CP 13a and 13b Redox Titrations: Manganate(VII) and Iodine–Thiosulfate
A focused revision guide to Core Practicals 13a and 13b for Edexcel International A Level Chemistry: the iron(II) standard, the manganate(VII) titration read at the top of the meniscus, the choice of sulfuric acid, the self-indicating end point, the 1 : 5 calculation, the iodine–thiosulfate titration with starch added near the end, and the errors examiners expect you to evaluate.
Redox Titrations: Manganate(VII) and Iodine–Thiosulfate
Titrate iron(II) with self-indicating manganate(VII), watch autocatalysis in the ethanedioate titration, then find copper(II) by iodine and thiosulfate with starch added near the end.
© Dr. Mohammed Al-Fatah – onlinelearningsystem.net
Recap: Oxidation Numbers and Half-Equations
Three quick questions on the redox ideas this practical builds on: oxidation numbers, balancing a half-equation and naming the reducing agent.
What This Practical Is Testing
Core Practicals 13a and 13b finds the concentration of a reducing agent or an oxidising agent by titration. Part A titrates iron(II) ions with potassium manganate(VII). Part B titrates iodine, liberated from an oxidising agent, with sodium thiosulfate.
The burette technique, the rough titre and the rule for concordant titres (within 0.20 cm³ of each other) come from Core Practical 3 (concentration of hcl by titration) and Core Practical 4 (preparation of a standard solution).
What is new is the mole ratio, taken from the balanced redox equation, and an end point with no added indicator.
On the Edexcel International course this practical is examined through the Unit 6 practical skills paper (WCH16), which asks for methods, apparatus choices, data processing, uncertainty and evaluation.
Specification points 16.15 to 16.17 ask for structured and unstructured redox titration calculations with iron(II) and manganate(VII) and with thiosulfate and iodine, and for a discussion of measurement uncertainty.
Key idea: In a redox titration the mole ratio comes from the balanced redox equation, never from a 1 : 1 guess: 1 MnO₄⁻ reacts with 5 Fe²⁺, and 1 I₂ reacts with 2 S₂O₃²⁻.
Safety and Apparatus
The glassware is the same as for an acid–base titration. Two extra pieces matter here: the volumetric flask and balance used to make the iron(II) standard, and the starch solution for the iodine end point.
| Item | What it is for | Precision or note |
|---|---|---|
| 50 cm³ burette | Delivers the titrant: KMnO₄(aq) in part A, Na₂S₂O₃(aq) in part B | ±0.05 cm³ per reading; KMnO₄ is read at the top of the meniscus |
| 25.0 cm³ pipette and filler | Measures the Fe²⁺(aq) or Cu²⁺(aq) being analysed | ±0.06 cm³ |
| 250 cm³ volumetric flask | Makes the iron(II) standard solution | ±0.3 cm³ |
| Balance reading to 0.01 g | Weighs the ammonium iron(II) sulfate | ±0.005 g per reading |
| 250 cm³ conical flask and white tile | Holds the mixture; the tile shows the faint pink or the last trace of blue-black | swirl after every addition |
| 10 cm³ measuring cylinder | Adds the excess dilute sulfuric acid, or the excess KI | ±0.1 cm³; only an excess is needed, so a cylinder is precise enough |
| Starch solution | Indicator for iodine, added near the end point of part B | freshly made, about 2 cm³ |
The apparatus for a redox titration: the burette is read at the top of the meniscus when it holds manganate(VII).
Safety: Potassium manganate(VII) is an oxidising agent and stains skin and clothing. Dilute sulfuric acid (1.0 mol dm⁻³) is an irritant. Iodine solution stains and irritates; copper(II) salts are harmful if swallowed. Wear eye protection throughout.
