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Redox titrations: conditions, ratios and endpoints

Key idea: A redox titre measures the sample amount through a balanced electron-transfer ratio. For acidified manganate(VII) added to iron(II), the end-point is the first persistent faint pink colour.

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  • GCE A-Level H2 Chemistry 2027
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A redox titration uses an electron-transfer reaction to connect a known titrant amount to an unknown sample amount. The burette technique is familiar; the reaction conditions, mole ratio and endpoint observation come from the particular redox system.

Use titration calculations for the volume-to-amount step. If balancing is the difficulty, revisit redox equations from half-equations.

Check your understanding 1 Choose a redox endpoint

Iron(II) is titrated with purple manganate(VII) in acid. Which observation should end the titration?

Endpoint observation
Show answer

While iron(II) remains, added purple manganate(VII) reacts. Stop at the first faint pink that persists for the method’s stated time; deep purple means excess titrant.

A direct manganate(VII) titration

In acid solution, manganate(VII) ions oxidise iron(II) ions:

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

One mole of MnO₄⁻ reacts with five moles of Fe²⁺. This ratio follows the balanced reaction, rather than an assumption that all titrations are 1:1.

The H⁺ ions are reactants. Insufficient acidity can change the manganese product and invalidate this equation. Use the acid and conditions specified by the method; sulfuric acid is commonly used for this system. Other acids can interfere through side reactions. These are supervised laboratory techniques: wear eye protection and follow the supplied handling instructions for the acid and oxidising titrant.

Why no separate indicator is needed here

When potassium manganate(VII) is in the burette, each small addition of its purple solution is consumed while enough iron(II) remains. Near completion, add it dropwise with swirling. The endpoint is the first faint pink colour that persists for the time specified by the method: a small excess of manganate(VII) remains.

Recognise the manganate(VII) endpoint

Manganate(VII): transient purple, then persistent pale pinkManganate(VII) is added to iron(II) in adequately acidified solution, normally using the specified sulfuric acid method. Local purple is consumed before the endpoint. The first persistent pale pink signals a slight excess of manganate(VII); deep purple is an overshoot.Before endpointlocal colourAfter mixingcolour disappearsEndpoint after mixingfirst persistent pale pink
Manganate(VII): transient purple, then persistent pale pink
  1. A purple manganate(VII) addition enters the acidified iron(II) sample.
  2. After mixing, purple disappears while iron(II) remains; Mn²⁺ forms.
  3. The first faint pink that remains after mixing marks a small excess of MnO₄⁻.

This schematic is for manganate(VII) added to iron(II) in adequately acidified solution, normally using the specified sulfuric acid method. Colours, drop size, mixing and playback timing are illustrative, not experimental evidence or the required persistence time. Use the method’s observation interval; deep purple is an overshoot.

The equivalence point is the stoichiometric completion of the reaction. The endpoint is the observation used to stop. A deep purple excess is an overshoot, not a more convincing endpoint. Do not apply this colour rule to a different redox titrant.

Worked example 1

Determine an iron(II) concentration

Problem

A 25.0 cm³ iron(II) sample, acidified as specified by the method, requires 16.40 cm³ of 0.0200 mol dm⁻³ potassium manganate(VII). Use the equation above to calculate the iron(II) concentration.
Show full solution

  1. Find the titrant amount

    Method

    Convert the titre to dm³ and use n = cV.

    Reason

    The burette measures the volume of the known solution.

    Working

    Answer

    n(MnO₄⁻) = 0.0200(0.01640) = 3.280 × 10⁻⁴ mol

  2. Apply the reaction ratio

    Method

    Multiply by five to find the iron(II) amount.

    Reason

    The equation requires five iron(II) ions per manganate(VII) ion.

    Working

    Answer

    n(Fe²⁺) = 5(3.280 × 10⁻⁴) = 1.640 × 10⁻³ mol

  3. Report the sample concentration

    Method

    Divide by the 0.0250 dm³ sample volume.

    Reason

    The calculated iron(II) amount came from this aliquot.

    Working

    Answer

    c(Fe²⁺) = (1.640 × 10⁻³)/0.0250 = 0.0656 mol dm⁻³

Check your understanding 2 Use the iron(II) ratio

A 20.0 cm³ acidified iron(II) aliquot needs 15.00 cm³ of 0.0200 mol dm⁻³ manganate(VII). Find the titrant amount, iron(II) amount and iron(II) concentration.

Show the steps Optional working
Unit: mol
Unit: mol
Unit: mol dm⁻³
Show answer

The titrant amount is 0.0200(0.01500) = 3.00 × 10⁻⁴ mol. Iron(II) amount is five times this, 1.50 × 10⁻³ mol, so concentration is 0.00150/0.0200 = 0.0750 mol dm⁻³.

Check your understanding 3 Transfer the method to hydrogen peroxide

An acidified 25.0 cm³ hydrogen peroxide sample requires 20.20 cm³ of 0.0200 mol dm⁻³ manganate(VII) solution. The reaction is:

2MnO₄⁻ + 5H₂O₂ + 6H⁺ → 2Mn²⁺ + 5O₂ + 8H₂O

Calculate the hydrogen peroxide concentration.

Show the steps Optional working
Unit: mol
Unit: mol dm⁻³
Show answer
n(MnO₄⁻) = 0.0200(0.02020) = 4.040 × 10⁻⁴ mol
n(H₂O₂) = (5/2)(4.040 × 10⁻⁴) = 1.010 × 10⁻³ mol
c(H₂O₂) = (1.010 × 10⁻³)/0.0250 = 0.0404 mol dm⁻³

Check your understanding 4 Explain the mole ratio and show your method

Show the full method for the 25.0 cm³ peroxide aliquot requiring 20.20 cm³ of 0.0200 mol dm⁻³ manganate(VII), using the reaction equation above. Explain why multiplying the titrant amount by five would be wrong.

Show answer
n(MnO₄⁻) = 0.0200(0.02020) = 4.040 × 10⁻⁴ mol
n(H₂O₂) = (5/2)(4.040 × 10⁻⁴) = 1.010 × 10⁻³ mol
c(H₂O₂) = (1.010 × 10⁻³)/0.0250 = 0.0404 mol dm⁻³

The ratio is now five moles of peroxide to two moles of manganate(VII). Multiplying by five would double the correct peroxide amount. Keep the measurement method, but obtain the ratio afresh from the reaction.

In an indirect redox analysis, the sample first produces another substance that is then titrated, which adds an extra reaction step to the calculation.

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