Redox Titrations with Transition metal Ions
A concise revision guide to redox titrations: manganate(VII) with iron(II) and with ethanedioate, the mole ratios, the end points and worked calculations.
- 3.1.2.5ix
- 3.1.7vi
- 3.2.5.5v
What these spec points say
- 3.1.2.5ix use balanced equations to calculate concentrations and volumes for reactions in solutions
- 3.1.7vi combine half-equations to give an overall redox equation
- 3.2.5.5v perform calculations for the redox titrations of Fe2+ and C2O42- with MnO4- and similar redox reactions
Before You Start
This page brings together redox equations and titration calculations. Half-equations and E⦵ values are covered in 3.1.11 Electrode potentials and electrochemical cells. The redox reactions of the transition metals are covered in the redox chemistry page.
- Amount in moles = concentration × volume in dm³, or n = cV.
- Divide a volume in cm³ by 1000 to convert it to dm³.
- A mean titre uses concordant titres only: titres within 0.10 cm³ of each other.
- An oxidising agent gains electrons. A reducing agent loses electrons.
- The electrons lost must equal the electrons gained. This fixes the mole ratio.
GCSE Recap: Titrations and Electron Transfer
Three quick questions on the GCSE ideas this page builds on.
Manganate(VII) and Iron(II)
Potassium manganate(VII), KMnO₄, is a strong oxidising agent in acidic solution. It oxidises iron(II) ions to iron(III) ions and is reduced to manganese(II) ions.
| Change | Half-equation | Change in oxidation state |
|---|---|---|
| Reduction | MnO₄⁻(aq) + 8H⁺(aq) + 5e⁻ → Mn²⁺(aq) + 4H₂O(l) | Mn: +7 to +2 |
| Oxidation | Fe²⁺(aq) → Fe³⁺(aq) + e⁻ | Fe: +2 to +3 |
Combining the half-equations
- MnO₄⁻ gains 5 electrons. Each Fe²⁺ loses 1 electron.
- Multiply the iron(II) half-equation by 5, so that both involve 5 electrons.
- Add the two half-equations and cancel the electrons.
MnO₄⁻(aq) + 8H⁺(aq) + 5Fe²⁺(aq) → Mn²⁺(aq) + 5Fe³⁺(aq) + 4H₂O(l)
Key idea: 1 mol of MnO₄⁻ reacts with 5 mol of Fe²⁺. Moles of Fe²⁺ = 5 × moles of MnO₄⁻.
A Self-Indicating Titration
No indicator is added. The manganate(VII) solution acts as its own indicator, so the titration is self-indicating.
| Ion | Colour in dilute solution |
|---|---|
| MnO₄⁻ | deep purple |
| Mn²⁺ | very pale pink: looks colourless |
| Fe²⁺ | very pale green: looks almost colourless |
| Fe³⁺ | very pale yellow: looks almost colourless |
- The MnO₄⁻ solution goes in the burette. The reducing agent and the acid go in the conical flask.
- Each purple drop is decolourised as MnO₄⁻ is reduced to Mn²⁺.
- Once all the reducing agent has reacted, the next drop of MnO₄⁻ stays in the flask.
- The end point is the first permanent pale pink colour.
A manganate(VII) titration: purple MnO₄⁻ from the burette is decolourised in the flask until one drop in excess gives the first permanent pale pink colour.
Practical points
- The deep purple colour hides the bottom of the meniscus. Read the top of the meniscus for both readings.
- Oxygen in the air slowly oxidises Fe²⁺ to Fe³⁺. Titrate iron(II) solutions soon after making them.
- Stand the flask on a white tile and add the MnO₄⁻ drop by drop near the end point.
- Titration technique is part of the assessed practical work: see Required Practicals.
Exam wording: “The colour change at the end point is from colourless to the first permanent pale pink, caused by a slight excess of MnO₄⁻ ions.”
Quick Check: Reasoning About the End Point
Pick the one accurate statement in each set.
