Oxidising agents, reducing agents and colour evidence
Name agents by what they do to the other species, and explain what happens to the agent itself.
Extended only (Supplement) · Chemistry 0620, 2026–2028.
Before you begin
- Identify oxidation and reduction through electrons and oxidation numbers.
- Read ionic charges and distinguish an observation from an explanation.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
A species gains electrons during a reaction. What happens to it?
What you will learn
- Identify an oxidising agent as a species that is reduced and a reducing agent as one that is oxidised.
- Interpret the required acidified manganate(VII) and iodide colour evidence with its stated conditions.
An agent's name describes its effect on the other substance
An oxidising agent causes another substance to be oxidised and is itself reduced, usually by accepting electrons. A reducing agent causes another substance to be reduced and is itself oxidised by donating electrons. In Zn+Cu²⁺→Zn²⁺+Cu, zinc is the reducing agent because it donates electrons to copper(II) ions; Cu²⁺ is the oxidising agent because it accepts them. Do not name an agent solely from the word oxidised attached to itself.
For 2Fe²⁺+Cl₂→2Fe³⁺+2Cl⁻, iron(II) ions donate electrons and are the reducing agent, while chlorine accepts electrons and is the oxidising agent. Chloride is a product in this stated reaction, not the oxidising reactant. An element or ion's role can depend on the reaction: use the actual change and conditions rather than assigning one permanent role to every substance.
Describe the required observations and connect them to the change
Acidified aqueous potassium manganate(VII) is purple. In the stated acidic test with a suitable reducing agent, the purple manganate(VII) ions are reduced to manganese(II) ions, and the purple colour disappears to give a colourless result at the usual dilute test level. The reducing sample is oxidised. Keep the acidified condition: in different conditions manganese-containing products and colours can differ, so a brown result must not simply be reported as the expected acidic test.
Aqueous potassium iodide is colourless. A suitable oxidising agent oxidises iodide ions to iodine, producing a brown solution under the usual aqueous test conditions. Iodide is the reducing agent and the added oxidising agent is reduced. Iodine can also give a blue-black colour with starch as supporting identification, but brown iodine formation is the specified iodide observation here. These are explanations of supervised observations, not instructions for home mixing; absence of a colour change alone does not prove that every possible redox reaction is absent.
Pause and explain
In Zn+Cu²⁺→Zn²⁺+Cu, which is the oxidising agent?
Worked example
A sample turns acidified purple potassium manganate(VII) colourless in the stated test. Identify the sample's role, the manganate(VII) ion's role and the changes of each.
Show the worked solution
- Manganate(VII) ions are reduced to manganese(II) ions, explaining loss of the purple colour in the stated acidic conditions.
- They accept electrons and therefore act as the oxidising agent.
- The reacting sample supplies electrons, is oxidised and acts as the reducing agent.
Answer The sample is a reducing agent and is oxidised; manganate(VII) is the oxidising agent and is reduced.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
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.
Identify both agents in 2Fe²⁺+Cl₂→2Fe³⁺+2Cl⁻, justifying each by electron transfer.
Give me a hint
Name what each reactant does to the other and what happens to itself.
Compare my reasoning
- Fe²⁺ loses electrons to become Fe³⁺ and is oxidised.
- By supplying those electrons it reduces chlorine, so Fe²⁺ is the reducing agent.
- Cl₂ gains electrons to become Cl⁻, is reduced and acts as the oxidising agent.
Fe²⁺ is the reducing agent; Cl₂ is the oxidising agent.
Look for these in your work
- The agents are reactants in this equation.
- Their own changes and electron roles are explained.
A suitable sample removes the purple colour of acidified aqueous potassium manganate(VII). State the expected observation and identify which reactant is oxidised and which reduced.
Give me a hint
Preserve the acidic condition and distinguish the sample from the coloured reagent.
Compare my reasoning
- Purple manganate(VII) becomes colourless in the stated dilute acidic test as manganese(II) forms.
- Manganate(VII) is reduced and is the oxidising agent.
- The sample is oxidised and acts as the reducing agent.
Purple to colourless: manganate(VII) is reduced; the reducing sample is oxidised.
Look for these in your work
- The observation includes its required condition.
- Agent names agree with their electron roles.
In Cl₂+2I⁻→2Cl⁻+I₂, identify the agents, explain a colourless-to-brown observation for aqueous iodide, and assess the claim that brown colour means iodide was reduced.
Give me a hint
Follow iodine's oxidation number from −1 to zero and chlorine's from zero to −1.
Compare my reasoning
- Iodide increases from −1 to zero, loses electrons and is oxidised; it is the reducing agent.
- Chlorine decreases from zero to −1, gains electrons and is the oxidising agent.
- Brown aqueous iodine is consistent with iodide oxidation under the stated test conditions; it does not show iodide reduction.
I⁻ is the reducing agent and is oxidised to brown iodine; Cl₂ is the oxidising agent and is reduced.
Look for these in your work
- Both agents follow the signed changes.
- The colour observation is correctly connected to iodine formation.
Common mistakes
- Calling the oxidised substance the oxidising agent merely because the words resemble each other.
- Reporting brown manganese oxide as the expected colourless acidic manganate(VII) test.
- Calling potassium iodide itself brown before iodine forms.
- Treating an observation without conditions as proof of one unique substance.
What can you explain now?
An oxidising sample turns colourless aqueous potassium iodide brown. What happens to iodide, and what is its role?
Compare with the explanation
Iodide is oxidised to iodine and acts as the reducing agent.
Iodide donates electrons to the reacting oxidising agent, which is reduced; the brown colour is iodine evidence in the stated test.
After trying it yourself, choose your next review. This is your self-assessment.
Your review choice appears on Today. Sign in to sync it across devices.
Make it stick
Explain both colour tests from observation to electron transfer, then identify agents in a fresh equation. Include what happens to the agent itself.
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.