Method: Step by Step (Part A, Iron(II) with Manganate(VII))
The standard solution of iron(II) is made from ammonium iron(II) sulfate hexahydrate, (NH₄)₂Fe(SO₄)₂·6H₂O. The manganate(VII) is in the burette and no indicator is added.
| Step | What you do | Why |
|---|---|---|
| Make the standard | Weigh 9.80 g of (NH₄)₂Fe(SO₄)₂·6H₂O, dissolve in dilute sulfuric acid and make up to 250 cm³ in a volumetric flask. | The acid stops the Fe²⁺ hydrolysing and slows its oxidation by air; the concentration is 0.100 mol dm⁻³. |
| Fill the burette | Rinse the burette with 0.0200 mol dm⁻³ KMnO₄(aq), fill it and read the top of the meniscus. | The solution is too dark to see the bottom of the meniscus; read the same point every time. |
| Pipette the sample | Pipette 25.0 cm³ of the iron(II) solution into a conical flask on a white tile. | An accurately known volume of the solution being analysed. |
| Acidify | Add about 10 cm³ of 1.0 mol dm⁻³ sulfuric acid from a measuring cylinder. | H⁺ is a reactant in the reduction of MnO₄⁻; an excess makes sure Mn²⁺ forms and not brown MnO₂. |
| Rough titration | Add the KMnO₄(aq) with swirling until the first permanent pale pink appears. | Shows roughly where the end point is, so the accurate titrations can be done dropwise near it. |
| Accurate titrations | Repeat, adding dropwise near the end point, and stop at the first pale pink that lasts about 30 s. | One drop of excess MnO₄⁻ colours the colourless solution: the titration is self-indicating. |
| Concordant results | Repeat until two titres agree within 0.20 cm³ of each other; record every reading to the nearest 0.05 cm³. | Concordant titres show the end point was judged consistently. |
| Calculate | Take the mean of the concordant titres only and use the 1 : 5 ratio. | The rough titre is not used; the ratio comes from the balanced equation. |
The manganate(VII) titration method from filling the burette to concordant titres.
Technique point: The pale green of iron(II) and the pale yellow of iron(III) are so faint at this dilution that the flask looks colourless until the end point. Swirl constantly: the purple must be gone before the next drop is judged.
Check: Iron in Steel Wire
Order the method for a different sample analysed with the same titration.
The Chemistry: Half-Equations, the Acid and the End Point
Manganate(VII) is reduced from +7 to +2 and gains five electrons; each iron(II) ion loses one. Five iron(II) ions are therefore needed for each manganate(VII) ion.
MnO₄⁻(aq) + 8H⁺(aq) + 5e⁻ → Mn²⁺(aq) + 4H₂O(l)
Fe²⁺(aq) → Fe³⁺(aq) + e⁻ (× 5)
MnO₄⁻(aq) + 8H⁺(aq) + 5Fe²⁺(aq) → Mn²⁺(aq) + 4H₂O(l) + 5Fe³⁺(aq)
Why no indicator
The titration is self-indicating. MnO₄⁻ is intensely purple and Mn²⁺ is almost colourless at this concentration, so each drop is decolourised while iron(II) remains. The first drop in excess turns the whole solution pale pink.
Why the pink fades. Left standing, excess MnO₄⁻ slowly reacts with the Mn²⁺ formed to give brown MnO₂, so the pink does not last for ever. Read the first pink that stays for about 30 s; do not wait for a permanent colour.
Why sulfuric acid, and not another acid
Hydrochloric acid cannot be used: manganate(VII) oxidises chloride ions to chlorine, some titrant is used up on the acid, and the titre is too high.
Nitric acid is itself an oxidising agent. It oxidises some iron(II) to iron(III) before the titration, so the titre is too low.
Too little acid of any kind reduces manganate(VII) only to manganese(IV) oxide, a brown precipitate, and the 1 : 5 ratio no longer holds. Dilute sulfuric acid supplies plenty of H⁺ and takes no part in any other redox reaction.
Exam wording: “Dilute sulfuric acid is used because it provides H⁺ ions but is not oxidised by manganate(VII); hydrochloric acid would be oxidised to chlorine, using up extra manganate(VII) and making the titre too high.”
