).
Oxidation number (Extended): track e⁻ transfer using assigned charges.
Test for oxidising agent: turns moist KI paper from white to BROWN-PURPLE.
Test for reducing agent: turns acidified KMnO₄ from PURPLE to COLOURLESS.

What you’ll learn

Mapped to the Cambridge IGCSE 0620 syllabus (2026-2028).

8.2 — Define oxidation and reduction in terms of oxygen, hydrogen, electrons.
8.2 — Identify oxidising and reducing agents in a reaction.
8.2 — Use oxidation numbers to identify what is oxidised/reduced (Extended).
8.2 — Describe tests for oxidising and reducing agents.

Three definitions of oxidation/reduction

Gain or loss of oxygen, hydrogen, or (best) electrons.

Definition 1: Oxygen.

Oxidation = GAIN of oxygen.
Reduction = LOSS of oxygen.

Examples: $\text{C} + \text{O}_{2} \to \text{CO}_{2}$ — carbon is OXIDISED. $\text{Fe}_{2}\text{O}_{3} + 3\text{CO} \to 2\text{Fe} + 3\text{CO}_{2}$ — iron oxide is REDUCED to iron.

Definition 2: Hydrogen.

Oxidation = LOSS of hydrogen.
Reduction = GAIN of hydrogen.

Examples: $\text{H}_{2}\text{S} + \text{Cl}_{2} \to 2\text{HCl} + \text{S}$ — H₂S is OXIDISED (loses H). Cl₂ is REDUCED (gains H).

Definition 3: Electrons (best — works for any redox).

OIL: Oxidation Is Loss of electrons.
RIG: Reduction Is Gain of electrons.

Examples:

$\text{Mg} \to \text{Mg}^{2+} + 2\text{e}^{-}$ . Mg is OXIDISED (loses 2 e⁻).
$\text{Cl}_{2} + 2\text{e}^{-} \to 2\text{Cl}^{-}$ . Cl₂ is REDUCED (gains 2 e⁻).

Definition	Oxidation is…	Reduction is…
1. Oxygen	Gain of oxygen	Loss of oxygen
2. Hydrogen	Loss of hydrogen	Gain of hydrogen
3. Electrons (best)	Loss of electrons (OIL)	Gain of electrons (RIG)

Why electrons are the BEST definition. Some reactions don't involve oxygen or hydrogen — but they still involve electron transfer. Definition 3 covers all cases.

Worked. $\text{Zn} + \text{CuSO}_{4} \to \text{ZnSO}_{4} + \text{Cu}$ .

Zn → Zn²⁺ + 2e⁻ (loses electrons → oxidised).
Cu²⁺ + 2e⁻ → Cu (gains electrons → reduced).
Zn is the reducing AGENT. Cu²⁺ is the oxidising AGENT.

Oxidation Is Loss, Reduction Is Gain — the electrons lost by one species are exactly the electrons gained by the other.

Oxygen: oxidise = gain O.
Hydrogen: oxidise = lose H.
Electrons: OIL RIG.
Electron definition is the most general.

Oxidising and reducing agents

Oxidising agent: reduced (gains e⁻). Reducing agent: oxidised (loses e⁻). They swap.

Oxidising agent. A substance that OXIDISES another substance. To oxidise something else, it must gain electrons (be reduced) itself.

Strong oxidising agents: chlorine (Cl₂), oxygen, fluorine, KMnO₄ (in acid), $\text{H}_{2}\text{O}_{2}$ , $\text{HNO}_{3}$ .

Reducing agent. A substance that REDUCES another substance. To reduce something else, it must lose electrons (be oxidised) itself.

Strong reducing agents: alkali metals, hydrogen, carbon, KI, sulfur dioxide.

Worked. $\text{Cl}_{2} + 2\text{KI} \to 2\text{KCl} + \text{I}_{2}$ .

Cl₂ → Cl⁻: gains electrons → REDUCED. Cl₂ is the OXIDISING AGENT.
2I⁻ → I₂: loses electrons → OXIDISED. I⁻ (in KI) is the REDUCING AGENT.

