A LevelChemistryAQARevision NotesAdvanced Inorganic ChemistryTransition MetalsRedox Titrations

Redox Titrations (AQA A Level Chemistry): Revision Note

Exam code: 7405

Written by: Stewart Hird

Reviewed by: Caroline Carroll

Updated on 7 July 2026

Redox Titrations

Redox titration involves an oxidising agent being titrated against a reducing agent
Electrons are transferred from one species to another
In acid-base titrations, indicators are used to show the endpoint of a reaction; however, redox titrations using transition metal ions naturally change colour when changing oxidation state, so indicators are not always necessary
They are said to be 'self-indicating'
Two common redox titrations are manganate(VII) against Fe²⁺ and against C₂O₄^2-

Manganate(VII) Titrations

Potassium manganate(VII) is an oxidising agent and is a deep purple colour
In acidic solutions, it is reduced to the almost colourless manganese(II) ion (the ion is actually pink, but in low concentrations it is effectively colourless)
The reduction equation for the manganate(VII) ion is

MnO₄^- (aq) + 8H⁺(aq) + 5e^- → Mn²⁺(aq) + 4H₂O (l)

purple colourless

The potassium ion is a spectator ion, so it can be left out of the equations
By convention, the potassium manganate(VII) solution is placed in the burette, so that as it reacts with the reducing agent, the solution becomes colourless
At the endpoint, the manganate(VII) ion becomes in excess, so the first appearance of a permanent colour change marks the endpoint
The colour seen is pink; if the colour is purple, then the endpoint has been overshot, and there is too much manganate(VII) in excess
Analysis of iron in iron(II) sulfate tablets is typical manganate(VII) titration

Ethandioate Titrations

Manganate(VII) can also be used to analyse solutions of ethandioate ions by titration
This titration is carried out warm at around 60 ^oC
The equation for the reaction is:

2MnO₄^- + 5C₂O₄^2- + 16H⁺ $\to$ 2Mn²⁺ + 10CO₂ + 8H₂O

Mn²⁺ions autocatalyse the reaction between MnO₄^-and C₂O₄^2-, so the reaction speeds up and then slows down as no more Mn²⁺is produced

Related topics

Worked Example

A health supplement tablet containing iron(II) sulfate was analysed by titration. A tablet weighing 2.25 g was dissolved in dilute sulfuric acid and titrated against 0.100 mol dm^-³ KMnO₄.The titration required 26.50 cm³ for complete reaction. Calculate the percentage by mass of iron in the tablet.

Answer

Step 1: Write the balanced equation for the reaction

oxidation: Fe²⁺(aq) → Fe³⁺(aq) + e^-

reduction: MnO₄^- (aq) + 8H⁺(aq) + 5e^- → Mn²⁺(aq) + 4H₂O (l)

overall: MnO₄^- (aq) + 8H⁺(aq) + 5Fe²⁺(aq) → Mn²⁺(aq) + 4H₂O (l) + 5Fe³⁺(aq)

Step 2: Determine the amount of MnO₄^- used in the titration

moles of MnO₄^-= 0.0265 dm³ x 0.100 mol dm^-³= 0.00265 mol

Step 3: Determine the amount of iron in the reaction

From the equation for the reaction, we know the reacting ratio MnO₄^-: Fe²⁺= 1: 5

∴ moles of Fe²⁺= 0.00265 mol MnO₄^- x 5 = 0.01325 mol

Step 4: Convert moles into the mass of iron

Mass of iron = 0.01325 mol x 55.85 g mol^-1 = 0.740 g

Step 5: Find the percentage of iron in the tablet

∴ % Fe in the tablet = (0.740/ 2.25) x 100 = 32.9%

Worked Example

A 1.26 g sample of impure ethanedioic acid, H₂C₂O₄, was dissolved in water and made up to 250.0 cm³ in a volumetric flask.

A 25.00 cm³ aliquot of this solution required 20.00 cm^³of 0.0200 mol dm⁻³ acidified potassium manganate(VII) solution for complete reaction.

Calculate the molar mass (M_r) of the ethanedioic acid.

Answer

Step 1: Write the balanced equation for the reaction

2MnO₄^- + 5C₂O₄^2- + 16H⁺ $\to$ 2Mn²⁺ + 10CO₂ + 8H₂O

Step 2: Calculate the moles of manganate(VII)

Convert the titre into dm³: 20.00 cm³ = 0.02000 dm³

moles=concentration × volume

=0.0200 × 0.02000

=4.00 × 10⁻⁴ mol

Step 3: Calculate the moles of ethanedioic acid in the aliquot

From the equation: 2mol MnO₄^-= 5 mol C₂O₄^2-

moles of C₂O₄^2-= 4.00 × 10⁻⁴× 5/2

= 1.00 × 10⁻³ mol

Step 4: Calculate the moles in the whole flask

The solution was made up to 250.0 cm³^, which is 10 times the aliquot volume

Total moles=1.00×10⁻³× 10 = 1.00 × 10⁻² mol

Step 5: Calculate the molar mass

$M_{r} = \frac{mass}{moles}$

$= \frac{1.26}{0.0100}$

= 126

Examiner Tips and Tricks

Always show your working in redox titration problems, as marks can be awarded for the steps even if the final answer is wrong.

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Redox Titrations (AQA A Level Chemistry): Revision Note

Redox Titrations

Manganate(VII) Titrations

Ethandioate Titrations

Worked Example

Worked Example

Examiner Tips and Tricks

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Author: Stewart Hird

Reviewer: Caroline Carroll