Check: Choosing the Conditions
Choose the correct statements about the acid, the meniscus, the indicator and the fading pink in new situations.
Worked Calculation: the Iron(II) Standard
The table gives a complete set of results for 25.0 cm³ portions of the iron(II) solution titrated with 0.0200 mol dm⁻³ KMnO₄. Every reading is to the nearest 0.05 cm³, taken at the top of the meniscus.
| Titration | Rough | 1 | 2 | 3 |
|---|---|---|---|---|
| Final reading / cm³ | 25.40 | 25.10 | 25.00 | 25.15 |
| Initial reading / cm³ | 0.00 | 0.10 | 0.05 | 0.15 |
| Titre / cm³ | 25.40 | 25.00 | 24.95 | 25.00 |
Titres 1, 2 and 3 are concordant (within 0.20 cm³ of each other), so the mean titre is (25.00 + 24.95 + 25.00) ÷ 3 = 24.98 cm³. The rough titre is not used.
Step 1: moles of MnO₄⁻ = 0.0200 × 24.98 ÷ 1000 = 4.996 × 10⁻⁴ mol
Step 2: moles of Fe²⁺ = 5 × 4.996 × 10⁻⁴ = 2.498 × 10⁻³ mol (in 25.0 cm³)
Step 3: [Fe²⁺] = 2.498 × 10⁻³ ÷ 0.0250 = 0.0999 mol dm⁻³
Answer: 0.0999 mol dm⁻³ (3 s.f., the precision of the KMnO₄ concentration).
From the mean titre to the concentration of iron(II): four steps, the 1 : 5 ratio, and the check against the mass weighed out.
Compare with the mass weighed out. 9.80 g of (NH₄)₂Fe(SO₄)₂·6H₂O (Mr 392.1) is 9.80 ÷ 392.1 = 0.02499 mol in 250 cm³, which is 0.100 mol dm⁻³.
The titration agrees with the standard to within its own uncertainty (about 0.6%), so no iron(II) had been oxidised by air.
Unstructured calculations: Write the route before the numbers: titre → moles of titrant → ratio from the equation → moles in the pipetted volume → scale to the whole solution → concentration, mass or percentage. Carry four figures and round only the answer.
Worked Example: Iron in an Iron Tablet
The same titration finds the mass of iron in a tablet. Five tablets are crushed and dissolved in 100 cm³ of dilute sulfuric acid; 10.0 cm³ portions are titrated with 0.00500 mol dm⁻³ KMnO₄, giving a mean titre of 12.50 cm³.
Step 1: moles of MnO₄⁻ = 0.00500 × 12.50 ÷ 1000 = 6.25 × 10⁻⁵ mol
Step 2: moles of Fe²⁺ in 10.0 cm³ = 5 × 6.25 × 10⁻⁵ = 3.125 × 10⁻⁴ mol
Step 3: moles of Fe²⁺ in 100 cm³ = 3.125 × 10⁻⁴ × 10 = 3.125 × 10⁻³ mol
Step 4: mass of iron = 3.125 × 10⁻³ × 55.8 = 0.1744 g in five tablets
Answer: 0.1744 ÷ 5 = 0.0349 g = 34.9 mg of iron per tablet (3 s.f.).
The iron tablet calculation as a flow: scale from the 10.0 cm³ portion to the whole 100 cm³ before dividing by five tablets.
Why a dilute titrant. The tablet solution is about three times weaker than the standard above and the portion is only 10.0 cm³, so 0.00500 mol dm⁻³ KMnO₄ keeps the titre in the useful 10 to 25 cm³ range. A titre under 10 cm³ carries a large percentage uncertainty.
Watch the scaling: Students lose marks by dividing by five before scaling up from the 10.0 cm³ portion to the 100 cm³ solution, or by forgetting the × 10 altogether. Write the volume beside every mole value.
