Redox Titration with manganate (VII)

MnO4 is a strong oxidising agent and so it be used in titrations to determine the amount of a reducing agent present.

MnO4 has a very intense purple colour and is reduced to Mn2+ which is a very pale pink and almost colourless at low concentrations. In a typical titration, MnO4 is added from the burette and is quickly reduced to Mn2+ (i.e., the purple colour disappears almost instantaneously). At the end-point, the supply of reducing agent is exhausted so the MnO4 is no longer reduced and this can be seen by the permanent apperance of a pale pink / purple colour in the mixture. This means that MnO4 is self-indicating.

 

Question: 6.96 g of FeCl2.xH2O is dissolved in distilled water to form 250 cm3 of solution. A 25 cm3 sample of this solution is titrated with 35.0 cm3 of 0.02 mol/dm3 acidified KMnO4. Determine the value of x.

Step 1

We know that moles = concentration x volume:

n MnO4 = 0.02 x (35 / 1000) =  7.00 x 10-4 mol

Step 2

The balanced symbol equation isn’t given so we need to work it out.

You should be able to see that this is a redox reaction with Fe2+ oxidised to Fe3+ and MnO4 reduced to Mn2+. Write out each of the half-equations seperately and then balance in terms of the central ion, O, H and charge. We then need to combine the reactions by multiplying the oxidation reaction by 5, and adding both sides:

Oxidation

Fe2+ –> Fe3+ + e

Reduction

MnO4 + 8 H+ + 5e –> Mn2+ + 4 H2O

Total

5 Fe2+ + MnO4 + 8 H+ + 5e –> 5 Fe3+ + Mn2+ + 4 H2O

Step 3

The overall balanced symbol equation tells us that Fe2+ (i.e., FeCl2.xH2O) reacts with MnO4 (i.e., KMnO4) in a 5:1 ratio.

n FeCl2.xH2O = 5 x 7.00 x 10-4 = 0.0035 mol

Step 4

The total is 10x bigger than the sample (i.e., 250 / 25):

n FeCl2.xH2O = 10 x 0.0035 mol = 0.035 mol in total

Step 5

We have the mass of FeCl2.xH2O (6.96 g) and can assume that it is pure. This means that the Mr of FeCl2.xH2O is:

Mr FeCl2.xH2O = 6.96 / 0.035 = 198.8

Step 6

Finally, we subtract the Mr of FeCl2 (126.8) to obtain the Mr of the xH2O part of the formula

Mr xH2O = 198.8 – 126.8 = 72

The Mr of H2O is 18, so clearly x = 4