Electron transfer and oxidation-number changes
Use the wider redox definition when oxygen is absent, and calculate oxidation numbers from charge accounting.
Exam scope and assessment notes
Extended only (Supplement) · Chemistry 0971, 2026–2028. Cambridge states that IGCSE 0971 has the same subject content as IGCSE 0620, with 9–1 grading. This StudyForA pathway maps the 2026–2028 syllabus only. This 9–1 graded qualification is available only in a limited number of administrative zones. Confirm that your centre can enter you for this code and exam series with Cambridge International. Cambridge has published a separate syllabus for 2029. StudyForA has not reconciled that version yet; this page supports exams in 2026, 2027 and 2028 only.
Before you begin
- Read ionic charges and ionic half-equations.
- Use formula atom ratios and distinguish oxidation numbers from coefficients.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which half-equation shows electron loss?
What you will learn
- Identify electron loss as oxidation and electron gain as reduction.
- Calculate oxidation numbers and use their increases or decreases to recognise redox.
Track electrons and signed changes
Oxidation is loss of electrons and an increase in oxidation number; reduction is gain of electrons and a decrease in oxidation number. In Zn+Cu²⁺→Zn²⁺+Cu, zinc goes from 0 to +2 and loses two electrons, while copper goes from +2 to 0 and gains two. This is redox even though no oxygen is present. A more positive number is an increase: −1 to 0 is oxidation, not reduction.
The half-equations Zn→Zn²⁺+2e⁻ and Cu²⁺+2e⁻→Cu show the electrons lost and gained. Multiply half-equations when necessary so transferred electron counts match before adding them; electrons cancel in the overall equation. Electrons written on the product side indicate loss, while electrons on the reactant side indicate gain.
Make oxidation numbers add up to the total charge
An uncombined element has oxidation number zero, including H₂, O₂ and Cl₂. A monatomic ion's oxidation number equals its charge, so Mg²⁺ is +2 and Cl⁻ is −1. The sum of oxidation numbers, including each formula's atom counts, is zero for a neutral compound and equals the total charge for a polyatomic ion. Do not give every atom in an ion the whole ion charge.
For the examples used here, oxygen is assigned −2 in ordinary oxides and hydrogen +1 in the illustrated compounds; these values are not universal rules for every possible substance. In SO₄²⁻, sulfur's value x satisfies x+4(−2)=−2, giving +6. In NH₄⁺, nitrogen's x satisfies x+4(+1)=+1, giving −3. Compare the same element before and after a reaction; unchanged oxidation numbers do not demonstrate redox.
Pause and explain
What does Cl⁻ becoming Cl₂ show about chlorine?
Worked example
Find sulfur's oxidation number in SO₂ and SO₄²⁻ using oxygen −2, then decide whether changing SO₂ to sulfate oxidises or reduces sulfur.
Show the worked solution
- Neutral SO₂ requires x+2(−2)=0, so sulfur is +4.
- SO₄²⁻ requires x+4(−2)=−2, so sulfur is +6.
- The increase from +4 to +6 is oxidation of sulfur; another species must be reduced in the complete redox reaction.
Answer Sulfur is +4 in SO₂ and +6 in SO₄²⁻; the increase is oxidation.
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Using O=−2 and K=+1, calculate chlorine's oxidation number in neutral KClO₃. State whether Cl⁻ changing to this compound is oxidation or reduction.
Give me a hint
Set the atom-weighted sum equal to zero, then compare chlorine's values.
Compare my reasoning
- The neutral compound requires +1+x+3(−2)=0.
- Solving gives x=+5 for chlorine.
- Cl⁻ starts at −1 and increases to +5, so chlorine is oxidised.
Chlorine is +5 in KClO₃; the increase from −1 is oxidation.
Look for these in your work
- All three oxygens contribute to the sum.
- The signed increase gives the correct change.
Identify both changes in 2Fe²⁺+Cl₂→2Fe³⁺+2Cl⁻, and write matched half-equations.
Give me a hint
Each iron loses one electron; two chlorine atoms together gain two.
Compare my reasoning
- Iron goes +2→+3 and is oxidised; chlorine goes 0→−1 and is reduced.
- The matched oxidation half-equation is 2Fe²⁺→2Fe³⁺+2e⁻.
- Reduction is Cl₂+2e⁻→2Cl⁻; the two electrons cancel on addition.
Fe²⁺ is oxidised and Cl₂ reduced; 2Fe²⁺→2Fe³⁺+2e⁻ and Cl₂+2e⁻→2Cl⁻.
Look for these in your work
- Electron loss and gain are matched.
- The overall atoms and charge are conserved.
Compare H⁺+OH⁻→H₂O with Zn+Cu²⁺→Zn²⁺+Cu. Using the illustrated H=+1 and O=−2 assignments, explain which is redox rather than deciding from the absence of oxygen gas.
Give me a hint
A reaction is redox when oxidation numbers change with simultaneous electron loss and gain.
Compare my reasoning
- H stays +1 and O stays −2 during neutralisation, so that ionic equation has no oxidation-number change.
- In displacement, Zn changes 0→+2 while Cu changes +2→0.
- The displacement is redox through electron transfer; the stated neutralisation is not redox despite forming a new substance.
The zinc displacement is redox; H⁺+OH⁻→H₂O is not redox by the oxidation-number test.
Look for these in your work
- Each conclusion follows from before-and-after numbers.
- Chemical change is not assumed to mean redox.
Common mistakes
- Assigning a non-zero oxidation number to an uncombined element.
- Ignoring the atom multiplier or an ion's total charge.
- Calling −1 to 0 a decrease because both look less positive than +1.
- Leaving uncancelled electrons in an ordinary overall redox equation.
What can you explain now?
Using H=+1, find nitrogen's oxidation number in NH₄⁺.
Compare with the explanation
−3
The four hydrogens contribute +4 and the whole ion is +1, so nitrogen must contribute −3, not −4.
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Make it stick
Solve oxidation numbers in a neutral compound and a polyatomic ion, then match electron-loss and electron-gain half-equations for a reaction containing no oxygen.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.