Skip to main content
EnglandChemistrySyllabus dot point

How are redox titrations carried out and used to determine the amount of a substance?

Redox titrations with manganate(VII) and thiosulfate-iodine, constructing redox equations from half-equations, disproportionation, and using titration data to calculate amounts and concentrations.

An Eduqas A-Level Chemistry PI1.2 answer on redox titrations with manganate(VII) and iodine-thiosulfate, building redox equations from half-equations, disproportionation, and redox titration calculations.

Generated by Claude Opus 4.811 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

Jump to a section
  1. What this topic is asking
  2. Building redox equations from half-equations
  3. The manganate(VII) titration
  4. The iodine-thiosulfate titration
  5. Disproportionation
  6. Redox titration calculations
  7. Examples in context
  8. Try this

What this topic is asking

Eduqas topic PI1.2 covers redox reactions in practice: the two key redox titrations (manganate(VII) and iodine-thiosulfate), building balanced redox equations by combining half-equations, the idea of disproportionation, and using titration data to calculate amounts and concentrations. It applies the electrode-potential theory of PI1.1 to quantitative analysis.

Building redox equations from half-equations

A balanced redox equation is the sum of a reduction and an oxidation half-equation, scaled so the electrons cancel. For example, the manganate(VII)/iron(II) reaction combines MnO4−+8H++5e−→Mn2++4H2O\text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} with 5×(Fe2+→Fe3++e−)5 \times (\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + \text{e}^-) to give the overall 1:51:5 equation.

The manganate(VII) titration

The iodine-thiosulfate titration

Disproportionation

Redox titration calculations

Examples in context

Example 1. Finding the purity of iron tablets. Dissolving an iron(II) supplement and titrating against manganate(VII) determines the iron content, a standard application of the manganate titration in quality control.

Example 2. Measuring chlorine in pool water. The iodine-thiosulfate method liberates iodine in proportion to the oxidising chlorine present, then measures it with thiosulfate, allowing pool operators to check disinfectant levels.

Try this

Q1. State the colour change at the end point of a manganate(VII) titration. [1 mark]

  • Cue. The first permanent (faint) pink colour, as excess purple manganate(VII) is no longer decolourised.

Q2. Deduce the oxidation state of manganese in the manganate(VII) ion, MnO4−\text{MnO}_4^-. [1 mark]

  • Cue. Oxygen is −2-2, so x+4(−2)=−1x + 4(-2) = -1, giving x=+7x = +7.

Exam-style practice questions

Practice questions written in the style of WJEC Eduqas exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Eduqas 20195 marksA 25.0 cm325.0\ \text{cm}^3 sample of iron(II) sulfate solution required 24.00 cm324.00\ \text{cm}^3 of 0.0200 mol dm−30.0200\ \text{mol dm}^{-3} potassium manganate(VII) for complete oxidation in acid. The equation is MnO4−+8H++5Fe2+→Mn2++4H2O+5Fe3+\text{MnO}_4^- + 8\text{H}^+ + 5\text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} + 5\text{Fe}^{3+}. Calculate the concentration of the iron(II) ions.
Show worked answer →

Moles of MnO4−=0.0200×24.001000=4.80×10−4 mol\text{MnO}_4^- = 0.0200 \times \dfrac{24.00}{1000} = 4.80 \times 10^{-4}\ \text{mol} (1).

The 1:51:5 ratio gives moles of Fe2+=5×4.80×10−4=2.40×10−3 mol\text{Fe}^{2+} = 5 \times 4.80 \times 10^{-4} = 2.40 \times 10^{-3}\ \text{mol} (2).

Concentration =2.40×10−325.0/1000=0.0960 mol dm−3= \dfrac{2.40 \times 10^{-3}}{25.0/1000} = 0.0960\ \text{mol dm}^{-3} (2).

Markers reward the moles of manganate, the 1:51:5 ratio and the final concentration.

Eduqas 20214 marks(a) Explain what is meant by disproportionation. (b) Chlorine reacts with cold dilute sodium hydroxide: Cl2+2NaOH→NaCl+NaClO+H2O\text{Cl}_2 + 2\text{NaOH} \rightarrow \text{NaCl} + \text{NaClO} + \text{H}_2\text{O}. Show, using oxidation states, that chlorine has disproportionated.
Show worked answer →

(a) Disproportionation is a reaction in which the same element is simultaneously oxidised and reduced (its atoms end up in both a higher and a lower oxidation state) (1).

(b) In Cl2\text{Cl}_2 chlorine is 00. In NaCl it is −1-1 (reduced) and in NaClO it is +1+1 (oxidised) (2). Because chlorine goes from 00 to both −1-1 and +1+1 in the same reaction, it has disproportionated (1).

Related dot points

Sources & how we know this