Tip. Each agent does the OPPOSITE of its name to itself. Oxidising agent is REDUCED; reducing agent is OXIDISED.

An agent always does the opposite of its name to itself: the oxidiser is reduced, the reducer is oxidised.

Oxidising agent: reduced itself.
Reducing agent: oxidised itself.
Strong oxidisers: Cl₂, O₂, KMnO₄, H₂O₂.
Strong reducers: alkali metals, H₂, C, KI.

Oxidation numbers (Extended)

Track electrons by assigning each atom an oxidation number. Increase = oxidised. Decrease = reduced.

Oxidation number rules.

Pure element (e.g. $\text{O}_{2}$ , $\text{Fe}$ ): oxidation number 0.
Monatomic ion: equal to its charge (e.g. Na⁺ has oxidation no. $+1$ , O²⁻ has $-2$ ).
Hydrogen: usually $+1$ (with non-metals); $-1$ in metal hydrides (e.g. NaH).
Oxygen: usually $-2$ (with most elements); $-1$ in peroxides ( $\text{H}_{2}\text{O}_{2}$ ); $+2$ in $\text{OF}_{2}$ .
Group 1 metals: $+1$ . Group 2: $+2$ . Aluminium: $+3$ .
Sum of oxidation numbers in a neutral compound = 0; in an ion = ion's charge.

Rule	Oxidation number	Example
Pure element	0	O₂, Fe
Monatomic ion	equal to its charge	Na⁺ = +1, O²⁻ = −2
Hydrogen	usually +1; −1 in metal hydrides	H₂O = +1, NaH = −1
Oxygen	usually −2; −1 in peroxides; +2 in OF₂	H₂O = −2, H₂O₂ = −1
Group 1 / Group 2 / Al	+1 / +2 / +3	Na⁺, Mg²⁺, Al³⁺
Sum in a species	0 in a neutral compound; = charge in an ion	MnO₄⁻ sums to −1

Worked: assign oxidation numbers in MnO₄⁻.

O is $-2$ each (×4 = $-8$ ).
Sum = $-1$ (the charge of the ion).
Mn must be: $-1 - (-8) = +7$ .

The four oxygens contribute −8, so manganese must be +7 for the ion to carry an overall −1 charge.

Identifying oxidation/reduction.

Oxidation: oxidation number INCREASES.
Reduction: oxidation number DECREASES.

Worked. $\text{MnO}_{4}^{-} + \ldots \to \text{Mn}^{2+}$ .

Mn: $+7 \to +2$ . Decrease → REDUCED.
MnO₄⁻ is the OXIDISING AGENT (it accepts electrons).

Worked. $2\text{Fe}^{2+} \to 2\text{Fe}^{3+} + 2\text{e}^{-}$ .

Fe: $+2 \to +3$ . Increase → OXIDISED.

Naming compounds. Roman numerals in compound names give the oxidation number of the metal: e.g. iron(III) chloride means $\text{Fe}^{3+}$ (oxidation number $+3$ ).

Pure element: 0.
O usually $-2$ , H usually $+1$ .
Sum = 0 (neutral) or ion charge.
Increase: oxidised. Decrease: reduced.

Tests for oxidising and reducing agents

KI paper goes brown for oxidisers; acidified KMnO₄ goes colourless for reducers.

Test for an OXIDISING agent.

Use moist potassium iodide ( $\text{KI}$ ) paper.
An oxidising agent will OXIDISE I⁻ to I₂.
I₂ is brown-purple.
Result: white KI paper turns BROWN/PURPLE.

Test for a REDUCING agent.

Use acidified potassium manganate(VII) ( $\text{KMnO}_{4}$ ) — purple solution.
A reducing agent will REDUCE MnO₄⁻ ( $+7$ ) to Mn²⁺ ( $+2$ ).
Mn²⁺ is colourless.
Result: purple turns COLOURLESS.

Moist KI paper goes white→brown for an oxidiser; acidified KMnO₄ goes purple→colourless for a reducer.