Planning the Concentrations
Planning questions ask you to choose a titrant concentration so that the titre falls between about 20 and 25 cm³: large enough for a small percentage uncertainty, small enough to fit one burette filling.
Step 1: moles of Fe²⁺ in 25.0 cm³ of 0.0999 mol dm⁻³ = 2.50 × 10⁻³ mol
Step 2: moles of MnO₄⁻ needed = 2.50 × 10⁻³ ÷ 5 = 5.00 × 10⁻⁴ mol
Step 3: for a 25.0 cm³ titre, [KMnO₄] = 5.00 × 10⁻⁴ ÷ 0.0250 = 0.0200 mol dm⁻³
Answer: 0.0200 mol dm⁻³ KMnO₄, which is the concentration used above. For the copper(II) titration, 2.00 × 10⁻³ mol of S₂O₃²⁻ at 0.100 mol dm⁻³ gives 20.0 cm³, again in range.
If the sample is unknown, run a rough titration first. A rough titre under 10 cm³ means dilute the titrant; one over 50 cm³ means dilute the sample, or use a smaller pipette, and record the dilution factor.
Why 20 to 25 cm³: The ±0.10 cm³ on a titre is fixed. On 25.00 cm³ it is 0.40%; on 5.00 cm³ it is 2.0%. The percentage uncertainty falls as the titre grows, up to the 50 cm³ limit of the burette.
Check: Redox Titration Calculations
Four unstructured calculations on new data: work each one out on paper, then flip the card.
Method: Step by Step (Part B, Iodine with Thiosulfate)
Thiosulfate ions reduce iodine to iodide and are oxidised to tetrathionate. The iodine is made in the flask by the oxidising agent being analysed, here copper(II) ions, reacting with an excess of potassium iodide.
2Cu²⁺(aq) + 4I⁻(aq) → 2CuI(s) + I₂(aq)
I₂(aq) + 2S₂O₃²⁻(aq) → 2I⁻(aq) + S₄O₆²⁻(aq)
| Step | What you do | Why |
|---|---|---|
| Pipette the sample | Pipette 25.0 cm³ of the copper(II) solution into a conical flask. | An accurately known volume of the oxidising agent. |
| Liberate the iodine | Add about 10 cm³ of 1 mol dm⁻³ KI(aq): the solution turns brown and a white precipitate of CuI forms. | Iodide in excess, so the iodine formed depends only on the Cu²⁺ present. |
| Titrate to pale straw | Run in 0.100 mol dm⁻³ Na₂S₂O₃(aq) from the burette, swirling, until the brown fades to pale straw. | Most of the iodine is removed before the starch goes in. |
| Add starch | Add about 2 cm³ of freshly made starch solution: the mixture turns blue-black. | Starch shows the last traces of iodine; added early it binds too much iodine and the end point is late. |
| Finish dropwise | Add thiosulfate drop by drop until the blue-black just disappears. | The end point looks cream or off-white, because the CuI precipitate stays in the flask. |
| Repeat | Repeat until two titres agree within 0.20 cm³ of each other; titrate each portion promptly. | Iodine is volatile: a delay loses iodine and lowers the titre. |
The iodine–thiosulfate titration of copper(II) in five tiles: liberate iodine with excess iodide, titrate to pale straw, add starch, finish dropwise to an off-white end point.
Exam wording: “Add thiosulfate until the solution is pale straw, then add starch: the blue-black colour disappears at the end point. Starch is added near the end point because iodine bound to starch is released slowly.”
Worked Calculation: Copper(II) by Iodine–Thiosulfate
25.0 cm³ of the copper(II) solution, treated with excess KI, needs a mean titre of 20.00 cm³ of 0.100 mol dm⁻³ Na₂S₂O₃. The calculation chains the two equations: thiosulfate gives iodine, iodine gives copper(II).
