Redox Titrations

Redox titrations use electron-transfer reactions instead of acid-base reactions to determine unknown concentrations. The most commonly examined redox titrations at A-Level involve potassium manganate(VII) and iodine-thiosulfate systems. Both rely on clear colour changes to indicate the endpoint.

Potassium Manganate(VII) Titrations

Potassium manganate(VII) (KMnO₄) is a strong oxidising agent. In acidic solution, the purple MnO₄⁻ ion is reduced to the almost colourless Mn²⁺ ion:

MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

This reaction is self-indicating: the MnO₄⁻ is intensely purple, while Mn²⁺ is very pale pink (effectively colourless in dilute solution). During the titration, each drop of KMnO₄ solution is decolourised by the reducing agent in the flask. The endpoint is reached when one drop of MnO₄⁻ is no longer decolourised — the solution turns a permanent pale pink/purple colour.

No separate indicator is needed.

The Acidic Conditions

The H⁺ ions are provided by adding excess dilute sulfuric acid to the flask before titrating. Hydrochloric acid is not used because Cl⁻ would be oxidised by MnO₄⁻, interfering with the results. Nitric acid is not used because it is itself an oxidising agent.

Redox Titration Procedure: Step by Step

flowchart TD
    A["Pipette 25.0 cm³ of Fe²⁺ solution into conical flask"] --> B["Add excess dilute H₂SO₄"]
    B --> C["Fill burette with standard KMnO₄ solution"]
    C --> D["Record initial burette reading"]
    D --> E["Add KMnO₄ dropwise, swirling"]
    E --> F{"Purple colour persists for 30 seconds?"}
    F -- No --> E
    F -- Yes --> G["Record final burette reading"]
    G --> H["Titre = final − initial"]
    H --> I["Repeat until concordant (within 0.10 cm³)"]
    I --> J["Calculate mean concordant titre"]

Iron(II) Determination with KMnO₄

The most common application is determining the concentration of iron(II) ions (Fe²⁺) in solution.

Oxidation half-equation: Fe²⁺ → Fe³⁺ + e⁻ Reduction half-equation: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

Multiplying the iron half-equation by 5 and combining:

MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺

The mole ratio is MnO₄⁻ : Fe²⁺ = 1 : 5

Worked Example 1 — Basic KMnO₄ Titration

25.0 cm³ of an iron(II) sulfate solution was acidified with excess dilute sulfuric acid and titrated with 0.0200 mol dm⁻³ KMnO₄ solution. The mean titre was 23.6 cm³. Calculate the concentration of Fe²⁺.

Step 1: Moles of MnO₄⁻ = 0.0200 × (23.6/1000) = 4.72 × 10⁻⁴ mol

Step 2: Mole ratio MnO₄⁻ : Fe²⁺ = 1 : 5 Moles of Fe²⁺ = 5 × 4.72 × 10⁻⁴ = 2.36 × 10⁻³ mol

Step 3: Concentration of Fe²⁺ = 2.36 × 10⁻³ / 0.0250 = 0.0944 mol dm⁻³

Worked Example 2 — Iron Tablet Analysis

A 1.39 g sample of an iron tablet was dissolved in dilute sulfuric acid and made up to 250 cm³ in a volumetric flask. 25.0 cm³ of this solution required 18.5 cm³ of 0.0100 mol dm⁻³ KMnO₄. Calculate the percentage by mass of iron in the tablet.

Step 1: Moles of MnO₄⁻ = 0.0100 × 0.0185 = 1.85 × 10⁻⁴ mol

Step 2: Moles of Fe²⁺ in 25.0 cm³ = 5 × 1.85 × 10⁻⁴ = 9.25 × 10⁻⁴ mol

Step 3: Moles of Fe²⁺ in 250 cm³ = 9.25 × 10⁻⁴ × 10 = 9.25 × 10⁻³ mol (Because 250/25.0 = 10, the original solution is 10 times the aliquot)

Step 4: Mass of Fe = 9.25 × 10⁻³ × 56.0 = 0.518 g

Step 5: Percentage = (0.518 / 1.39) × 100 = 37.3%

Worked Example 3 — Harder: Finding Mᵣ of a Salt

0.725 g of an iron(II) salt was dissolved in acid and titrated with 0.0200 mol dm⁻³ KMnO₄. The titre was 21.0 cm³. If the salt contains one Fe²⁺ per formula unit, find the Mᵣ of the salt.