Worked qualitative. A student adds chlorine water to a strip of moist KI paper. Paper turns brown. Conclusion: chlorine acts as an oxidising agent.

Worked qualitative. A student adds SO₂ gas to acidified KMnO₄. Purple fades to colourless. Conclusion: SO₂ acts as a reducing agent.

Cambridge tip. Always specify the FULL test: reagent, observation, and what the colour change tells you.

Oxidiser: KI paper turns BROWN (I₂ formed).
Reducer: KMnO₄ goes from PURPLE to COLOURLESS.
Quote both reagent and observation.

Quick recap

Three definitions: oxygen, hydrogen, electrons (OIL RIG).
Oxidising agent: reduced. Reducing agent: oxidised.
Oxidation no. up: oxidised. Down: reduced.
Test oxidiser: moist KI paper turns brown.
Test reducer: acidified KMnO₄ turns colourless.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Oxidation — loss of electrons (or gain of oxygen, or loss of hydrogen).
Reduction — gain of electrons (or loss of oxygen, or gain of hydrogen).
Oxidising agent — substance that oxidises another, being reduced itself.
Reducing agent — substance that reduces another, being oxidised itself.
Oxidation number — charge assigned to an atom to track electron transfer.

How it’s examined

Redox is examined every Paper 4 (6-8 marks): identify oxidising/reducing agents, half-equations, oxidation numbers, redox tests. Examiner reports flag students confusing 'oxidiser' (substance that does the oxidising) with 'oxidised' (substance that gets oxidised).

Sources: Cambridge IGCSE Chemistry 0620 syllabus 2026-2028 (8.2); 0620/42 Oct/Nov 2024 — Q13 (oxidation numbers, agents); 0620 Examiner Reports 2022-2024. Last reviewed 2026-05-07.

Take this whole topic with you

Step-by-step worked examples — Redox

Step-by-step solutions to past-paper-style questions on redox, written exactly the way a tutor would explain them at the board.

