Oxidising and reducing agents: reagent evidence
Key idea: An oxidising agent turns aqueous potassium iodide brown by forming iodine. A reducing agent turns acidified potassium manganate(VII) from purple to colourless.
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Guided lessons · coming soon
Guided lesson
Guided steps
- Recall what happens to a substance that gains electrons.
- Read what each reagent's colour change tells you about the unknown.
- Read how to do both tests on fresh portions.
- Follow how the brown colour shows an oxidising agent.
- Decide what happened to the manganate and what B is.
- Correct the learner's reasoning, then compare your answer.
- Decide what each test result shows about P, Q and R.
- Decide why the first tube stayed purple and what D is.
- Read when a test names a substance and when it only shows what it does.
- Decide what the same result shows for the gas and for the solution.
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ContinueAn unknown can take electrons from a reagent or give electrons to it. A colour change makes the reagent’s reaction visible. Work back from what happened to the reagent to what the unknown did.
Recall electron transfer
Check your understanding 1 Recall which way the electrons move
An unknown X accepts electrons from another substance. Before using the colour tests, decide what happens to X and which agent it is.
Show answer
X gains electrons and is reduced. Its partner loses electrons and is oxidised, so X is the oxidising agent.
Two test reagents
Each test reagent is chosen because it changes colour when it is oxidised or reduced. The unknown is then the species that did that to it.
| Test for | Add to the unknown | Positive result | What changed colour, and why |
|---|---|---|---|
| an oxidising agent | aqueous potassium iodide | colourless to brown | iodide ions from the reagent are oxidised to iodine, which is brown in solution |
| a reducing agent | acidified aqueous potassium manganate(VII) | purple to colourless | purple manganate(VII) ions from the reagent are reduced to a colourless product |
Read the table from right to left to make the inference:
- In the iodide test the reagent was oxidised, so the unknown oxidised it: the unknown is an oxidising agent and is itself reduced.
- In the manganate(VII) test the reagent was reduced, so the unknown reduced it: the unknown is a reducing agent and is itself oxidised.
The brown colour comes from iodine formed from iodide ions; the loss of purple comes from manganate(VII) ions being used up. Say which substance changed colour, not that the unknown changed colour.
Use only a few drops of the manganate(VII) reagent. If there is too much, some purple ions can be left over even after a reducing agent has reacted, so a tube that stays purple does not on its own mean that nothing happened.
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Practical task 1 Carry out both tests on fresh portionsOptional
Put about 1 cm³ of the unknown in each of two test tubes. Add a few drops of aqueous potassium iodide to one and a few drops of acidified aqueous potassium manganate(VII) to the other. Record the colour of each mixture before and after mixing. Wear eye protection; acidified potassium manganate(VII) is an oxidiser and stains skin.
Worked example 1
From a colour change to an agent
Problem
A student tests colourless solution A on two fresh portions. No other source of iodine is present.
| Reagent added to A | Before | After |
|---|---|---|
| aqueous potassium iodide | colourless | brown |
| acidified aqueous potassium manganate(VII) | purple | purple |
What do the results show about A?
Show solution step by step
Say what changed colour
Method
Find what made the brown colour.Reason
The brown colour is iodine, and the iodine can only have come from the iodide ions in the reagent.Working
Answer
Iodide ions were oxidised to iodine.Infer the unknown's role
Method
Ask what A did to the reagent.Reason
A caused the iodide ions to be oxidised, which is what an oxidising agent does.Working
Answer
A is an oxidising agent, so A is itself reduced.Use the other result
Working
Answer
The manganate(VII) stayed purple, so this test gives no sign that A is a reducing agent.
Quick check
Guided practice 2
What does decolourising show?
Problem
Solution B is added to acidified aqueous potassium manganate(VII). The purple colour disappears and the mixture becomes colourless. With aqueous potassium iodide, B gives no colour change. Decide what happened to the manganate(VII) ions and what kind of agent B is.
Decide from the evidence
Hints
Hint 1
Whose colour?
The purple colour belongs to the manganate(VII) ions, not to B.
Hint 2
Agent rule
Whatever reduced the manganate(VII) is a reducing agent, and a reducing agent is oxidised.
Show solution step by step
Find what changed colour
Method
Identify what lost its colour.Reason
Purple manganate(VII) ions become a colourless product when they are reduced.