Step 1: moles of S₂O₃²⁻ = 0.100 × 20.00 ÷ 1000 = 2.00 × 10⁻³ mol
Step 2: moles of I₂ = 2.00 × 10⁻³ ÷ 2 = 1.00 × 10⁻³ mol
Step 3: moles of Cu²⁺ = 2 × 1.00 × 10⁻³ = 2.00 × 10⁻³ mol, so Cu²⁺ : S₂O₃²⁻ = 1 : 1
Step 4: [Cu²⁺] = 2.00 × 10⁻³ ÷ 0.0250 = 0.0800 mol dm⁻³
Answer: 0.0800 mol dm⁻³ (3 s.f.). The 1 : 1 ratio between Cu²⁺ and S₂O₃²⁻ is a result, not an assumption: each Cu²⁺ gives half an I₂, and each I₂ needs two S₂O₃²⁻.
Other oxidising agents analysed the same way
| Oxidising agent | Reaction with excess iodide | Ratio to S₂O₃²⁻ |
|---|---|---|
| Chlorate(I) in bleach | ClO⁻(aq) + 2I⁻(aq) + 2H⁺(aq) → Cl⁻(aq) + I₂(aq) + H₂O(l) | 1 ClO⁻ : 2 S₂O₃²⁻ |
| Hydrogen peroxide | H₂O₂(aq) + 2I⁻(aq) + 2H⁺(aq) → I₂(aq) + 2H₂O(l) | 1 H₂O₂ : 2 S₂O₃²⁻ |
| Iodate(V), a primary standard | IO₃⁻(aq) + 5I⁻(aq) + 6H⁺(aq) → 3I₂(aq) + 3H₂O(l) | 1 IO₃⁻ : 6 S₂O₃²⁻ |
| Copper(II) | 2Cu²⁺(aq) + 4I⁻(aq) → 2CuI(s) + I₂(aq) | 1 Cu²⁺ : 1 S₂O₃²⁻ |
Standardising thiosulfate. Sodium thiosulfate solutions slowly decompose, so their concentration is found by titrating the iodine liberated from a weighed amount of potassium iodate(V), KIO₃, which is a primary standard. Bleach is diluted about ten-fold first so that the titre falls in range.
Two-stage ratios: Always go through iodine: oxidising agent → I₂ → S₂O₃²⁻. Write the moles of I₂ on its own line; most ratio errors come from skipping it.
Check: Copper in Brass
Chain the two equations and scale up for a different copper sample.
Errors, Uncertainty and Improvements
The burette and pipette uncertainties are the same as in any titration; the redox chemistry adds its own sources of error, and examiners want the direction of each one.
| Source of error | Effect on the result | Improvement |
|---|---|---|
| Iron(II) solution left standing in air | Some Fe²⁺ oxidised to Fe³⁺: titre too low, [Fe²⁺] too low | Make the solution fresh, in dilute sulfuric acid, and titrate promptly |
| Too little acid in the flask | Brown MnO₂ forms, the ratio changes and the end point is unclear | Add an excess of dilute sulfuric acid; if brown appears, discard the run |
| Hydrochloric acid used to acidify | Some MnO₄⁻ oxidises Cl⁻: titre too high | Use dilute sulfuric acid only |
| Bottom of the manganate(VII) meniscus read | Guesswork through a dark solution: random error, poor concordance | Read the top of the meniscus for every KMnO₄ reading |
| Pink taken as permanent only after minutes | Overshoot while waiting: titre too high | Stop at the first pink that lasts about 30 s |
| Starch added at the start | End point late and gradual: titre too high | Add starch only when the solution is pale straw |
| Iodine lost by evaporation | Less iodine to titrate: titre too low | Titrate as soon as the iodine is liberated; keep the flask cool and covered |
The errors that matter in a redox titration and which way each one moves the titre.
Percentage uncertainty. Each burette reading is ±0.05 cm³, so a titre carries ±0.10 cm³: for 24.98 cm³, 0.10 ÷ 24.98 × 100 = 0.40%. The 25.0 cm³ pipette (±0.06 cm³) adds 0.06 ÷ 25.0 × 100 = 0.24%. Total about 0.64%.