1Define oxidation and reduction (both definitions)
Getting started• definition, OIL RIG
Question
Define oxidation and reduction in terms of (a) oxygen and (b) electrons.
Step-by-step solution
1. Step 1
  (a) Oxygen: oxidation is the GAIN of oxygen; reduction is the LOSS of oxygen.
2. Step 2
  (b) Electrons: oxidation is the LOSS of electrons; reduction is the GAIN of electrons.
3. Step 3
  Remember the mnemonic OIL RIG: Oxidation Is Loss (of electrons), Reduction Is Gain.
Answer
Oxidation = gain of oxygen / loss of electrons. Reduction = loss of oxygen / gain of electrons (OIL RIG).
Examiner tip
Cambridge accepts both definitions. The electron definition is the Extended one and is needed for half-equations and agents.
2Identify oxidation and reduction using oxygen
Getting started• oxygen
Question
In the reaction $2\text{CuO} + \text{C} \to 2\text{Cu} + \text{CO}_{2}$ , state which species is oxidised and which is reduced.
Step-by-step solution
1. Step 1
  Carbon GAINS oxygen (C → CO₂), so carbon is oxidised.
2. Step 2
  Copper(II) oxide LOSES oxygen (CuO → Cu), so the copper oxide is reduced.
3. Step 3
  Because both happen together, this is a redox reaction. Carbon is the reducing agent here.
Answer
Carbon is oxidised (gains oxygen); copper(II) oxide is reduced (loses oxygen).
Examiner tip
On the oxygen definition, simply track which substance gains and which loses oxygen.
3Work out an oxidation number
Building confidence• Adapted from 0620/42 May/Jun 2024 Q17• oxidation number
Question
Work out the oxidation state of manganese in $\text{KMnO}_{4}$ and of sulfur in $\text{SO}_{4}^{2-}$ .
Step-by-step solution
1. Step 1
  In $\text{KMnO}_{4}$ : K is $+1$ , each O is $-2$ , and the compound is neutral (sum $=0$ ).
  $(+1) + \text{Mn} + 4(-2) = 0$
2. Step 2
  Solve for Mn.
  $\text{Mn} = +7$
3. Step 3
  In $\text{SO}_{4}^{2-}$ : each O is $-2$ and the ion charge is $-2$ (sum $=-2$ ).
  $\text{S} + 4(-2) = -2 \;\Rightarrow\; \text{S} = +6$
Answer
Mn is $+7$ in $\text{KMnO}_{4}$ ; S is $+6$ in $\text{SO}_{4}^{2-}$ .
Examiner tip
Set the sum of oxidation numbers equal to the overall charge: 0 for a neutral compound, the ion charge for an ion.
4Write half-equations for a displacement
Building confidence• half-equations, displacement
Question
Zinc displaces copper from copper(II) sulfate: $\text{Zn} + \text{CuSO}_{4} \to \text{ZnSO}_{4} + \text{Cu}$ . Write the oxidation and reduction half-equations and state what is oxidised.
Step-by-step solution
1. Step 1
  The sulfate ion is a spectator. Zinc loses 2 electrons (oxidation).
  $\text{Zn} \to \text{Zn}^{2+} + 2e^{-}$
2. Step 2
  Copper(II) ions gain 2 electrons (reduction).
  $\text{Cu}^{2+} + 2e^{-} \to \text{Cu}$
3. Step 3
  Zn is oxidised (loses electrons) and Cu²⁺ is reduced (gains electrons). Zn is the reducing agent; Cu²⁺ is the oxidising agent.
Answer
Oxidation: $\text{Zn} \to \text{Zn}^{2+} + 2e^{-}$ . Reduction: $\text{Cu}^{2+} + 2e^{-} \to \text{Cu}$ . Zinc is oxidised.
Examiner tip
Electrons go on the left of a reduction half-equation and the right of an oxidation half-equation. Charges must balance on both sides.
5Colour change of acidified manganate(VII)
Stretch• colour change, oxidising agent
Question
A few drops of acidified potassium manganate(VII) are added to aqueous iron(II) sulfate. Describe and explain the colour change, and identify the oxidising agent.
Step-by-step solution
1. Step 1
  Acidified $\text{KMnO}_{4}$ is purple. It acts as an oxidising agent, so it is itself reduced: Mn goes from $+7$ (in $\text{MnO}_{4}^{-}$ ) to $+2$ (in $\text{Mn}^{2+}$ ).
2. Step 2
  As $\text{MnO}_{4}^{-}$ is reduced to colourless $\text{Mn}^{2+}$ , the purple colour fades — the solution turns from purple to colourless (pale pink at the end point).
3. Step 3
  The iron(II) is oxidised: $\text{Fe}^{2+} \to \text{Fe}^{3+} + e^{-}$ . Because manganate(VII) does the oxidising, manganate(VII) is the oxidising agent (and is itself reduced).
Answer
Purple → colourless. Acidified potassium manganate(VII) is the oxidising agent (it oxidises Fe²⁺ to Fe³⁺ and is reduced from Mn(+7) to Mn(+2)).
Examiner tip
Purple → colourless signals manganate(VII) acting as an oxidising agent. The oxidising agent is the species that GETS reduced.
6Colour change as iodide is oxidised
Stretch• colour change, half-equations
Question
Chlorine is bubbled through colourless aqueous potassium iodide: $\text{Cl}_{2} + 2\text{KI} \to 2\text{KCl} + \text{I}_{2}$ . Describe the colour change, write the two half-equations, and identify the oxidising and reducing agents.
Step-by-step solution
1. Step 1
  Iodide ions are oxidised to iodine (lose electrons): the colourless solution turns brown (orange-brown) as $\text{I}_{2}$ forms.
  $2\text{I}^{-} \to \text{I}_{2} + 2e^{-}$
2. Step 2
  Chlorine is reduced (gains electrons): Cl goes from $0$ to $-1$ .
  $\text{Cl}_{2} + 2e^{-} \to 2\text{Cl}^{-}$
3. Step 3
  Chlorine does the oxidising, so $\text{Cl}_{2}$ is the oxidising agent (itself reduced). Iodide is the reducing agent (itself oxidised). This is also a displacement: the more reactive halogen, chlorine, displaces iodine.
Answer
Colourless → brown. Oxidation: $2\text{I}^{-} \to \text{I}_{2} + 2e^{-}$ ; reduction: $\text{Cl}_{2} + 2e^{-} \to 2\text{Cl}^{-}$ . $\text{Cl}_{2}$ = oxidising agent; $\text{I}^{-}$ = reducing agent.
Examiner tip
Colourless → brown means I⁻ has been oxidised to I₂ — a standard test that an oxidising agent is present.

