Redox Titration Calculations (Edexcel A Level Chemistry): Revision Note

Redox Titration Calculations

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

• You could be asked why other acids are not suitable for this redox titration in the exam so make sure you understand the suitability of dilute sulfuric acid

Table explaining why other acids are not suitable for the redox titration

Indicator and end point

• Potassium manganate(VII) titrations are self-indicating

  • Potassium manganate(VII) solution is purple

• The potassium manganate(VII) solution is titrated into the conical flask from the burette

• The manganate(VII) ions react with the analyte to form manganese(II) ions

  • The manganese(II) ions, Mn²⁺(aq), have a very pale pink colour

  • But, the concentration is so low that the solution looks colourless

• When all of the analyte ions have reacted with the manganate(VII) ions, a pale pink colour appears in the flask due to an excess of manganate(VII) ions

  • When this colour remains with swirling, the endpoint has been reached

Redox titration colour change for potassium permanganate and iron(II) ions

   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 =  

  • The formula of a sample of hydrated ethanedioic acid

Answer:

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

  • 1 : 5 ratio of MnO₄^- : Fe²⁺

  • Number of moles of MnO₄^- (aq)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 = 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

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) 1.26 x 10^-3 moles 

  • Number of moles of I₂ (aq) and ClO^- (aq) in 25.0 cm³ 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