Rank halide ions as reducing agents
Compare halide-ion reducing strength and predict which ions are oxidised by halogens or concentrated sulfuric acid.
AS Level; Cambridge International Chemistry 9701, 2025–2027 syllabus. State conditions for halogen reactions, distinguish halogen oxidising strength from halide-ion reducing strength, and balance redox equations. Chlorine, bromine, iodine and concentrated acids are hazardous; examples teach chemistry, not practical instructions. Written practice is self-checked
Before you begin
- Use molecular halogen formulas X₂, oxidation numbers, ionic equations and electron-transfer ideas.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What happens when a halide ion acts as a reducing agent?
What you will learn
- Rank F⁻, Cl⁻, Br⁻ and I⁻ as reducing agents
- Distinguish ion reducing strength from molecule oxidising strength
- Use electron loss to explain the trend
Halide ions can donate electrons to form halogens
A halide ion is oxidised by losing an electron: 2X⁻ → X₂ + 2e⁻. A stronger reducing agent donates electrons more readily.
Reducing strength increases down the group: F⁻ < Cl⁻ < Br⁻ < I⁻. Iodide is the strongest reducing agent in this set, while fluoride is the weakest.
Oxidising and reducing trends point in opposite directions
Halogen molecules become weaker oxidising agents down the group, while their halide ions become stronger reducing agents down the group. Keep the species named in each comparison.
This distinction predicts that bromide and especially iodide can reduce concentrated sulfuric acid, while chloride usually undergoes acid–base reaction without redox in that test.
Pause and explain
Which is the strongest reducing agent in the halide-ion series?
Worked example
Which is the stronger reducing agent, Br⁻ or I⁻? Explain using electron transfer.
Show the worked solution
- Iodide is lower in Group 17.
- Halide-ion reducing strength increases down the group.
- I⁻ loses electrons more readily to form I₂.
Answer I⁻ is the stronger reducing agent because it is oxidised to I₂ more readily than Br⁻ is oxidised to Br₂.
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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.
Write the oxidation half-equation for bromide ions forming bromine.
Give me a hint
Two bromide ions provide two atoms and release two electrons.
Compare my reasoning
- Bromine is diatomic Br₂.
- Two Br⁻ ions make one Br₂ molecule.
- The total charge balances with two electrons on the product side.
2Br⁻ → Br₂ + 2e⁻.
Look for these in your work
- Diatomic Br₂ formed
- Two ions used
- Charge balanced
Rank Cl⁻, Br⁻ and I⁻ in increasing reducing strength.
Give me a hint
Reducing strength increases down Group 17.
Compare my reasoning
- Chloride is highest of the three in the group.
- Bromide is below chloride.
- Iodide is strongest.
Cl⁻ < Br⁻ < I⁻ in reducing strength.
Look for these in your work
- Correct order
- Trend direction correct
- Property named
Explain why chlorine can oxidise iodide even though iodide is the stronger reducing agent.
Give me a hint
Pair the gain of electrons by Cl₂ with the loss of electrons by I⁻.
Compare my reasoning
- Cl₂ is a stronger oxidising agent than I₂.
- I⁻ is a stronger reducing agent than Cl⁻.
- The electron transfer is favorable: Cl₂ + 2I⁻ → 2Cl⁻ + I₂.
Cl₂ accepts electrons while I⁻ donates them; Cl₂ is the stronger oxidising agent and I⁻ the stronger reducing agent in the pair.
Look for these in your work
- Both roles correctly assigned
- Electron directions stated
- Overall reaction supports the explanation
Common mistakes
- Using the halogen oxidising-agent order for halide ions
- Calling an electron acceptor a reducing agent
- Forgetting the diatomic product X₂
- Claiming chloride is a strong reducing agent compared with iodide
What can you explain now?
What happens when a halide ion acts as a reducing agent?
Compare with the explanation
It loses electrons
It is oxidised to a halogen molecule.
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Make it stick
What is the down-group order for halide reducing strength, and how does it differ from halogen oxidising strength?
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