Tests for Oxidising and Reducing Agents
Use aqueous potassium iodide and acidified potassium manganate(VII) to distinguish oxidising and reducing agents, and separate colour observations from electron-transfer explanations.
On this page
These two redox tests are easiest to remember as reagent → observation → conclusion. The colour change is the macroscopic evidence; electron transfer explains it at particle level.
1. Definition
An oxidising agent causes another substance to be oxidised and is itself reduced (it gains electrons). A reducing agent causes another substance to be reduced and is itself oxidised (it loses electrons).
2. Key Ideas
- Test for an oxidising agent: add aqueous potassium iodide. A new brown colour indicates iodine formation when the sample’s colour does not mask the result.
- Test for a reducing agent: add acidified potassium manganate(VII). The purple reagent is decolourised.
- The reagent that detects the sample reacts in the opposite way: iodide ions are oxidised, while manganate(VII) ions are reduced.
Name the complete reagent, state the initial and final colours, then conclude whether an oxidising or reducing agent is present.
3. Detailed Explanations
A. Testing for an oxidising agent
Add aqueous potassium iodide, KI(aq), to a fresh portion of the sample. Use a separate fresh portion for the manganate(VII) test; mixing both reagents together would let them react with each other.
- Macroscopic observation: the colourless solution turns brown as iodine forms. Starch solution may be added to confirm iodine: a blue-black colour forms.
- Particle explanation: the sample accepts electrons from iodide ions, so the sample is an oxidising agent and the iodide ions are oxidised.
- Symbolic representation:
2I⁻(aq) → I₂(aq) + 2e⁻
The electron half-equation explains why the test works; the brown colour is the observation to record.
B. Testing for a reducing agent
Add a small amount of acidified potassium manganate(VII), KMnO₄(aq), to a fresh portion of the sample, following the laboratory instructions. The reagent is commonly acidified with dilute sulfuric acid.
- Macroscopic observation: the purple solution is decolourised.
- Particle explanation: the sample donates electrons to manganate(VII) ions. The sample is therefore a reducing agent, while the manganate(VII) ions are reduced to manganese(II) ions.
- Symbolic representation in acidic solution:
MnO₄⁻(aq) + 8H⁺(aq) + 5e⁻ → Mn²⁺(aq) + 4H₂O(l)
The acid condition matters: the half-equation uses H⁺ and produces Mn²⁺. Dilute manganese(II) solutions are effectively colourless in this test, though concentrated solutions can be pale pink. Without suitable acidic conditions, a brown manganese(IV) oxide product may form instead; that is not the prescribed purple-to-colourless result.
These tests establish oxidising or reducing behaviour under the test conditions, not the unique identity of a chemical. Several substances can give the same positive result. A colourless sample does not necessarily lack an oxidising agent, and a sample that gives no immediate change may need the specified conditions or time.
Potassium manganate(VII) solution is an oxidising reagent and can stain skin and clothing. Wear eye protection, use the small quantities directed, and rinse any spill immediately as instructed by your teacher.
4. Common Mistakes
- Writing “oxidising agent is oxidised” (wrong): oxidising agent is reduced.
- Reversing the test reagents: potassium iodide tests for an oxidising agent; acidified potassium manganate(VII) tests for a reducing agent.
- Writing only “turns colourless”: include the initial colour—purple to colourless.
- Calling potassium manganate(VII) “potassium manganate”: the oxidation state, (VII), is part of its name.
5. Exam Tips
- Reagent.
- Observation (colour change).
- Conclusion (oxidising/reducing agent present).
- Keep observation separate from inference. “Purple to colourless” is the observation; “a reducing agent is present” is the inference.
- If a question gives a half-equation, locate the electrons: left side means reduction; right side means oxidation.
6. Worked Examples
Modelled example 1
Identify the reducing agent (manganate(VII) test)
Problem
Study the worked solution
Record only what is seen
Method
State purple to colourless.Reason
An observation must report the reagent’s initial and final appearance without embedding the conclusion.Working
Acidified potassium manganate(VII) is decolourised.Make the inference
Method
Conclude that X contains a reducing agent.Reason
This prescribed reagent is reduced by electron donors.Working
Conclusion: reducing agent present.Explain electron transfer
Method
Make X donate electrons to manganate(VII) ions.Reason
Electron donation means X is oxidised and acts as the reducing agent, while MnO₄⁻ is reduced.Working
MnO₄⁻ forms Mn²⁺ in acidic solution.
Common misconception 2
Complete an exam answer
Learner response
Make the report precise
View solution step by step
Correct reagent and observation
Method
Name acidified potassium manganate(VII) and state purple to colourless.Reason
The learner correctly included (VII), but omitted “acidified”. “Clear” describes transparency, so a clear solution can still be coloured; state the actual purple-to-colourless change.Working
Add acidified potassium manganate(VII); it changes from purple to colourless.Add the inference
Method
Conclude that a reducing agent is present.Reason
The observation alone does not state what property of the sample was detected.Working
This supports the presence of a reducing agent under the test conditions.
Challenge 3
Identify the oxidising agent (iodide + starch test)
Second-reagent transfer
Link confirmation evidence to electron transfer
Hints
Hint 1: use the iodide half-equation
View solution step by step
Interpret the colour evidence
Method
Identify brown iodine and use starch to confirm it.Reason
Iodine is brown in solution and forms a blue-black complex with starch.Working
I₂ formed from I⁻.Identify iodide oxidation
Method
State that iodide loses electrons.Reason
Formation of iodine follows 2I⁻ → I₂ + 2e⁻.Working
I⁻ is oxidised.Infer Y's agent role
Method
Call Y an oxidising agent.Reason
Y causes iodide oxidation by accepting its electrons, so Y is reduced.Working
Oxidising agent present.
7. Mind Stretchers
Mind stretcher 1: Use the half-equation properlyExtension
Question: In the manganate(VII) test, explain why the sample is a reducing agent even though the purple reagent is the substance being reduced.
Show Answer
The sample supplies electrons to MnO₄⁻. Donating electrons means the sample is oxidised and acts as the reducing agent. The MnO₄⁻ ions accept those electrons, so they are reduced.
Mind stretcher 2: Stronger justification sentenceExtension
Question: Improve this sentence: “Solution Y is an oxidising agent because it turns KI brown.”
Show Answer
“Solution Y is an oxidising agent because it oxidises I⁻ to iodine, I₂ (brown), and iodine is confirmed by a blue-black colour with starch solution.”
Try independently: Two initially colourless samples, A and B, are tested using separate fresh portions. A forms a brown colour with aqueous KI, confirmed blue-black with starch. B decolourises acidified potassium manganate(VII). A learner says, “A contains chlorine, and B contains sulfur dioxide.” Which conclusions are supported, and which go beyond the evidence?
Show answer and reasoning
A shows oxidising behaviour: it oxidises iodide to iodine. B shows reducing behaviour: it donates electrons to acidified manganate(VII). The results do not uniquely identify chlorine or sulfur dioxide; other oxidising and reducing substances can give the same results. Further identifying evidence is needed.
Practise and check
Practise the two reagents, their colour changes, and the electron-transfer explanation for each conclusion.
Open the Redox Chemistry topic checkSyllabus and review details
- SEC G3 Pure Chemistry 2027 · 2027
Content structure and subject content, PDF pages 9–24
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