Why Dilute Sulfuric Acid Is Used
The reduction of MnO₄⁻ to Mn²⁺ needs 8 mol of H⁺ for each mole of MnO₄⁻. An excess of dilute sulfuric acid in the flask supplies them.
| Acid | Suitable? | Reason | Effect on the titre |
|---|---|---|---|
| Dilute sulfuric acid | Yes | supplies H⁺; SO₄²⁻ is not oxidised or reduced | correct |
| Hydrochloric acid | No | MnO₄⁻ oxidises Cl⁻ to toxic Cl₂ | too large |
| Nitric acid | No | it is an oxidising agent and oxidises some Fe²⁺ | too small |
| Ethanoic acid | No | a weak acid: it cannot supply enough H⁺ | too large |
E⦵ values explain the problem with chloride ions. E⦵ for MnO₄⁻/Mn²⁺ is +1.51 V and E⦵ for Cl₂/Cl⁻ is +1.36 V, so MnO₄⁻ can oxidise Cl⁻ to Cl₂.
Too little acid
MnO₄⁻(aq) + 4H⁺(aq) + 3e⁻ → MnO₂(s) + 2H₂O(l)
- With too little acid, or a weak acid, MnO₄⁻ is only reduced as far as manganese(IV) oxide, MnO₂.
- MnO₂ is a brown solid. The mixture turns brown and cloudy, which hides the end point.
- Each MnO₄⁻ now gains 3 electrons, not 5, so more MnO₄⁻ is needed and the titre is too large.
Check: Equations, Acid and End Point
Build the equation, explain the choice of acid and describe the end point of a manganate(VII) titration, in questions that do not repeat the examples on this page.
Worked Example: Percentage of Iron in a Salt
A 3.36 g sample of an iron(II) salt was dissolved in dilute sulfuric acid and made up to 250 cm³. A 25.0 cm³ portion needed a mean titre of 24.10 cm³ of 0.0100 mol dm⁻³ KMnO₄.
Calculate the percentage by mass of iron in the salt. (Ar of Fe = 55.8)
Step 1. Moles of MnO₄⁻ = (0.0100 × 24.10) / (1000) = 2.41 × 10⁻⁴ mol.
Step 2. Moles of Fe²⁺ in 25.0 cm³ = 5 × 2.41 × 10⁻⁴ = 1.205 × 10⁻³ mol.
Step 3. Moles of Fe²⁺ in 250 cm³ = 10 × 1.205 × 10⁻³ = 1.205 × 10⁻² mol.
Step 4. Mass of iron = 1.205 × 10⁻² × 55.8 = 0.67239 g, which is 0.672 g to 3 significant figures.
Answer. Percentage of iron = (0.67239) / (3.36) × 100 = 20.0%
The iron(II) salt calculation in five steps: moles of MnO₄⁻, moles of Fe²⁺ in the portion, moles in the whole flask, mass of iron and percentage by mass.
The same route answers other questions
- Percentage purity: convert the moles in the whole flask into a mass of the pure compound, then divide by the mass of the sample.
- Molar mass: divide the mass of the sample by the moles of Fe²⁺ in the whole flask.
- Concentration: divide the moles in the portion by the volume of the portion in dm³.
Exam focus: Keep the unrounded value in the calculator between steps. Round only the final answer, to the smallest number of significant figures in the data.
Manganate(VII) and Ethanedioate
Ethanedioate ions, C₂O₄²⁻, are oxidised to carbon dioxide by acidified MnO₄⁻. Each C₂O₄²⁻ ion loses 2 electrons.
| Change | Half-equation | Change in oxidation state |
|---|---|---|
| Reduction | MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O | Mn: +7 to +2 |
| Oxidation | C₂O₄²⁻ → 2CO₂ + 2e⁻ | C: +3 to +4 |
- Multiply the MnO₄⁻ half-equation by 2 and the C₂O₄²⁻ half-equation by 5, so that both involve 10 electrons.
- Add the half-equations and cancel the electrons.
2MnO₄⁻(aq) + 16H⁺(aq) + 5C₂O₄²⁻(aq) → 2Mn²⁺(aq) + 10CO₂(g) + 8H₂O(l)
Key idea: 2 mol of MnO₄⁻ react with 5 mol of C₂O₄²⁻. Moles of C₂O₄²⁻ = (5) / (2) × moles of MnO₄⁻.