Model Answers — Redox

High-scoring sample answers for redox on the Cambridge IGCSE 0620 paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
Paper 4 short-answer style1 mark
Define oxidation in terms of electrons. (1 mark)
Model answer
Oxidation is the loss of electrons.
Why this scores
One mark for 'loss of electrons'. The mnemonic is OIL — Oxidation Is Loss.
Question 2
Paper 4 structured style2 marks
Deduce the oxidation state of iron in $\text{Fe}_{2}\text{O}_{3}$ , and use it to name the compound. (2 marks)
Model answer
Each oxygen is $-2$ , so the three oxygens contribute $-6$ ; the compound is neutral, so the two iron atoms must total $+6$ , giving each iron an oxidation state of $+3$ . The compound is therefore iron(III) oxide.
Why this scores
One mark for the oxidation state ( $+3$ ); one mark for the Roman-numeral name iron(III) oxide. The Roman numeral is the oxidation state of the metal.
Question 3
Paper 4 structured style3 marks
For the reaction $\text{Mg} + \text{Cu}^{2+} \to \text{Mg}^{2+} + \text{Cu}$ , state which species is oxidised and which is reduced, giving a reason in each case. (3 marks)
Model answer
Magnesium is oxidised because it loses two electrons to form $\text{Mg}^{2+}$ (its oxidation state rises from $0$ to $+2$ ). The copper(II) ion is reduced because it gains two electrons to form copper metal (its oxidation state falls from $+2$ to $0$ ). Both changes occur together, so the reaction is redox.
Why this scores
Three marks: Mg oxidised (loses electrons); Cu²⁺ reduced (gains electrons); a valid reason tied to electron transfer or oxidation-state change.
Question 4
Paper 4 structured style4 marks
Explain what is meant by an oxidising agent, and identify the oxidising agent in $\text{Cl}_{2} + 2\text{KBr} \to 2\text{KCl} + \text{Br}_{2}$ . Give a reason. (4 marks)
Model answer
An oxidising agent is a substance that oxidises another species by accepting (gaining) electrons from it, and so is itself reduced. In this reaction, bromide ions are oxidised to bromine ( $\text{Br}$ goes from $-1$ to $0$ , losing electrons), while chlorine is reduced ( $\text{Cl}$ goes from $0$ to $-1$ , gaining electrons). Because chlorine accepts the electrons and is itself reduced, $\text{Cl}_{2}$ is the oxidising agent.
Why this scores
Four marks: definition (accepts electrons / oxidises another); is itself reduced; identifies Cl₂; reason via the oxidation-state change of Cl (0 → −1).
Question 5
Paper 4 (Extended) structured style5 marks
Acidified potassium manganate(VII) is added to a solution containing a reducing agent. Describe and explain the colour change observed, and state what this tells you about the manganate(VII). (5 marks)
Model answer
The solution changes from purple to colourless. Acidified potassium manganate(VII) is purple because of the manganate(VII) ion, $\text{MnO}_{4}^{-}$ , in which manganese has an oxidation state of $+7$ . Because a reducing agent is present, the manganate(VII) acts as an oxidising agent and is itself reduced: the manganese is reduced from $+7$ in $\text{MnO}_{4}^{-}$ to $+2$ in the colourless $\text{Mn}^{2+}$ ion, so the purple colour disappears. The disappearance of the purple colour therefore shows that the manganate(VII) is being reduced and is behaving as an oxidising agent.
Why this scores
Five marks across: purple → colourless; purple is MnO₄⁻ / Mn is +7; Mn reduced to +2; Mn²⁺ is colourless; manganate(VII) acts as an oxidising agent (is itself reduced).
Question 6
Paper 4 (Extended) structured style6 marks
Bromine is added to aqueous potassium iodide: $\text{Br}_{2} + 2\text{KI} \to 2\text{KBr} + \text{I}_{2}$ . Describe the colour change, write a half-equation for the oxidation and a half-equation for the reduction, and identify the oxidising and reducing agents. (6 marks)
Model answer
The colourless potassium iodide solution turns brown (orange-brown) as iodine is formed. The iodide ions are oxidised (they lose electrons), shown by the half-equation $2\text{I}^{-} \to \text{I}_{2} + 2e^{-}$ . The bromine is reduced (it gains electrons), shown by $\text{Br}_{2} + 2e^{-} \to 2\text{Br}^{-}$ . Because bromine accepts the electrons and is itself reduced, bromine ( $\text{Br}_{2}$ ) is the oxidising agent; because iodide donates the electrons and is itself oxidised, the iodide ion ( $\text{I}^{-}$ ) is the reducing agent.
Why this scores
Six marks: colourless → brown; correct oxidation half-equation (I⁻ → I₂, electrons on right, balanced); correct reduction half-equation (Br₂ → Br⁻, electrons on left, balanced); Br₂ as oxidising agent; I⁻ as reducing agent; reasoning tied to gain/loss of electrons.

