Redox Titration Calculations (Edexcel IAL Chemistry)

Redox Titration - Calculations

Spec point — 16.15

Redox Titrations

• In a titration, the concentration of a solution is determined by titrating with a solution of known concentration.
• In redox titrations, an oxidising agent is titrated against a reducing agent
• Electrons are transferred from one species to the other
• Indicators are sometimes used to show the endpoint of the titration
• However, most transition metal ions naturally change colour when changing oxidation state
• There are two common redox titrations you should know about manganate(VII) titrations and iodine-thiosulfate titrations

Potassium manganate(VII) titrations

• In these redox titrations the manganate(VII) is the oxidising agent and is reduced to Mn²⁺(aq)

• The iron is the reducing agent and is oxidised to Fe³⁺(aq) and the reaction mixture must be acidified, to excess acid is added to the iron(II) ions before the reaction begins

• The choice of acid is important, as it must not react with the manganate(VII) ions, so the acid normally used is dilute sulfuric acid

  • As it does not oxidise under these conditions and does not react with the manganate(VII) ions

Worked Example Equations

Find the stoichiometry for the reaction and complete the two half equations:

MnO₄^- (aq) + 5e^- + 8H⁺ (aq) → Mn²⁺ (aq) + 4H₂O (l)

Fe²⁺ (aq) → Fe³⁺ (aq) + e^-

Answers:

Balance the electrons:

MnO₄^- (aq) + 5e^- + 8H⁺ (aq) → Mn²⁺ (aq) + 4H₂O (l)

5Fe²⁺ (aq) → 5Fe³⁺ (aq) + 5e^-

Add the two half equations:

MnO₄^- (aq) + 8H⁺ (aq) + 5Fe²⁺ (aq) → Mn²⁺ (aq) + 4H₂O (l) + 5Fe³⁺ (aq)

• Manganate(VII) titrations can be used to determine:

  • The percentage purity of iron supplements

• Percentage purity = $\frac{\mathrm{mass} \mathrm{of} \mathrm{sample}}{\mathrm{mass} \mathrm{of} \mathrm{impure} \mathrm{sample}}\times100$

  • The formula of a sample of hydrated ethanedioic acid

Worked Example Analysis of iron tablets

An iron tablet, weighing 0.960 g was dissolved in dilute sulfuric acid. An average titre of 28.50 cm³ of 0.0180 mol dm^-3 potassium manganate(VII) solution was needed to reach the endpoint.

What is the percentage by mass of iron in the tablet?

Answer:

• MnO₄^- (aq) + 8H⁺ (aq) + 5Fe²⁺ → Mn²⁺ (aq) + 5Fe³⁺ (aq) + 4H₂O (l)

  • 1 : 5 ratio of MnO₄^- : Fe²⁺
  • Number of moles of MnO₄^- (aq)$=0.0180\times\frac{28.50}{1000}=$5.13 x 10^-4 moles
  • Moles of iron(II) = 5 x 5.13 x 10^-4 = 2.565 x 10^-3 moles
  • Mass of iron(II) = 55.8 x 2.565 x 10^-3 = 0.143127 g
  • Percentage by mass = $=\frac{0.143127}{0.960}\times100=$14.9%

Iodine-Thiosulfate Titrations

• A redox reaction occurs between iodine and thiosulfate ions:

2S₂O₃^2– (aq) + I₂ (aq) → 2I^–(aq) + S₄O₆^2–(aq)

• The light brown/yellow colour of the iodine turns paler as it is converted to colourless iodide ions
• When the solution is a straw colour,starch is added to clarify the end point
• The solution turns blue/black until all the iodine reacts, at which point the colour disappears.
• This titration can be used to determine the concentration of an oxidizing agent, which oxidizes iodide ions to iodine molecules
• The amount of iodine is determined from titration against a known quantity of sodium thiosulfate solution

Worked Example Analysis of household bleach

Chlorate(I) ions, ClO^-, are the active ingredient in many household bleaches.

10.0 cm³ of bleach was made up to 250.0 cm³.