What is different in this titration
- The reaction is slow at room temperature, because MnO₄⁻ and C₂O₄²⁻ are both negative ions and repel each other.
- The flask is warmed to about 60 °C before the titration, to speed up the reaction.
- The first drops of MnO₄⁻ are decolourised slowly. Later drops are decolourised quickly.
- The Mn²⁺ ions formed catalyse the reaction. This is autocatalysis.
- Bubbles of carbon dioxide are seen. The end point is still the first permanent pale pink colour.
- Ethanedioic acid, H₂C₂O₄, and its salts react in the same 2 : 5 ratio.
The two manganate(VII) titrations compared: iron(II) at room temperature in a 1 : 5 ratio, and ethanedioate at about 60 °C in a 2 : 5 ratio.
Worked Example: Water of Crystallisation in Ethanedioic Acid
1.56 g of hydrated ethanedioic acid, H₂C₂O₄·xH₂O, was dissolved in water and made up to 250 cm³. A 25.0 cm³ portion, acidified and warmed to 60 °C, needed 19.80 cm³ of 0.0250 mol dm⁻³ KMnO₄. Find x.
Step 1. Moles of MnO₄⁻ = (0.0250 × 19.80) / (1000) = 4.95 × 10⁻⁴ mol.
Step 2. Moles of H₂C₂O₄ in 25.0 cm³ = (5) / (2) × 4.95 × 10⁻⁴ = 1.2375 × 10⁻³ mol.
Step 3. Moles in 250 cm³ = 10 × 1.2375 × 10⁻³ = 1.2375 × 10⁻² mol.
Step 4. Molar mass of the hydrated acid = (1.56) / (1.2375 × 10⁻²) = 126.1 g mol⁻¹.
Answer. x = (36.1) / (18.0) = 2.01, so x = 2 and the formula is H₂C₂O₄·2H₂O.
Check: Manganate(VII) Titration Calculations
Work through calculations for manganate(VII) with iron(II) and with ethanedioate, using data that is not used on this page.
Similar Redox Titrations
A question may use a reducing agent that is not Fe²⁺ or C₂O₄²⁻. The reductant and its oxidation product are given. Count the electrons lost to find the ratio.
| Reducing agent | Half-equation | Ratio MnO₄⁻ : reducing agent |
|---|---|---|
| Iron(II), Fe²⁺ | Fe²⁺ → Fe³⁺ + e⁻ | 1 : 5 |
| Ethanedioate, C₂O₄²⁻ | C₂O₄²⁻ → 2CO₂ + 2e⁻ | 2 : 5 |
| Hydrogen peroxide, H₂O₂ | H₂O₂ → O₂ + 2H⁺ + 2e⁻ | 2 : 5 |
| Iron(II) ethanedioate, FeC₂O₄ | FeC₂O₄ → Fe³⁺ + 2CO₂ + 3e⁻ | 3 : 5 |
Iron(II) ethanedioate: both ions react
- In FeC₂O₄ the Fe²⁺ ion loses 1 electron and the C₂O₄²⁻ ion loses 2, a total of 3 electrons.
- Each MnO₄⁻ gains 5 electrons, so 3 mol of MnO₄⁻ react with 5 mol of FeC₂O₄: 15 electrons each way.
- Moles of FeC₂O₄ = (5) / (3) × moles of MnO₄⁻.
3MnO₄⁻ + 24H⁺ + 5FeC₂O₄ → 3Mn²⁺ + 5Fe³⁺ + 10CO₂ + 12H₂O
Worked example: purity of iron(II) ethanedioate
1.35 g of impure iron(II) ethanedioate, FeC₂O₄, was dissolved in an excess of dilute sulfuric acid and made up to 250 cm³. A 25.0 cm³ portion needed 26.40 cm³ of 0.0200 mol dm⁻³ KMnO₄. Find the percentage purity.
