Redox Titrations (Oxford AQA International A Level Chemistry)

Redox Titrations

• The oxidation states of transition element ions can change during redox reactions

  • A species will either be oxidised or reduced, depending on what reaction is occurring

• To find the concentration of specific ions in solution, a titration can be performed

• There are two specified redox titrations:

  • Ethanedioate or oxalate (C₂O₄^2-) and permanganate (MnO₄^-) in acid solution given suitable data

  • Iron(II) (Fe²⁺) and permanganate (MnO₄^-) in acid solution given suitable data

Potassium manganate(VII) titrations

• In these redox titrations:

  • Manganate(VII) is the oxidising agent

  • The manganate(VII) ions, MnO₄^- (aq), are reduced to manganese(II) ions, Mn²⁺(aq)

• The reaction mixture must be acidified

  • So, excess acid is added before the reaction begins

• Dilute sulfuric acid is normally used to acidify potassium manganate(VII) solution because:

  • It does not oxidise under these conditions

  • It 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:

  • Hydrochloric acid can be oxidised to chloride ions by manganate(VII) ions

  • Nitric acid is an oxidising agent and may oxidise the substance being analysed

  • Ethanoic acid is a weak acid and [H⁺] may be insufficient

  • Concentrated sulfuric acid may oxidise the substance being analysed

Indicator and endpoint

• 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

The endpoint in manganate(VII) titrations

Reaction of MnO₄^- & C₂O₄^2- in acid

• The reaction of manganate(VII) ions with ethanedioate ions is another example of a redox reaction

  • The ethanedioate ions, C₂O₄^2- (aq), are oxidised by manganate(VII) ions, MnO₄^- (aq)

  • Ethanedioate ions are sometimes called oxalate ions

• A titration reaction can be carried out to find the concentration of the toxic ethanedioate ions

• Like the iron(II) / manganate(VII) titration, the endpoint is the first permanent pink colour appears in the flask

  • At this point, the MnO₄^– is very slightly in excess

• The two half-reactions that are involved in this redox reaction are as follows:

C₂O₄^2– (aq) → 2CO₂ (g) + 2e^– 

• The C₂O₄^2– (aq) loses 2 electrons to form 2CO₂ (g)

• The oxidation number of carbon changes from +3 in C₂O₄^2– (aq) to +4 in CO₂ (g)

• Since there is an increase in oxidation number, this is the oxidation reaction

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

• The oxidation number of manganese changes from +7 in MnO₄^– (aq) to +2 in Mn²⁺ (aq)

• Since there is a decrease in oxidation number, this is the reduction reaction

• The half equations are combined to get the overall equation:

Oxidation:   C₂O₄^2– (aq) → 2CO₂ (g) + 2e^–

Reduction:   MnO₄^– (aq) + 8H⁺ (aq) + 5e^– → Mn²⁺ (aq) + 4H₂O (l) 

• Both half equations must have the same number of electrons, so:

  • The oxidation half equation is multiplied by 5 

  • The reduction half equation is multiplied by 2

Oxidation:   5C₂O₄^2– (aq) → 10CO₂ (g) + 10e^–

Reduction:   2MnO₄^– (aq) + 16H⁺ (aq) + 10e^– → 2Mn²⁺ (aq) + 8H₂O (l) 

• The reactants and products from each half equation can be combined together:

5C₂O₄^2– (aq) + 2MnO₄^– (aq) + 16H⁺ (aq) + 10e^– → 2Mn²⁺ (aq) + 8H₂O (l) + 10CO₂ (g) + 10e^– 

• Any species that appear on both sides of the overall equation can be cancelled out

  • In this case, there are 10e^– on both sides, which can be cancelled:

5C₂O₄^2– (aq) + 2MnO₄^– (aq) + 16H⁺ (aq) → 2Mn²⁺ (aq) + 8H₂O (l) + 10CO₂ (g)  

• This is an example of an autocatalysis reaction

• This means that the reaction is catalysed by one of the products as it forms

• In this reaction, the Mn²⁺ ions formed act as the autocatalyst

• The more Mn²⁺ formed, the faster the reaction gets, which then forms even more Mn²⁺ ions and speeds the reaction up even further

• Transition element ions can act as autocatalysts because they can change their oxidation states during a reaction

Reaction of MnO₄^– & Fe²⁺ in acid

• The concentration of Fe²⁺ ions can be determined by titrating a known volume of Fe(II) ions with a known concentration of MnO₄^– ions

• During the reaction of MnO₄^– with Fe²⁺, the purple colour of the manganate(VII) ions disappears

• The end-point is when all of the Fe²⁺ ions have reacted with the MnO₄^– ions, and the first trace of a permanent pink colour appears in the flask

  • At this point, the MnO₄^– is very slightly in excess

• The two half-reactions that are involved in this redox reaction are as follows:

Fe²⁺ (aq) → Fe³⁺ (aq) + e^–

• The Fe²⁺ (aq) loses an electron to form Fe³⁺ (aq)

• The oxidation number of iron changes from +2 in Fe²⁺ (aq) to +3 in Fe³⁺ (aq)

• So, this is the oxidation reaction

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

• The oxidation number of manganese changes from +7 in MnO₄^– (aq) to +2 in Mn²⁺ (aq)

• Since there is a decrease in oxidation number, this is the reduction reaction

• The half equations are combined to get the overall equation:

Oxidation:   Fe²⁺ (aq) → Fe³⁺ (aq) + e^–

Reduction:   MnO₄^– (aq) + 8H⁺ (aq) + 5e^– → Mn²⁺ (aq) + 4H₂O (l) 

• Both half equations must have the same number of electrons, so:

  • The oxidation half equation is multiplied by 5

  • The reduction half equation does not need any changes

Oxidation:   5Fe²⁺ (aq) → 5Fe³⁺ (aq) + 5e^–

Reduction:   2nO₄^– (aq) + 8H⁺ (aq) + 5e^– → Mn²⁺ (aq) + 4H₂O (l) 

• The reactants and products from each half equation can be combined together:

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

• Any species that appear on both sides of the overall equation can be cancelled out

  • In this case, there are 5e^– on both sides, which can be cancelled:

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