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Tests for Oxidising and Reducing Agents

Learn K324 / 6092 oxidant and reductant tests using aqueous potassium iodide and acidified potassium manganate(VII), observations and ionic equations.

  • SEC G3 Pure Chemistry 2027
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Learning objectives

  • describe the use of aqueous potassium iodide and acidified potassium manganate(VII) in testing for oxidising and reducing agents from the resulting colour changes.

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 colourless solution turning brown shows that iodine formed.
  • 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.
Write all three marking points

Name the complete reagent, state the initial and final colours, then conclude whether an oxidising or reducing agent is present.

3. Detailed Explanations

Evidence pathway for oxidising-agent and reducing-agent testsTwo test pathways. An oxidising agent changes aqueous potassium iodide from colourless to brown iodine, confirmed blue-black with starch, because iodide loses electrons. A reducing agent decolourises purple acidified potassium manganate seven because manganate seven ions gain electrons to form manganese two ions.Test for an oxidising agent1. Add aqueous KIcolourless solution2. Observe iodinebrown solution3. Confirmstarch: blue-black2I⁻(aq) → I₂(aq) + 2e⁻ | iodide is oxidisedTest for a reducing agent1. Add acidified KMnO₄purple solution2. Observesolution decolourises3. ConcludeThe sample donatedelectrons, so it is areducing agent.MnO₄⁻(aq) + 8H⁺(aq) + 5e⁻ → Mn²⁺(aq) + 4H₂O(l)
Read each test across three levels: record the visible colour change, identify the ions that transfer electrons, then use the half-equation to justify the conclusion.

A. Testing for an oxidising agent

Add aqueous potassium iodide, KI(aq), to the sample.

  • 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 acidified potassium manganate(VII), KMnO₄(aq), to the sample.

  • 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)

Laboratory Warning

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

3-part answer
  1. Reagent.
  2. Observation (colour change).
  3. 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)

Core

Problem

A purple solution of acidified KMnO₄(aq) turns colourless when added to Solution X. State the observation, conclusion and electron-transfer explanation.
Study the worked solution

  1. 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.

  2. 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.

  3. 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

Find and correct the mistake

Learner response

A student writes, “I added potassium manganate(VII) and it went clear.” Identify every imprecise part and rewrite the response with reagent, observation and conclusion.

Restore the three marking points

Complete reagent
Precise observation
Conclusion

View solution step by step

  1. Correct reagent and observation

    Method

    Name acidified potassium manganate(VII) and state purple to colourless.

    Reason

    “Potassium manganate” omits both the acid condition and (VII), while “clear” is less precise than colourless.

    Working

    Add acidified potassium manganate(VII); it changes from purple to colourless.

  2. 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

    …showing that a reducing agent is present.

Challenge 3

Identify the oxidising agent (iodide + starch test)

Minimal support

Second-reagent transfer

Solution Y forms a brown colour when aqueous KI is added; after starch solution is added, the mixture becomes blue-black. State what each observation shows and explain why Y is an oxidising agent.

Link confirmation evidence to electron transfer

Brown product
Blue-black result
Y's electron role

Hints

Hint 1: use the iodide half-equation
2I⁻ → I₂ + 2e⁻ places electrons on the product side.
View solution step by step

  1. 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⁻.

  2. Identify iodide oxidation

    Method

    State that iodide loses electrons.

    Reason

    Formation of iodine follows 2I⁻ → I₂ + 2e⁻.

    Working

    I⁻ is oxidised.

  3. 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.”

8. Quiz

Quiz Time!

Test the two 6092 reagents, their colour changes, electron-transfer explanations and exam conclusions.

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