Redox Titrations: Iron(II) and Manganate(VII)
A concise revision guide to manganate(VII) titrations: the half-equations and the 1 : 5 ratio with iron(II), why the titration is self-indicating, the method step by step, and structured and unstructured calculations.
- 16.15-a
- 16.15-b
- 16.15-c
- 16.17i-a
- 16.17i-b
What these spec points say
- 16.15-a be able to carry out structured titration calculations involving redox reactions
- 16.15-b be able to carry out unstructured titration calculations involving redox reactions
- 16.15-c be able to carry out redox titration calculations involving iron(II) ions and potassium manganate(VII)
- 16.17i-a CORE PRACTICALS 13a and 13b Carry out redox titrations with iron(II) ions
- 16.17i-b CORE PRACTICALS 13a and 13b Carry out redox titrations with potassium manganate(VII)
Before You Start
This page uses two earlier skills: building a redox equation from half-equations (Redox and Half-Equations; Topic 8 Redox Chemistry and Groups 1, 2 and 7) and titration calculations.
- 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.
- The mole ratio comes from the balanced equation.
- An oxidising agent gains electrons. A reducing agent loses electrons.
GCSE Recap: Concentration and Electron Transfer
Three quick questions on the GCSE ideas this page builds on.
The Reaction Between 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 itself reduced to manganese(II) ions.
| Change | Half-equation | Change in oxidation number |
|---|---|---|
| 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
- The manganate(VII) half-equation gains 5 electrons. The iron(II) half-equation 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⁺. Every calculation with iron(II) on this page depends on this 1 : 5 ratio.
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 |
- Each drop of purple MnO₄⁻ is decolourised as it reacts with Fe²⁺ in the flask.
- Once all the Fe²⁺ has reacted, the next drop of MnO₄⁻ has nothing to react with.
- The end point is the first permanent pale pink colour in the flask.
A manganate(VII) titration: purple MnO₄⁻ in the burette is decolourised by Fe²⁺ in the flask until one drop in excess gives the first permanent pale pink colour.
Exam wording: “The end point is the first permanent pale pink colour, caused by a slight excess of MnO₄⁻ ions. The colour change is from colourless to pale pink.”
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 is added to the flask to supply them.
| Acid | Suitable? | Reason |
|---|---|---|
| Dilute sulfuric acid | Yes | supplies H⁺; the SO₄²⁻ ion is not oxidised or reduced |
| Hydrochloric acid | No | MnO₄⁻ oxidises Cl⁻ to Cl₂, so the titre is too large |
| Nitric acid | No | it is an oxidising agent and oxidises some Fe²⁺, so the titre is too small |
| Concentrated sulfuric acid | No | it is an oxidising agent and may oxidise some Fe²⁺ |
| Ethanoic acid | No | a weak acid: it does not supply enough H⁺ |
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⁻. See the electrochemical series.
Too little acid
MnO₄⁻(aq) + 4H⁺(aq) + 3e⁻ → MnO₂(s) + 2H₂O(l)
- With too little acid, MnO₄⁻ is only reduced as far as manganese(IV) oxide, MnO₂.
- MnO₂ is a brown solid, so the mixture turns brown and cloudy and the end point is hidden.
- Each MnO₄⁻ now gains 3 electrons, not 5. The ratio is no longer 1 : 5 and the result is wrong.
- More MnO₄⁻ is needed to oxidise the same amount of Fe²⁺, so 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.
The Method Step by Step
- Weigh the sample that contains iron(II) and dissolve it in an excess of dilute sulfuric acid.
- Transfer the solution and all the washings to a 250 cm³ volumetric flask. Make up to the mark with distilled water and invert to mix.
- Use a pipette to transfer 25.0 cm³ of this solution to a conical flask.
- Add about 20 cm³ of dilute sulfuric acid to the conical flask. This is an excess.
- Fill a burette with standard potassium manganate(VII) solution and record the initial reading.
- Add the manganate(VII) with swirling until the first permanent pale pink colour appears. Record the final reading.
- Repeat until two titres are concordant, then calculate the mean titre.
Practical points
- The MnO₄⁻ goes in the burette, so that the flask changes from colourless to pink at the end point.
- 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, so that the pale pink colour is easy to see.
- Add the manganate(VII) drop by drop near the end point.
- This titration is part of the assessed practical work: see Core Practicals 12, 13a and 13b.
Remember: A standard solution has an accurately known concentration. Here the standard solution is the potassium manganate(VII) in the burette.
Quick Check: What Each Error Does to the Titre
Pick the one accurate statement in each set about a slip made during the titration.
Structured Calculation: Iron in Iron Tablets
Crushed iron tablets with a mass of 7.85 g were dissolved in dilute sulfuric acid and made up to 250 cm³. A 25.0 cm³ portion needed a mean titre of 22.50 cm³ of 0.0200 mol dm⁻³ KMnO₄.
Calculate the percentage by mass of iron(II) in the tablets. (Ar of Fe = 55.8)
Step 1. Moles of MnO₄⁻ = (0.0200 × 22.50) / (1000) = 4.50 × 10⁻⁴ mol.
Step 2. Moles of Fe²⁺ in 25.0 cm³ = 5 × 4.50 × 10⁻⁴ = 2.25 × 10⁻³ mol.
Step 3. Moles of Fe²⁺ in 250 cm³ = 10 × 2.25 × 10⁻³ = 2.25 × 10⁻² mol.
Step 4. Mass of Fe²⁺ = 2.25 × 10⁻² × 55.8 = 1.2555 g, which is 1.26 g to 3 significant figures.