Step 1. Moles of MnO₄⁻ = (0.0200 × 26.40) / (1000) = 5.28 × 10⁻⁴ mol.
Step 2. Moles of FeC₂O₄ in 25.0 cm³ = (5) / (3) × 5.28 × 10⁻⁴ = 8.80 × 10⁻⁴ mol.
Step 3. Moles of FeC₂O₄ in 250 cm³ = 10 × 8.80 × 10⁻⁴ = 8.80 × 10⁻³ mol.
Answer. Percentage purity = (1.26544) / (1.35) × 100 = 93.7%
Remember: Assume that the impurities do not react with MnO₄⁻. If they did, the titre and the calculated purity would be too high.
Quick Check: Finding the Vanadium in an Ore
Work out each stage yourself, then flip the card to compare.
Common Exam Points
Say
- “The end point is the first permanent pale pink colour.”
- “Dilute sulfuric acid is added in excess to provide the H⁺ ions needed to reduce MnO₄⁻ to Mn²⁺.”
- “Moles of Fe²⁺ = 5 × moles of MnO₄⁻” and “moles of C₂O₄²⁻ = (5) / (2) × moles of MnO₄⁻”.
- “Both the Fe²⁺ and the C₂O₄²⁻ in FeC₂O₄ are oxidised, so the ratio is 3 : 5.”
Do not say
- “The end point is purple” (a purple flask means the end point has been overshot).
- “Hydrochloric acid is used to acidify” (chloride ions are oxidised by MnO₄⁻).
- “An indicator is added” to a manganate(VII) titration (it is self-indicating).
Watch for
- Scaling up: a 25.0 cm³ portion from a 250 cm³ flask is one tenth of the sample.
- The 2 : 5 ratio for ethanedioate, not 1 : 5.
- Whether the question wants the mass of iron, of Fe²⁺ ions or of the whole compound.
- Warming the flask for ethanedioate, but not for iron(II).
- Titres recorded to 2 decimal places, ending in 0 or 5.
Check: Redox Titrations
Bring the whole page together: method, choice of acid, sources of error and calculations for manganate(VII) with iron(II) and ethanedioate, in contexts not used on this page.
FAQs
Use these quick answers to check redox titrations with manganate(VII).
Why is no indicator needed in a manganate(VII) titration?
MnO₄⁻ is purple and the Mn²⁺ formed is almost colourless, so the titration is self-indicating. Each drop is decolourised while some reducing agent remains. The end point is the first permanent pale pink colour in the flask.
Why is dilute sulfuric acid used to supply the H⁺ ions?
Sulfate ions are neither oxidised nor reduced. Chloride ions from hydrochloric acid are oxidised to chlorine by MnO₄⁻, nitric acid is itself an oxidising agent, and ethanoic acid is too weak. Too little acid gives brown MnO₂.
What is the mole ratio of manganate(VII) to ethanedioate?
It is 2 : 5. Each MnO₄⁻ gains five electrons and each C₂O₄²⁻ loses two: 2MnO₄⁻ + 16H⁺ + 5C₂O₄²⁻ → 2Mn²⁺ + 10CO₂ + 8H₂O. With Fe²⁺, which loses one electron, the ratio is 1 : 5.
Why is the ethanedioate titration warmed to about 60 °C?
The reaction between MnO₄⁻ and C₂O₄²⁻ is slow at room temperature, so the flask is warmed to speed it up. The Mn²⁺ ions formed then catalyse the reaction, so later drops of manganate(VII) are decolourised much more quickly.
How does iron(II) ethanedioate react with manganate(VII)?
Both ions are oxidised. Fe²⁺ loses one electron and C₂O₄²⁻ loses two, so each FeC₂O₄ supplies three electrons. Each MnO₄⁻ accepts five, so the ratio is 3MnO₄⁻ : 5FeC₂O₄, not 1 : 5.
Copyright notice: This OLS revision content, including the explanations, layout, diagrams, tables and embedded learning structure, is authored for Online Learning System by Dr. Mohammed Al-Fatah. It may not be copied, reproduced, redistributed or adapted without written permission.
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