Step 1: Moles of MnO₄⁻ = 0.0200 × 0.0210 = 4.20 × 10⁻⁴ mol

Step 2: Moles of Fe²⁺ = 5 × 4.20 × 10⁻⁴ = 2.10 × 10⁻³ mol

Step 3: Since 1 Fe²⁺ per formula unit, moles of salt = 2.10 × 10⁻³ mol

Step 4: Mᵣ = mass / moles = 0.725 / 2.10 × 10⁻³ = 345 g mol⁻¹

This could correspond to Mohr's salt, (NH₄)₂Fe(SO₄)₂·6H₂O, which has Mᵣ = 392.

Iodine-Thiosulfate Titrations

This is a two-step process used to determine the concentration of an oxidising agent.

Step 1: Reaction with Excess Iodide

The oxidising agent reacts with excess potassium iodide (KI) solution to produce iodine (I₂):

Oxidising agent + 2I⁻ → I₂ + reduced form

The iodine turns the solution brown/yellow.

Step 2: Titration with Sodium Thiosulfate

The iodine produced is then titrated with sodium thiosulfate (Na₂S₂O₃) solution:

I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻

The mole ratio is I₂ : S₂O₃²⁻ = 1 : 2

The Starch Indicator

As thiosulfate is added, the brown iodine colour fades to yellow. When the solution is pale yellow, starch indicator is added. Starch forms an intense blue-black complex with iodine. The endpoint is when the blue-black colour disappears and the solution becomes colourless.

Starch is added late (when the solution is pale yellow) because adding it too early when the iodine concentration is high causes the starch-iodine complex to coagulate, making the endpoint difficult to see.

Why Add Starch Late?

When starch is added	What happens	Effect on accuracy
Too early (dark brown)	Blue-black complex is very intense and sticky	Endpoint unclear, overshooting likely
At the right time (pale yellow)	Sharp colour change: blue-black → colourless	Accurate endpoint
Too late (almost colourless)	Little colour change visible	Endpoint may be missed

Iodine-Thiosulfate Worked Examples

Worked Example 4 — Copper(II) Determination

Excess KI was added to 25.0 cm³ of copper(II) sulfate solution. The iodine liberated required 22.5 cm³ of 0.100 mol dm⁻³ Na₂S₂O₃ for titration. Find the concentration of Cu²⁺.

Reaction 1: 2Cu²⁺ + 4I⁻ → 2CuI + I₂ Reaction 2: I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻

Step 1: Moles of S₂O₃²⁻ = 0.100 × 0.0225 = 2.25 × 10⁻³ mol

Step 2: Moles of I₂ = 2.25 × 10⁻³ / 2 = 1.125 × 10⁻³ mol (1:2 ratio)

Step 3: From equation 1, 2Cu²⁺ produces 1 I₂, so moles Cu²⁺ = 2 × 1.125 × 10⁻³ = 2.25 × 10⁻³ mol

Step 4: [Cu²⁺] = 2.25 × 10⁻³ / 0.0250 = 0.0900 mol dm⁻³

Worked Example 5 — Vitamin C Determination

Vitamin C (ascorbic acid, C₆H₈O₆) can be determined by reaction with excess I₂ solution, then back-titrating the unreacted I₂ with thiosulfate. 10.0 cm³ of 0.0500 mol dm⁻³ I₂ was added to a vitamin C tablet dissolved in water. The excess I₂ required 5.50 cm³ of 0.100 mol dm⁻³ Na₂S₂O₃.

