Redox titrations rely on the changes below to determine the end point.

Iron(II)/Manganate(VII) (Fe²⁺/MnO₄⁻)

The purple manganate(VII) solution is added to the almost colourless iron(II) solution.

At the end point, the iron(II) solution is slightly pink from the manganate(VII).

No indicator is needed because manganate(VII) is purple.


Iodine/Thiosulphate (I₂/S₂O₃²⁻)

Sodium thiosulphate solution is added to an aqueous solution of iodine.

Starch is used as an indicator, because otherwise the end point is very difficult to identify.

In the presence of starch, the iodine solution is blue-black, which changes to colourless at the end point from the loss of iodine.


Example

Iron(II) ethanedioate dihydrate can be analysed by titration using potassium manganate(VII) in acidic solution.
In this reaction, manganate(VII) ions oxidise iron(II) ions and ethanedioate ions.

A 1.381g sample of impure FeC₂O₄.2H₂O was dissolved in an excess of dilute sulphuric acid and made up to 250ml of solution.
25ml of this solution decolourised 22.35ml of a 0.0193 mol.dm^–3 solution of potassium manganate(VII).

Use the half-equations given below to calculate the reacting ratio of moles of manganate(VII) ions to moles of iron(II) ethanedioate.

\(MnO_4^- + 8H^+ + 5e^- ⟶ Mn^{2+} + 4H_2O \)

\(Fe^{2+} ⟶ Fe^{3+} + e^- \)

\(C_2O_4^{2-} ⟶ 2CO_2 + 2e^- \)

Then calculate the percentage by mass of FeC₂O₄.2H₂O in the original impure sample.

Solution

Combine the iron(II) and ethandioate half equations to form:

\(FeC_2O_4 ⟶ 2CO_2 + Fe^{3+} + 3e^- \)

Balance this equation with the manganate(VII) half equation by balancing the electrons (3x5):

\(3MnO_4^- + 24H^+ + 15e^- ⟶ 3Mn^{2+} + 12H_2O \)

\(5FeC_2O_4 ⟶ 10CO_2 + 5Fe^{3+} + 15e^- \)

Combine the two equations to form:

\(5FeC_2O_4 + 3MnO_4^- + 24H^+ ⟶ 3Mn^{2+} + 10CO_2 + 5Fe^{3+} + 12H_2O \)

The molar ratio of manganate(VII) to iron(II) ethandioate is 3:5.

Next, calculate the moles of manganate(VII) used in the titration:

\(0.0193 × \dfrac{22.35}{1000} = 0.00043 \)

Calculate the number of moles of the iron(II) salt in 25ml:

\(0.00043 × \dfrac{5}{3} = 0.00072 \)

Calculate the number of moles of the iron(II) salt in 250ml:

\(0.00072 × 10 = 0.0072 \)

Molar mass of FeC₂O₄ is 179.8

Calculate mass of pure iron(II) salt:

\(179.8 × 0.0072 = 1.295g \)

Calculate the % purity of iron(II) salt:

\(\dfrac{1.295}{1.391} × 100 = 93.2\% \)