Topic 15.4
Variable oxidation states
of transition metals
Redox titration
Aims
 The aim of this activity is for students to practice redox titration calculations
Teacher notes
Before tackling this activity students should know the equations for the redox titration of Fe2 with
MnO4 and Cr2O72 as outlined Topic 15.4. Students are given the relevant equations for the
calculations but unless they know the reacting ratios they will have to combine them and produce a full
equation before commencing the activity.
MnO4(aq)  5e  8H(aq) → Mn2(aq)  4H2O(l)
Cr2O72(aq)  14H(aq)  6e → 2Cr3(aq)  7H2O(l)
Fe2(aq) → Fe3(aq)  e
Answers to questions
1 A sample of iron(II) sulfate crystals FeSO47H2O had been left out in the air and some of the Fe2
ions had been converted to Fe3. A total of 4.2 g of the impure crystals were dissolved in a total of
250 cm3 of water and dilute sulfuric acid. Portions of 25 cm3 of this solution were titrated with a
solution of potassium dichromate(VI). The concentration of dichromate(VI) ions in this solution was
0.1 mol dm3. The average titre was 23.50 cm3. Use the following steps to find the percentage purity
of the original crystals.
a) Calculate the moles of dichromate ions used in each titration.
Moles of dichromate ions used in each titration  2.35 × 104 mol
b) Calculate the number of moles of Fe2 ions in each 25 cm3 sample.
Number of moles of Fe2 ions in each 25 cm3 sample  6 × 2.35  104 mol  1.41 × 103 mol
c) Calculate the number of moles of Fe2 ions in the original 4.2 g of iron(II) sulfate crystals.
Number of Fe2 ions in the original 4.2 g of iron(II) sulfate crystals 
10  1.41 × 103 mol  1.41 × 102 mol
d) What is the mass of this number of moles of Fe2 ions? (Mr(Fe)  56.)
Mass of this number of Fe2 ions  56  1.41 × 102  0.79 g
e) Calculate the mass of Fe2 ions in the 4.2 g sample if it had been 100% pure.
Mass of 1 mole of FeSO47H2O  278 g
4.2 g  0.015 mol of which 0.84 g is Fe2 ions.
f) Calculate the percentage purity of the crystals.
Percentage purity of the crystals  94%
AQA Chemistry A2 Stretch and challenge teacher notes © Nelson Thornes Ltd 2009
1
Topic 15.4
Variable oxidation states
of transition metals
2 An impure sample of iron of mass 2.55 g was dissolved in
dilute sulfuric acid and the solution made up to 250 cm3. The solution contained iron(II) ions
together with the impurities. Portions of 25 cm3 of this solution were titrated with potassium
manganate(VII) solution of concentration 0.02 mol dm3. The average titre was 28.50 cm3. Calculate
the percentage purity of the sample of iron.
The students are given no structure to help them answer this question. However, they should
follow a similar series of steps to the example above. The answer has also been given in a
structured format.
Moles of manganate ions used in each titration  5.7  104 mol
Number of moles of Fe2 ions in each 25 cm3 sample  5  5.7  104 mol  2.85  103 mol
Number of Fe2 ions from original 2.55 g of iron  10  2.85  103 mol  2.85  102 mol
Mass of this number of Fe2 ions  56  2.85  102  1.596 g
 1.596 
Percentage purity of crystals  
  100  62.6%
 2.55 
3 Copper(II) ions oxidise iodide ions to iodine. The iodine produced can be titrated with standard
thiosulfate solution, and, from the amount of iodine produced, the concentration of the copper(II)
ions in the solution can be calculated. The relevant equations for this process are given below:
2Cu2(aq)  4I(aq) → 2CuI(s)  I2(aq)
2S2O32(aq)  I2(aq) → S4O62(aq)  2I(aq)
A sample of 4.256 g of CuSO45H2O is dissolved and made up to 250 cm3 in a volumetric flask. A
25 cm3 portion is added to an excess of potassium iodide. The iodine formed required 18.00 cm3 of a
0.0950 mol dm3 solution of sodium thiosulfate for reduction. Calculate the percentage of copper in
the crystals.
Moles of thiosulfate ions  1.71 × 103 mol
1
From the equation: moles I2 
thiosulfate moles
2
Moles I2  0.855  103 mol
From the equation: moles Cu  2  I2 moles
Moles Cu  1.71  103 mol
Mass Cu  63.5  1.71  103 mol  0.109 g
 1.09 
Percentage of copper in crystals  
  100  25.6%
 4.256 
AQA Chemistry A2 Stretch and challenge teacher notes © Nelson Thornes Ltd 2009
2