25.0 cm³ of this solution had 10.0 cm³ of 1.0 mol dm^-3 potassium iodide and then acidified with 1.0 mol dm^-3 hydrochloric acid.

ClO^- (aq) + 2I^- (aq) + 2H⁺ (aq) → Cl^- (aq) + I₂ (aq) + H₂O (l)

This was titrated with 0.05 mol dm^-3 sodium thiosulfate solution giving an average titre of 25.20 cm³.

2S₂O₃^2- (aq) + I₂ (aq) → 2I^- (aq) + S₄O₆^2- (aq)

What is the concentration of chlorate(I) ions in the bleach?

Answer:

• One mole of ClO^- (aq) produces one mole of I₂ (aq) which reacts with two moles of 2S₂O₃^2- (aq)

  • Therefore, 1 : 2 ratio of ClO^- (aq) : S₂O₃^2- (aq)

  • Number of moles of S₂O₃^2- (aq) $=\frac{0.05\times25.20}{1000}=$1.26 x 10^-3 moles

  • Number of moles of I₂ (aq) and ClO^- (aq) in 25.0 cm³ $=\frac{1.26\times10^{-3}}{2}=$6.30 x 10^-4 moles

  • Number of moles of ClO^- (aq) in 250.0 cm³ = 6.30 x 10^-4 x 10 = 6.30 x 10^-3 moles

  • The 250.0 cm³ was prepared from 10.0 cm³ bleach

    • 10 cm³ bleach = 6.30 x 10^-3 moles of ClO^- ions
    • 1.0 dm³ bleach = 0.630 moles of ClO^- ions
    • Therefore, the concentration of ClO^- ions in the bleach is 0.630 mol dm^-3

Exam Hint General sequence for redox titration calculations

1. Write down the half equations for the oxidant and reductant
2. Deduce the overall equation
3. Calculate the number of moles of manganate(VII) or dichromate(VI) used
4. Calculate the ratio of moles of oxidant to moles of reductant from the overall redox equation
5. Calculate the number of moles in the sample solution of the reductant
6. Calculate the number of moles in the original solution of reductant
7. Determine either the concentration of the original solution or the percentage of reductant in a known quantity of sample

Redox Titration - Uncertainty

Spec point — 16.16

Percentage Uncertainties

• Percentage uncertainties are a way to compare the significance of an absolute uncertainty on a measurement
• This is not to be confused with percentage error, which is a comparison of a result to a literature value
• The formula for calculating percentage uncertainty is as follows:

Percentage uncertainty = $\frac{\mathrm{uncertainty}}{\mathrm{measured} \mathrm{value}}\times100$

Adding or subtracting measurements

• When you are adding or subtracting two measurements then you add together the absolute measurement uncertainties

• For example,

  • Using a balance to measure the initial and final mass of a container
  • Using a thermometer for the measurement of the temperature at the start and the end
  • Using a burette to find the initial reading and final reading

• In all these examples you have to read the instrument twice to obtain the quantity

• If each time you read the instrument the measurement is ‘out’ by the stated uncertainty, then your final quantity is potentially ‘out’ by twice the uncertainty

Reducing uncertainty in titrations

• In titrations, uncertainty can be reduced by adjusting the concentration of the known liquid, which is usually in the burette

  • This is to make it so that the burette readings are larger

• For example, if a burette had an uncertainty of ±0.1 cm³, the initial reading was 0.0 cm³ and the final reading was 5.0 cm³

  • The titre is 5.0 cm³
  • The percentage uncertainty is, therefore, $\frac{0.2}{5.0}\times100$ = 4%

• If the concentration of the chemical in the burette was diluted so that the initial reading was 0.0 cm³ and the final reading was 30.0 cm³

  • The titre is 30.0 cm³
  • The percentage uncertainty is, therefore, $\frac{0.2}{30.0}\times100$ = 0.67%

• Overall, to reduce the uncertainty in burette readings, you:

  • Dilute the chemical in the burette
  • This means that more will be required for the titration
  • Therefore, the overall uncertainty in the burette readings is reduced