Working
Answer
The manganate(VII) ions were reduced.Infer B's role
Working
Answer
B reduced the manganate(VII), so B is a reducing agent and was itself oxidised. The unchanged iodide test gives no evidence that B is an oxidising agent.
Quick check
If you chose “oxidising agent”, you may have read “manganate(VII) is an oxidising reagent” as a description of B. The reagent is the oxidising agent here; B is what it oxidised.
Spot the mistake 3
“It turned brown, so it was oxidised”
Learner response
Solution C is colourless. When aqueous potassium iodide is added, the mixture turns brown. A learner writes: “Solution C turned brown, so C was oxidised and is a reducing agent.” Find the mistake and write the correct conclusion.
Correct the reasoning
Show solution step by step
Find the first error
Method
Check which substance changed colour.Reason
The brown colour is iodine, formed from the iodide ions in the potassium iodide, not from C.
Working
Answer
First error: “C turned brown” — the reagent’s iodide was oxidised to brown iodine.
Write the correct conclusion
Working
Answer
C oxidised the iodide ions, so C is an oxidising agent, and C was itself reduced.
Common mistake
The brown colour is iodine made from the reagent, not a change in C. C oxidised the iodide, so C is an oxidising agent and was itself reduced.
Check your understanding 2 Three unknowns
Colourless solutions P, Q and R are tested on fresh portions. P turns aqueous potassium iodide brown and decolourises acidified potassium manganate(VII). Q leaves the iodide colourless and decolourises the manganate(VII). R gives no change with either. What does each result show?
Show answer
P oxidised the iodide and reduced the manganate(VII), so it acted as both an oxidising agent and a reducing agent. Q reduced the manganate(VII), so it is a reducing agent. R changed neither reagent, so these tests do not show it to be either; that does not mean R can never be oxidised or reduced.
Check your understanding 3 A purple result that needs another look
A student adds a large amount of acidified potassium manganate(VII) to colourless solution D, and the mixture stays purple. With only a few drops of the reagent on a fresh portion of D, the purple colour disappears. Why did the first tube stay purple, and what is D?
Show answer
In the first tube there was more manganate(VII) than D could reduce, so some purple was left over. With only a few drops, D reduced all of it and the mixture went colourless. D is a reducing agent, and is itself oxidised.
Naming a substance or describing its behaviour
Acidified potassium manganate(VII) also appears in the gas tests, where it answers a different question. Before you conclude, decide which question the test answered.
| Observation | What it shows |
|---|---|
| a gas decolourises acidified potassium manganate(VII) | the gas is sulfur dioxide: of the gases in the gas tests, only sulfur dioxide gives this result |
| a solution decolourises acidified potassium manganate(VII) | the solution contains a reducing agent; iron(II) ions and other reducing agents give the same result, so the substance is not named |
| a gas or solution turns aqueous potassium iodide from colourless to brown | an oxidising agent is present; chlorine gives this result, but chlorine is identified by bleaching damp litmus paper, not by the potassium iodide test |
These tests tell you what the unknown does to the reagent. A test names the unknown only when you know it is one of a short list, such as the six gases, and only one of them gives that result.
Check your understanding 4 Does the result name the unknown?
G is one of the six gases in the gas tests, and it decolourises acidified potassium manganate(VII). Colourless solution S gives the same change, and no other tests have been done on it. What does each result show, and which reagent would test a fresh portion of S for an oxidising agent?
Show answer
G is sulfur dioxide: it is the only one of the six gases that decolourises the reagent. S contains a reducing agent, but the result does not say which. To test S for an oxidising agent, add aqueous potassium iodide to a fresh portion: colourless to brown shows one.
Common mistakes
- Saying the unknown changed colour. The colour change belongs to the reagent: iodine forming, or manganate(VII) being used up.
- Mixing up the tests: potassium iodide detects an oxidising agent, and acidified manganate(VII) detects a reducing agent.
- Leaving out “acidified” when naming the manganate(VII) reagent.
- Naming a solution, such as iron(II), from a manganate(VII) result alone. Many reducing agents give the same result.
- Giving only the final colour instead of the change, such as “purple to colourless”.
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Syllabus and review details
- SEC G3 Combined Science Chemistry component 2027 · 2027
Chemistry content structure and subject content, PDF pages 26–36
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