Key Formulae — Redox

The formulae you need to memorise for redox on the Cambridge IGCSE 0620 paper, with every variable defined in plain English and a note on when to use it.

OIL RIG
$\text{Oxidation Is Loss (of electrons), Reduction Is Gain}$
When to use
Quick mnemonic for redox identification using electrons.
Oxidation-number rules (key ones)
$\text{Element: 0; \; Group I: +1; \; Group II: +2; \; H: +1; \; O: -2; \; sum = overall charge}$
When to use
Computing oxidation numbers in formulae and ions.

Key Definitions and Keywords — Redox

Definitions to memorise and the exact keywords mark schemes credit for redox answers — sharpened from recent examiner reports for the 2026 0620 sitting.

Oxidation
Examiner keyword
Gain of oxygen, or loss of electrons (an increase in oxidation state).
Reduction
Examiner keyword
Loss of oxygen, or gain of electrons (a decrease in oxidation state).
Oxidising agent
Examiner keyword
A substance that oxidises another by accepting electrons from it — and so is itself REDUCED.
Reducing agent
Examiner keyword
A substance that reduces another by donating electrons to it — and so is itself OXIDISED.
Oxidation number (state)
Examiner keyword
The charge an atom would have if all its bonds were ionic. Used to identify redox and to name compounds with Roman numerals (e.g. iron(II), iron(III)).

Common Mistakes and Misconceptions — Redox

The traps other students keep falling into on redox questions — taken from recent Cambridge IGCSE 0620 examiner reports and mark schemes — and how to avoid them.

✕Inverting OIL RIG (saying oxidation is gain of electrons).
0620/42 — recurring
Why it happens
Memory slip — the oxygen and electron definitions feel opposite.
How to avoid it
OIL = Oxidation Is Loss (of electrons). RIG = Reduction Is Gain. With oxygen it is the reverse: oxidation is gain of oxygen.
✕Calling the species that GETS oxidised the oxidising agent.
Why it happens
The naming feels backwards.
How to avoid it
An oxidising AGENT does the oxidising — so it itself gets REDUCED. The reducing agent is the one that gets oxidised.
✕Mixing up the manganate(VII) and iodide colour changes.
0620 Examiner Reports
Why it happens
Both are diagnostic tests, so they blur together.
How to avoid it
Acidified KMnO₄ (oxidising agent): purple → colourless. Iodide being oxidised: colourless → brown (I⁻ → I₂).
✕Putting electrons on the wrong side of a half-equation or not balancing charge.
Why it happens
Rushing without checking the charges.
How to avoid it
Oxidation: electrons on the RIGHT (lost). Reduction: electrons on the LEFT (gained). Always check the total charge balances both sides.

