Redox Titrations
Redox Titrations
- In a titration, the concentration of a solution is determined by titrating with a solution of known concentration.
- In redox titrations, an oxidizing 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 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)
Examiner Tip
General sequence for redox titration calculations
- Write down the half equations for the oxidant and reductant
- Deduce the overall equation
- Calculate the number of moles of manganate(VII) or dichromate(VI) used
- Calculate the ratio of moles of oxidant to moles of reductant from the overall redox equation
- Calculate the number of moles in the sample solution of the reductant
- Calculate the number of moles in the original solution of reductant
- Determine either the concentration of the original solution or the percentage of reductant in a known quantity of sample
Iodine-Thiosulfate Titrations
- A redox reaction occurs between iodine and thiosulfate ions:
2S2O32– (aq) + I2 (aq) → 2I–(aq) + S4O62– (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