Making the standard adds little: the balance (two readings, ±0.01 g on 9.80 g) gives 0.10% and the 250 cm³ flask (±0.3 cm³) gives 0.12%.
Chemical errors such as air oxidation are far larger than any of these, which is why the evaluation mark goes to them.
Validity. The specification asks you to discuss what measurement uncertainty means for the validity of a final result: a concentration quoted as 0.0999 mol dm⁻³ with 0.64% uncertainty is 0.0999 ± 0.0006 mol dm⁻³, so a fourth significant figure is not justified.
Exam focus: For every error, say which way the titre moves and why: “the titre is too high because some manganate(VII) is used up oxidising chloride ions”.
Check: Errors in Redox Titrations
Evaluate new situations: the wrong acid, early starch, the uncertainty of a small titre and the meniscus rule.
Common Mistakes
The same errors appear in students’ write-ups and exam answers year after year. Each one costs a mark that the corrected version earns.
Ratio. Using 1 : 1 for manganate(VII) and iron(II). The ratio is 1 : 5; for iodine and thiosulfate it is 1 : 2, and for copper(II) and thiosulfate 1 : 1.
End point. Describing the manganate(VII) end point as “purple to colourless”. The flask goes from colourless to the first permanent pale pink.
Indicator. Adding an indicator to a manganate(VII) titration, or calling the manganate(VII) “the indicator”.
Acid. Choosing hydrochloric or nitric acid, or saying only that sulfuric acid “provides H⁺” without saying why the others fail.
Starch. Adding it at the start of an iodine–thiosulfate titration, or calling the copper(II) end point “colourless” when the CuI makes it cream.
Scaling. Forgetting to scale from the pipetted portion to the whole volumetric flask before working out a mass or a percentage.
Meniscus. Reading the bottom of the manganate(VII) meniscus, which cannot be seen, or mixing top and bottom readings.
Common Exam Points
Say
“Manganate(VII) is self-indicating: the end point is the first permanent pale pink.” “Sulfuric acid is used because hydrochloric acid would be oxidised to chlorine by manganate(VII).” “Starch is added when the solution is pale straw; the blue-black colour disappears at the end point.”
Do not say
“The indicator is potassium manganate(VII).” (It is the titrant; no indicator is added.) “The end point is when the solution turns purple.” “MnO₄⁻ reacts with Fe²⁺ in a 1 : 1 ratio.” “The copper(II) end point is colourless.” (It is cream, because of the CuI.)
Watch for
Half-equations combined to give the ratio; iron in a tablet or the purity of a salt; two-stage iodine calculations (oxidising agent → I₂ → S₂O₃²⁻); readings to 0.05 cm³ and concordance within 0.20 cm³ of each other; the direction of every error.
FAQs
Short answers to the questions students most often ask about these two titrations.
Why is no indicator needed with potassium manganate(VII)?
Manganate(VII) is intensely purple and its product, Mn²⁺, is almost colourless. While iron(II) remains each drop is decolourised; the first drop in excess turns the whole solution pale pink, and that is the end point.
Why do we read the top of the meniscus?
Manganate(VII) solution is so dark that the bottom of the meniscus cannot be seen. Reading the top for both readings gives the correct titre, because the same offset is subtracted away.
Why sulfuric acid and not hydrochloric or nitric acid?
Sulfuric acid supplies the H⁺ the half-equation needs and takes no part in any other redox reaction. Manganate(VII) oxidises chloride to chlorine (titre too high); nitric acid oxidises iron(II) first (titre too low).
When exactly should I add the starch?
When the brown iodine has faded to pale straw. Added earlier, much of the iodine binds to the starch and is released only slowly, so the blue-black fades gradually and the end point is late.
How is this titration used in practical questions?
Expect Core Practicals 13a and 13b as a calculation with new data: a mean titre, the 1 : 5 or 1 : 2 ratio, scaling to the flask, then a concentration, mass per tablet or purity, plus one error and its direction.
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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