Answer. Percentage of iron(II) = (1.2555) / (7.85) × 100 = 16.0%
The iron tablet calculation in five steps: moles of MnO₄⁻, moles of Fe²⁺ in the portion, moles in the whole flask, mass of iron(II) and percentage by mass.
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 (here 3).
Unstructured Calculation: Water of Crystallisation
In an unstructured question the steps are not given. Plan the route before starting.
- Moles of MnO₄⁻ from the mean titre.
- Multiply by 5 for the moles of Fe²⁺ in the portion.
- Scale up to the whole volumetric flask.
- Use the mass of the sample to find what is asked: a molar mass, a value of x or a percentage purity.
Worked example
6.95 g of hydrated iron(II) sulfate, FeSO₄·xH₂O, was dissolved in dilute sulfuric acid and made up to 250 cm³. A 25.0 cm³ portion needed 25.00 cm³ of 0.0200 mol dm⁻³ KMnO₄. Find x.
Step 1. Moles of MnO₄⁻ = (0.0200 × 25.00) / (1000) = 5.00 × 10⁻⁴ mol.
Answer. x = (126.1) / (18.0) = 7.01, so x = 7 and the formula is FeSO₄·7H₂O.
Percentage purity
- Convert the moles of Fe²⁺ in the whole flask into a mass of the pure compound.
- Percentage purity = (mass of pure compound) / (mass of impure sample) × 100.
- Assume that the impurities do not react with MnO₄⁻.
Check: Manganate(VII) Titration Calculations
Work through structured and unstructured calculations for iron(II) and manganate(VII) titrations, with data not used on this page.
Other Reducing Agents
Manganate(VII) is used to analyse other reducing agents in the same way. Only the mole ratio changes.
| 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 |
2MnO₄⁻ + 16H⁺ + 5C₂O₄²⁻ → 2Mn²⁺ + 10CO₂ + 8H₂O
2MnO₄⁻ + 6H⁺ + 5H₂O₂ → 2Mn²⁺ + 5O₂ + 8H₂O
- Each of these reducing agents loses 2 electrons. Two MnO₄⁻ ions gain 10 electrons, so 5 are needed.
- The ethanedioate reaction is slow at room temperature, so the flask is warmed to about 60 °C first.
- It starts slowly because MnO₄⁻ and C₂O₄²⁻ are both negative ions, which repel each other.
- The Mn²⁺ ions formed catalyse this reaction (autocatalysis): the first drops decolourise slowly, later drops quickly.
- Ethanedioic acid, H₂C₂O₄, reacts in the same 2 : 5 ratio.
Worked example
A 25.0 cm³ portion of sodium ethanedioate solution, acidified and warmed to 60 °C, needed 21.60 cm³ of 0.0200 mol dm⁻³ KMnO₄. Find the concentration of the ethanedioate ions.
Step 1. Moles of MnO₄⁻ = (0.0200 × 21.60) / (1000) = 4.32 × 10⁻⁴ mol.
Step 2. Moles of C₂O₄²⁻ = (5) / (2) × 4.32 × 10⁻⁴ = 1.08 × 10⁻³ mol.
Answer. Concentration = (1.08 × 10⁻³) / (0.0250) = 0.0432 mol dm⁻³
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.
- Moles of FeC₂O₄ = (5) / (3) × moles of MnO₄⁻.
3MnO₄⁻ + 24H⁺ + 5FeC₂O₄ → 3Mn²⁺ + 5Fe³⁺ + 10CO₂ + 12H₂O
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⁻.”
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” (the titration 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 and hydrogen peroxide, not 1 : 5.
- Whether the question wants the mass of iron, of Fe²⁺ ions or of the whole compound.
- Titres recorded to 2 decimal places, ending in 0 or 5.
Check: Manganate(VII) Titrations
Bring the whole page together: method, sources of error, other reducing agents and calculations, in contexts not used on this page.
FAQs
Use these quick answers to check manganate(VII) titrations with iron(II).
Why is no indicator needed in a manganate(VII) titration?
Manganate(VII) ions are purple and the Mn²⁺ ions formed are almost colourless, so each drop is decolourised while Fe²⁺ remains. The end point is the first permanent pale pink colour, caused by one drop of excess MnO₄⁻.
Why is dilute sulfuric acid used and not hydrochloric acid?
The reaction needs H⁺ ions, and sulfate ions are neither oxidised nor reduced. Chloride ions from hydrochloric acid would be oxidised to chlorine by MnO₄⁻, making the titre too high. Nitric acid is itself an oxidising agent, so it is not used either.
Why do manganate(VII) and iron(II) react in a 1 : 5 ratio?
MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O needs five electrons, and each Fe²⁺ → Fe³⁺ + e⁻ supplies one. So 1 mol of MnO₄⁻ reacts with 5 mol of Fe²⁺: MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O.
What does a brown colour in the flask mean?
Too little acid was added. Without enough H⁺ ions, MnO₄⁻ is reduced only to brown manganese(IV) oxide, MnO₂, instead of Mn²⁺. The mole ratio is then wrong, so the titration is repeated with an excess of dilute sulfuric acid.
How do I calculate the moles of iron(II) from the titre?
Moles of MnO₄⁻ = concentration × titre in dm³, then multiply by 5. A titre of 22.50 cm³ of 0.0200 mol dm⁻³ KMnO₄ is 4.50 × 10⁻⁴ mol of MnO₄⁻, so the portion contains 2.25 × 10⁻³ mol of Fe²⁺.
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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