Redox (II) Redox titration calculations

Edexcel A level Chemistry (2017)

Topic 14: Redox (II) Redox titration calculations

Here are the two learning objectives:

14/18. To be able to carry out both structured and non-structured titration calculations including Fe2+/MnO4−, and I2/S2O32−

14/19. To be able to understand the methods used in redox titrations

Redox Titrations

Two examples of redox titrations are required for the Edexcel course.  You can see from the learning objectives the two redox reactions required are those between manganate(VII) and iron(II) and between iodine and thiosulphate. 

How to carry out these two redox titrations: methodology

Redox titrations rely on following changes in the redox reaction chemistry to determine the titration end point.

1. The Manganate(VII)/iron(II) titration

In a manganate(VII)—iron(II) titration, the purple manganate(VII) solution is added to the almost colourless iron(II) solution so that at the end point the iron(II) solution is slightly pink from the manganate(VII).

It is easier to follow a change from colourless to pink than it is a colour change from pink to colourless which would happen if the iron(II) solution was added to a solution of pink manganate(VII) ions.

No indicator is needed because the purple manganate(VII) is self indicating.

2. The Iodine/thiosulphate titration

In the case of the iodine—thiosulphate titration, an indicator is required since the end point is not easy to see.  In this titration, you usually add sodium thiosulphate solution from the burette to an aqueous solution of iodine in the conical flask.  Aqueous solutions of iodine are brown but fade to yellow then colourless at the end point so to follow the colour change and serve a sharp end point is impossible.

Instead starch solution is added to the iodine solution once it becomes golden coloured (see the video below).  This results in a deep blue–black starch–iodine complex.  Adding thiosulphate solution brings an endpoint with a sudden loss of the blue–black colour: a very sharp endpoint.   

You can watch a very brief iodine thiosulphate titration here.

Calculations using redox titration results.

I’m going to talk you through a non-structured calculation on the reasonable assumption that if you can carry out these kinds of calculation then the structured calculations you will find fairly straightforward. 

Here is a typical non–structured calculation question:

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.

MnO₄^—   +   8H⁺   +  5e^—   ⟶  Mn²⁺   +   4H₂O

Fe²⁺    ⟶     Fe³⁺   +   e^—

C₂O₄^2—   ⟶   2CO₂   +  2e^—

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

First, calculate the reacting molar ratio:

Each iron(II) ion provides one electron on oxidation and each ethandioate ion provides two electrons: a total of three moles of electrons per iron(II)ethandioate.

5 moles iron(II)ethandioate provide 15 electrons and these reduce three moles manganate(VII) ions.

So the overall equation has to be:

5FeC₂O₄  + 3MnO₄^— + 24H⁺ ⟶ 3Mn²⁺ + 10CO2 + 5Fe³⁺ + 12H₂O

Now you have the reacting ratio of manganate(VII) to iron(II)ethandioate as 3 moles to 5 moles.

Second, you calculate the moles of manganate(VII) actually used in the titration and work back from this value.

Use n=cV

Moles MnO₄^—  =  0.0193  × 22.35/1000   =   0.00043 moles

Third calculate the number of moles of the iron(II) salt in the 25ml of its solution:

Moles FeC₂O₄  =   0.00043 ×  5/3   =   0.00072 moles

Fourth, note that 25ml sample was a tenth of the total sample dissolved in the 250ml volumetric flask.  Therefore:

Total moles FeC₂O₄    =  0.0072moles  

Fifth, calculate the molar mass of the iron(II) salt:

Fe—                55.8

C—12 × 2  =   24

O—16 × 4  =   64  +

H2O—18 × 2= 36

Giving a total Mr (FeC₂O₄ ) of 179.8 g.mol^—1

Sixth, calculate the mass of pure iron(II) salt using mass =  molar mass ×  moles

Mass of FeC₂O₄   =   179.8  ×  0.0072  = 1.295g    

Lastly, calculate the percentage by mass of the pure iron(II) salt  since

% purity   =    pure mass   ×   100  /total mass

% purity   =    1.295  ×   100/1.391   =   93.2%

And there you have it.  Step by step you can work through the calculation and see that the stages are common to most types of calculation like this one.