Redox — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Definition

Oxidation is…

Reduction is…

1. Oxygen

Gain of oxygen

Loss of oxygen

2. Hydrogen

Loss of hydrogen

Gain of hydrogen

3. Electrons (best)

Loss of electrons (OIL)

Gain of electrons (RIG)

Rule

Oxidation number

Example

Pure element

O₂, Fe

Monatomic ion

equal to its charge

Na⁺ = +1, O²⁻ = −2

Hydrogen

usually +1; −1 in metal hydrides

H₂O = +1, NaH = −1

Oxygen

usually −2; −1 in peroxides; +2 in OF₂

H₂O = −2, H₂O₂ = −1

Group 1 / Group 2 / Al

+1 / +2 / +3

Na⁺, Mg²⁺, Al³⁺

Sum in a species

0 in a neutral compound; = charge in an ion

MnO₄⁻ sums to −1

Test for an OXIDISING agent.

Use moist potassium iodide ( $\text{KI}$ ) paper.
An oxidising agent will OXIDISE I⁻ to I₂.
I₂ is brown-purple.
Result: white KI paper turns BROWN/PURPLE.

Test for a REDUCING agent.

Use acidified potassium manganate(VII) ( $\text{KMnO}_{4}$ ) — purple solution.
A reducing agent will REDUCE MnO₄⁻ ( $+7$ ) to Mn²⁺ ( $+2$ ).
Mn²⁺ is colourless.
Result: purple turns COLOURLESS.

Moist KI paper goes white→brown for an oxidiser; acidified KMnO₄ goes purple→colourless for a reducer.

Worked qualitative. A student adds chlorine water to a strip of moist KI paper. Paper turns brown. Conclusion: chlorine acts as an oxidising agent.

Worked qualitative. A student adds SO₂ gas to acidified KMnO₄. Purple fades to colourless. Conclusion: SO₂ acts as a reducing agent.

Cambridge tip. Always specify the FULL test: reagent, observation, and what the colour change tells you.

Redox

Short Notes - Redox

Detailed Study Notes

At a glance

What you’ll learn

Three definitions of oxidation/reduction

Oxidising and reducing agents

Oxidation numbers (Extended)

Tests for oxidising and reducing agents

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Define oxidation and reduction (both definitions)

2Identify oxidation and reduction using oxygen

3Work out an oxidation number

4Write half-equations for a displacement

5Colour change of acidified manganate(VII)

6Colour change as iodide is oxidised

OIL RIG

Oxidation-number rules (key ones)

Oxidation

Reduction

Oxidising agent

Reducing agent

Oxidation number (state)

✕Inverting OIL RIG (saying oxidation is gain of electrons).

✕Calling the species that GETS oxidised the oxidising agent.

✕Mixing up the manganate(VII) and iodide colour changes.

✕Putting electrons on the wrong side of a half-equation or not balancing charge.

Redox — frequently asked questions

Redox

Short Notes - Redox

Detailed Study Notes

At a glance

What you’ll learn

Three definitions of oxidation/reduction

Oxidising and reducing agents

Oxidation numbers (Extended)

Tests for oxidising and reducing agents

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Define oxidation and reduction (both definitions)

2Identify oxidation and reduction using oxygen

3Work out an oxidation number

4Write half-equations for a displacement

5Colour change of acidified manganate(VII)

6Colour change as iodide is oxidised

OIL RIG

Oxidation-number rules (key ones)

Oxidation

Reduction

Oxidising agent

Reducing agent

Oxidation number (state)

✕Inverting OIL RIG (saying oxidation is gain of electrons).

✕Calling the species that GETS oxidised the oxidising agent.

✕Mixing up the manganate(VII) and iodide colour changes.

✕Putting electrons on the wrong side of a half-equation or not balancing charge.

Redox — frequently asked questions