Step 1: Total moles I₂ = 0.0500 × 0.0100 = 5.00 × 10⁻⁴ mol

Step 2: Moles S₂O₃²⁻ = 0.100 × 0.00550 = 5.50 × 10⁻⁴ mol Moles excess I₂ = 5.50 × 10⁻⁴ / 2 = 2.75 × 10⁻⁴ mol

Step 3: Moles I₂ that reacted with vitamin C = 5.00 × 10⁻⁴ − 2.75 × 10⁻⁴ = 2.25 × 10⁻⁴ mol

Step 4: The reaction is C₆H₈O₆ + I₂ → C₆H₆O₆ + 2HI (1:1 ratio) Moles vitamin C = 2.25 × 10⁻⁴ mol

Step 5: Mass vitamin C = 2.25 × 10⁻⁴ × 176.0 = 0.0396 g = 39.6 mg

Comparison of Redox Titration Systems

Feature	KMnO₄ titrations	I₂/S₂O₃²⁻ titrations
What is in the burette	KMnO₄ solution	Na₂S₂O₃ solution
What is in the flask	Fe²⁺ (or other reducing agent) + H₂SO₄	I₂ (from reaction with oxidising agent)
Indicator	Self-indicating (purple → colourless)	Starch (added late; blue-black → colourless)
Endpoint colour	Permanent pale pink	Colourless (from blue-black)
Key mole ratio	MnO₄⁻ : Fe²⁺ = 1 : 5	I₂ : S₂O₃²⁻ = 1 : 2
Acid used	Dilute H₂SO₄ only	Not always needed

Common Mistakes in Redox Titrations

Mistake	Effect	How to avoid
Using HCl instead of H₂SO₄ with KMnO₄	Cl⁻ oxidised, titre too high	Always use dilute H₂SO₄
Adding starch too early	Murky blue, hard to see endpoint	Add when solution is pale yellow
Forgetting scaling factor for aliquots	Answer is 10× too small	Multiply by (total volume / aliquot volume)
Wrong mole ratio	Factor of 2 or 5 error	Write balanced equation first
Not converting cm³ to dm³	1000× error	Always divide by 1000
Confusing burette and flask solutions	Inverted calculation	KMnO₄ is in the burette; Fe²⁺ is in the flask

Redox titrations are a favourite topic for A-Level examiners because they combine practical skills, mole calculations, and redox chemistry. Practice the systematic approach: find moles of the titrant, use the mole ratio, then calculate the unknown.

A-Level Deep Dive: Redox Titrations

Spec mapping

Edexcel 9CH0 specification Topic 3 — Redox I and Topic 14 — Redox II, sub-strand on redox titrations, requires students to understand and apply manganate(VII) titrations (KMnO $_4$ as self-indicator), iodine/thiosulfate titrations (starch indicator), and the corresponding stoichiometric calculations to determine concentrations of unknowns (refer to the official specification document for exact wording). Examined across Paper 2 (transition-metal context) and Paper 3 (synoptic practical). The data sheet provides standard electrode potentials for the relevant couples; the half-equations and 1 : 5 (Mn/Fe) and 2 : 1 (S $_2$ O $_3^{2-}$ /I $_2$ ) ratios must be derived or recalled.

Worked example with full mark scheme

Question (8 marks): A 25.00 cm $^3$ aliquot of an FeSO $_4$ solution is acidified with dilute H $_2$ SO $_4$ and titrated with 0.0200 mol dm $^{-3}$ KMnO $_4$ , requiring 24.50 cm $^3$ to reach the endpoint.

(a) Write the balanced ionic equation. (2)

(b) Calculate the concentration of FeSO $_4$ in mol dm $^{-3}$ . (3)

(d) State and justify the choice of indicator. (1)

Solution with mark scheme:

(a) From half-equations: $\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$ and $\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-$ .

Multiply Fe equation by 5 and combine:

$\text{MnO}_4^- + 5\text{Fe}^{2+} + 8\text{H}^+ \rightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}$ .

M1 A1.

(b) Step 1 — moles of KMnO $_4$ .

$n(\text{MnO}_4^-) = 0.02450 \times 0.0200 = 4.900 \times 10^{-4}\,\text{mol}$ .

M1.

Step 2 — moles of Fe $^{2+}$ .

1 : 5 ratio: $n(\text{Fe}^{2+}) = 5 \times 4.900 \times 10^{-4} = 2.450 \times 10^{-3}\,\text{mol}$ .