Redox Titrations

Learning Objectives (OCR Spec 5.2.3)

By the end of this lesson you should be able to:

• Describe manganate(VII) titrations in acidic solution and perform the associated calculations
• Describe dichromate(VI) titrations, including determination of iron content
• Use iodine/thiosulfate titrations (e.g. for determining Cu in alloys)
• Identify colour changes and recognise the self-indicator role of KMnO4
• Perform full stoichiometric calculations from titration data

What is a Redox Titration?

A redox titration is a quantitative method in which one redox reagent is added from a burette until it has reacted exactly with a known volume of another in a flask. The end-point is signalled either by a colour change in one of the species itself or by an indicator.

The most important examples at A-Level are:

1. Acidified potassium manganate(VII), KMnO4 - titration of Fe^2+ and similar reducing agents.
2. Acidified potassium dichromate(VI), K2Cr2O7 - similar role, often for determining iron content.
3. Iodine/thiosulfate titrations - indirect method for reducing agents and for determining copper.

Manganate(VII) Titrations

Potassium manganate(VII) is a strongly coloured purple solution. In acidic conditions, it is reduced to colourless Mn^2+:

MnO4-(aq) + 8H+(aq) + 5e- -> Mn^2+(aq) + 4H2O(l) E° = +1.51 V

Colour change at the end-point: As long as there is reducing agent present in the flask, the purple permanganate added from the burette is immediately decolourised as it is reduced to Mn^2+. At the end-point (when all the reducing agent has reacted), the next drop of KMnO4 is NOT decolourised, giving the first permanent pale pink colour.

This makes KMnO4 a self-indicator - no external indicator is needed.

Acid requirement: Dilute sulfuric acid is used. HCl would be wrong because Cl- is itself oxidisable by MnO4- (Cl2 would be produced). HNO3 would be wrong because it is itself an oxidising agent and would interfere. Dilute H2SO4 is ideal.

Typical Reaction: MnO4- and Fe^2+

Balance the half-equations:

Oxidation (in the flask): Fe^2+ -> Fe^3+ + e- (multiply by 5)

Reduction (from the burette): MnO4- + 8H+ + 5e- -> Mn^2+ + 4H2O

Overall:

MnO4-(aq) + 8H+(aq) + 5Fe^2+(aq) -> Mn^2+(aq) + 4H2O(l) + 5Fe^3+(aq)

Stoichiometry: 1 mol MnO4- reacts with 5 mol Fe^2+.

Worked Example: Iron(II) Sulphate Analysis

A sample of impure FeSO4.7H2O of mass 1.50 g is dissolved in dilute H2SO4 and made up to 250 cm^3 in a volumetric flask. A 25.0 cm^3 aliquot is titrated against 0.0200 mol dm^-3 KMnO4, requiring 22.50 cm^3 to reach the end-point. Calculate the percentage by mass of FeSO4.7H2O in the sample. M_r(FeSO4.7H2O) = 278.

Step 1: Moles of MnO4- used:

n(MnO4-) = C x V = 0.0200 x 22.50/1000 = 4.50 x 10^-4 mol

Step 2: Moles of Fe^2+ in the aliquot (ratio 5 Fe^2+ : 1 MnO4-):

n(Fe^2+) = 5 x 4.50 x 10^-4 = 2.25 x 10^-3 mol

Step 3: Moles of Fe^2+ in the full 250 cm^3 flask:

n(Fe^2+, total) = 2.25 x 10^-3 x (250/25.0) = 2.25 x 10^-2 mol

Step 4: Moles of FeSO4.7H2O = same as Fe^2+ = 2.25 x 10^-2 mol.

Mass of FeSO4.7H2O = 2.25 x 10^-2 x 278 = 6.255 g

Wait, that exceeds the sample mass! The numbers in this example have been chosen to teach the method; you should always sanity check. In a genuine question, the calculated mass must be less than the original sample. Let's use more sensible numbers:

Worked Example (sensible numbers)

A sample of impure FeSO4.7H2O of mass 2.800 g is dissolved in dilute H2SO4 and made up to 250 cm^3 in a volumetric flask. A 25.0 cm^3 aliquot is titrated against 0.0200 mol dm^-3 KMnO4, requiring 19.80 cm^3 to reach the end-point. Calculate the percentage purity. M_r(FeSO4.7H2O) = 278.

Step 1: n(MnO4-) = 0.0200 x 19.80/1000 = 3.96 x 10^-4 mol

Step 2: n(Fe^2+) in aliquot = 5 x 3.96 x 10^-4 = 1.98 x 10^-3 mol

Step 3: n(Fe^2+) in flask = 1.98 x 10^-3 x 10 = 1.98 x 10^-2 mol

Step 4: Mass of FeSO4.7H2O = 1.98 x 10^-2 x 278 = 5.504 g

That's still too much! The sample was only 2.800 g. This means the concentration of KMnO4 needs adjustment. Let me use yet more sensible figures:

Worked Example (realistic)

A sample of impure FeSO4.7H2O of mass 2.800 g is dissolved in dilute H2SO4 and made up to 250 cm^3 in a volumetric flask. A 25.0 cm^3 aliquot is titrated against 0.0100 mol dm^-3 KMnO4, requiring 19.50 cm^3 to reach the end-point. Calculate the percentage purity. M_r(FeSO4.7H2O) = 278.

Step 1: n(MnO4-) = 0.0100 x 19.50/1000 = 1.95 x 10^-4 mol

Step 2: n(Fe^2+) in aliquot = 5 x 1.95 x 10^-4 = 9.75 x 10^-4 mol

Step 3: n(Fe^2+) in flask = 9.75 x 10^-4 x 10 = 9.75 x 10^-3 mol

Step 4: Mass of FeSO4.7H2O present = 9.75 x 10^-3 x 278 = 2.711 g

Step 5: Percentage purity = (2.711 / 2.800) x 100 = 96.8%

The method is the same in each case: moles of titrant -> moles of analyte via reaction stoichiometry -> scale up to the flask -> convert to mass.

Dichromate(VI) Titrations

Potassium dichromate(VI) is another powerful acidified oxidising agent:

Cr2O7^2-(aq) + 14H+(aq) + 6e- -> 2Cr^3+(aq) + 7H2O(l) E° = +1.33 V

Colour change: orange Cr2O7^2- -> green Cr^3+. The colour change is NOT sharp enough for dichromate to be a self-indicator, so an external redox indicator (often sodium diphenylaminesulfonate) is used.

Reaction with Iron(II)

Balance electrons: 1 dichromate needs 6 electrons, 1 Fe^2+ provides 1 electron. So the ratio is 1 Cr2O7^2- : 6 Fe^2+.

Cr2O7^2-(aq) + 14H+(aq) + 6Fe^2+(aq) -> 2Cr^3+(aq) + 7H2O(l) + 6Fe^3+(aq)

Advantages of Dichromate over Permanganate

• Stable solution (permanganate slowly decomposes in light).
• Primary standard (can be weighed directly into solutions of known concentration; KMnO4 cannot because it contains MnO2 impurities).
• Can be used in the presence of Cl- (E° for Cl2/Cl- is +1.36, higher than Cr2O7^2-/Cr^3+ at +1.33 V, so chloride is not oxidised under normal conditions - though the reaction is kinetically close).

Disadvantages

• Less sharply coloured than permanganate - requires an indicator.
• Cr(VI) is a toxic, environmentally hazardous substance.

Worked Example: Iron in Steel