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Core Experiments and Their Chemistry

Redox titrations and electrochemical cells

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Overview

Recall questions on Redox titrations and electrochemical cells.

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Answer each one before you reveal it.

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Key term

Why is no indicator needed in a titration of iron(II) ions with potassium manganate(VII)?

Key term

Why is dilute sulfuric acid, rather than hydrochloric or nitric acid, used to acidify a manganate(VII) titration?

Key term

Suggest why a measured cell potential falls slowly during a long measurement.

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    Why is no indicator needed in a titration of iron(II) ions with potassium manganate(VII)?

    Model answer

    The manganate(VII) ion is intensely purple and is reduced to the almost colourless manganese(II) ion as it reacts. Once all the iron(II) has been oxidised, the next drop of titrant is not decolourised and the solution takes on a permanent pale pink colour, so the reagent is self-indicating.

  2. 2

    Why is dilute sulfuric acid, rather than hydrochloric or nitric acid, used to acidify a manganate(VII) titration?

    Model answer

    Manganate(VII) is a strong enough oxidising agent to oxidise chloride ions to chlorine, so hydrochloric acid would consume extra titrant and give a titre that is too large. Nitric acid is itself an oxidising agent and would oxidise some of the iron(II) before the titration, while a weak acid such as ethanoic acid would not supply enough hydrogen ions for the reduction to manganese(II).

  3. 3

    Write the ionic equation for the reaction between manganate(VII) ions and iron(II) ions in acidic solution.

    Model answer

    MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺. Both atoms and charges balance: the left-hand side carries a total charge of +17 and so does the right-hand side.

  4. 4

    A 25.0 cm³ portion of an iron(II) solution requires 24.60 cm³ of 0.0200 mol dm⁻³ potassium manganate(VII) for oxidation in acidic solution. Calculate the concentration of the iron(II) ions.

    Model answer

    Moles of MnO₄⁻ = 0.0200 × 0.02460 = 4.920 × 10⁻⁴ mol. The equation shows that five moles of Fe²⁺ react with each mole of MnO₄⁻, so moles of Fe²⁺ = 5 × 4.920 × 10⁻⁴ = 2.460 × 10⁻³ mol. Concentration = 2.460 × 10⁻³ / 0.0250 = 0.0984 mol dm⁻³.

  5. 5

    Why must an iron(II) solution for titration be prepared shortly before it is used?

    Model answer

    Iron(II) is slowly oxidised to iron(III) by oxygen from the air, particularly in neutral or alkaline solution. A solution left standing would contain less iron(II) than expected and give a titre that is too small, so it is made up in dilute sulfuric acid and used promptly.

  6. 6

    Why is the flask swirled particularly carefully near the end point of a manganate(VII) titration?

    Model answer

    The purple colour of a local excess of titrant is slow to disappear if the mixture is not well mixed, so it can be mistaken for the end point. Thorough swirling makes sure the titrant has reacted before the next drop is added.

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