Redox Titration - Iron(II) & Manganate(VII) (Edexcel International A Level Chemistry)

Revision Note

Richard

Author

Richard

Last updated

Core Practical 13a: Iron(II) & Manganate(VII) Titration

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 Fe2+(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 permanganate acts as its own indicator, as the purple potassium permanganate solution is added to the titration flask from the burette and reacts rapidly with the Fe2+(aq)
    • The burette used in this practical should be one with white numbering not black, as you would struggle to read the values for your titres against the purple colour of the potassium permanganate if black numbering was used

  • The manganese(II) ions, Mn2+(aq), have a very pale pink colour but they are present in such a low concentration that the solution looks colourless
  • As soon as all of the iron(II), Fe2+(aq), ions have reacted with the added manganate(VII) ions, Mn7+(aq), a pale pink tinge appears in the flask due to an excess of manganate(VII) ions, Mn7+(aq)

Redox titration colour change for potassium permanganate and iron(II) 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 = fraction numerator mass space of space sample over denominator mass space of space impure space sample end fraction space cross times space 100 
    • 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)equals space fraction numerator 0.0180 space cross times space 25.0 over denominator 1000 end fraction space equals5.13 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) = 56.0 x 2.565 x 10-3 = 0.14364 g
    • Percentage by mass = equals space fraction numerator 0.14364 over denominator 0.960 end fraction space cross times 100 equals space15.0%

You've read 0 of your 5 free revision notes this week

Sign up now. It’s free!

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

Did this page help you?

Richard

Author: Richard

Expertise: Chemistry

Richard has taught Chemistry for over 15 years as well as working as a science tutor, examiner, content creator and author. He wasn’t the greatest at exams and only discovered how to revise in his final year at university. That knowledge made him want to help students learn how to revise, challenge them to think about what they actually know and hopefully succeed; so here he is, happily, at SME.