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Redox titration technique (Core Practical 13)

Edexcel IAL Chemistry · WCH16 Practical Skills in Chemistry II

Core Practical 13 uses manganate(VII) and iodine-thiosulfate titrations. The equations and core method are in the Unit 5 lessons; Unit 6 asks about each step and what you would see.

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Manganate(VII) titrations

KMnO4 goes in the burette. Its colour is so dark that you read the top of the meniscus.
The flask contains the reducing agent (Fe2+ or ethanedioate) and excess dilute sulfuric acid, measured with a measuring cylinder because it is in excess.
Not hydrochloric acid: Cl- would be oxidised to chlorine, using extra manganate(VII). Not nitric acid: it is an oxidising agent itself.
Self-indicating: MnO4- is decolourised as it reacts. The end point is the first permanent pale pink, caused by a slight excess of MnO4-.
With ethanedioate, warm the flask to about 60 °C. The reaction starts slowly (high activation energy, both ions negative) and then speeds up because the Mn2+ formed is the catalyst: autocatalysis.

Iodine-thiosulfate titrations

1. Add an excess of potassium iodide to the oxidising agent (Cu2+, IO3- in acid, ClO-). Iodine is released.

2. Titrate the iodine with sodium thiosulfate from the burette until the brown colour fades to pale straw (yellow).

3. Add starch: the solution turns blue-black.

4. Continue dropwise until the blue-black colour just disappears.

Starch is added near the end because a large amount of iodine forms a complex with starch that does not release the iodine readily, giving a late end point.

Making the solutions

Make up an iron(II) salt (such as ammonium iron(II) sulfate) in dilute sulfuric acid with boiled, cooled deionised water, to reduce air oxidation of Fe2+.
Use the standard solution steps: weigh by difference, dissolve, transfer with washings, make up to the mark, invert to mix (January 2022 and October 2024 papers).

Worked example

Describe how to find the end point when titrating iron(II) ions with potassium manganate(VII).

Add MnO4- from the burette to the acidified Fe2+ solution, swirling.
Each drop is decolourised as long as Fe2+ remains.
The end point is the first drop that gives a permanent pale pink colour, from a slight excess of MnO4-.

Answer: Colourless (pale green) to the first permanent pale pink.

Key idea: MnO4-: in the burette, read the top of the meniscus, excess dilute H2SO4, self-indicating, end point first permanent pink; autocatalysis by Mn2+ with ethanedioate. Thiosulfate: excess KI, titrate to pale straw, add starch, blue-black to colourless.

Check yourself

Why is hydrochloric acid not used to acidify manganate(VII) titrations?
- It is too weak an acid
- It reacts with the iron(II) ions
- It turns the solution pink
- Chloride ions are oxidised to chlorine, using up extra manganate(VII)
Show answer
Chloride ions are oxidised to chlorine, using up extra manganate(VII). Manganate(VII) is a strong enough oxidising agent to oxidise Cl-.
What is the end point colour when manganate(VII) is added to acidified iron(II) ions?
- Colourless
- The first permanent pale pink
- Deep purple
- Blue-black
Show answer
The first permanent pale pink. A slight excess of MnO4- colours the solution pink.
Why is the top of the meniscus read in a burette containing potassium manganate(VII)?
- The top is more accurate for every solution
- Manganate(VII) does not form a meniscus
- To reduce the titre
- The solution is too dark to see the bottom of the meniscus
Show answer
The solution is too dark to see the bottom of the meniscus. Dark solutions hide the bottom of the meniscus.
Why is starch added only near the end point in an iodine-thiosulfate titration?
- A large amount of iodine forms a complex with starch that releases iodine slowly, giving an inaccurate end point
- Starch reacts with thiosulfate
- Starch is expensive
- Starch turns colourless in iodine
Show answer
A large amount of iodine forms a complex with starch that releases iodine slowly, giving an inaccurate end point. Add it when the solution is pale straw.
In the titration of ethanedioate with manganate(VII), the pink colour disappears slowly at first and then quickly. Why?
- The flask cools down
- The manganate(VII) is diluted
- Mn2+ ions formed in the reaction catalyse it
- Ethanedioate is used up
Show answer
Mn2+ ions formed in the reaction catalyse it. This is autocatalysis.
Why is a measuring cylinder acceptable for adding the dilute sulfuric acid to the flask in a manganate(VII) titration?
- The acid is in excess, so its exact volume does not affect the titre
- Sulfuric acid cannot be pipetted
- The acid does not react
- Measuring cylinders are more precise than pipettes
Show answer
The acid is in excess, so its exact volume does not affect the titre. The October 2024 report credited this idea.
Explain what causes the colour at the end point of a manganate(VII) titration.
Show answer
A slight excess of manganate(VII) ions (MnO4-) that have not reacted makes the solution pink.. Only a few candidates explained this in October 2024.

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Last updated 8 November 2026. Spotted a mistake? Tell us.