Redox Systems (Cambridge (CIE) A Level Chemistry)

The Permanganate & Oxalate Redox System

• 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 three particular redox titrations that need to be learnt:

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

  • Permanganate (MnO₄^-) and ethanedioate (C₂O₄^2-) in acid solution given suitable data

  • Copper (II) (Cu²⁺) and iodide (I^-) given suitable data

• The first redox titration involving transition element ions, that needs to be learned, is the titration of permanganate (MnO₄^-) and ethanedioate, sometimes known as oxalate (C₂O₄^2-) in acid solution given suitable data

Reaction of MnO₄^- & C₂O₄^2- in acid

• The reaction of MnO₄^– with ethanedioate, C₂O₄^2– is an example of a redox reaction in which the ethanedioate ions (C₂O₄^2–) get oxidised by manganate(VII) (MnO₄^–) ions

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

• As before, the endpoint is when all of the ethanedioate ions have reacted with the MnO₄^– ions, and 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

The Ferrous & Permanganate Redox System

• The second redox titration involving transition element ions, that needs to be learned, is the titration of permanganate (MnO₄^-) and iron(II) ions  (Fe²⁺) 

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) 

The Cupric & Iodide Redox Systems

• The third redox titration involving transition metal ions, that needs to be learnt, is the titration between copper(II) ions (Cu²⁺) - sometimes known as cupric ions - and iodide ions (I^-)

Reaction of Cu²⁺ & I^-

• The reaction of Cu²⁺ with I^- is an example of a redox reaction in which the copper ions (Cu²⁺) oxidise the iodide ions (I^-) and as a result are themselves reduced

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

2I^– (aq) → I₂ (aq) + 2e^–

• The 2I^– (aq) loses an electron each to form I₂ (aq)

  • The oxidation number of iodine changes from -1 in I^– (aq) to 0 in I₂ (aq)

  • So, this is the oxidation reaction

Cu²⁺ (aq) + e^– → Cu⁺ (aq) 

• The Cu²⁺ (aq) gains an electron to form Cu⁺ (aq)

  • The oxidation number of copper changes from +2 in Cu²⁺ (aq) to +1 in Cu⁺ (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:   2I^– (aq) → I₂ (aq) + 2e^–

Reduction:   Cu²⁺ (aq) + e^– → Cu⁺ (aq) 

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

  • The oxidation half equation does not need any changes

  • The reduction half equation is multiplied by 2

Oxidation:   2I^– (aq) → I₂ (aq) + 2e^–

Reduction:   2Cu²⁺ (aq) + 2e^– → 2Cu⁺ (aq) 

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

2I^– (aq) + 2Cu²⁺ (aq) + 2e^– → 2Cu⁺ (aq) + I₂ (aq) + 2e^–

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

  • In this case, there are 2e^– on both sides, which can be cancelled:

2I^– (aq) + 2Cu²⁺ (aq) → 2Cu⁺ (aq) + I₂ (aq) 

• When excess iodide ions are reacted with Cu(II), a precipitate of copper(I) iodide and iodine is formed:

  • 2Cu²⁺ (aq) + 4I^- (aq) →  I₂ (aq) + 2CuI (s)          Reaction 1

• A titration reaction can be carried out to find an unknown concentration of the copper(II) solution

• This is done by finding the amount of iodine which is liberated during the reaction, through a titration:

1. A known concentration of sodium thiosulfate solution is added to the mixture formed in Reaction 1 from a burette

   • The I₂ formed in Reaction 1 will react with the thiosulfate ions

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

   • As the iodine is used up, the brownish colour of the solution gets lighter

   • When most of the iodine colour is gone, starch is added, which turns deep blue/black with the remaining I₂ (aq)

   • Step 5: Titrate further until the blue/black colour disappears, i.e. when all of the iodine has reacted

     • By knowing the number of moles of thiosulfate ions added in the titration, you can use the molar ratios from the reaction equations and work backwards to calculate the number of moles of Cu(II)

   • Look at Reaction 2 It can be concluded that half the number of moles of I₂ reacts when compared to the moles of thiosulfate that react

   • Now look at Reaction 1

     • The number of moles of I₂ which react in Reaction 2, is the moles formed in Reaction 1

     • The number of moles of Cu(II) is twice that of I₂ (aq), i.e. the same number of moles as thiosulfate ions added in the titration

   • Divide the number of moles of Cu(II) by the volume in dm³ to get the concentration of Cu(II)

Calculations of Other Redox Systems

Calculations of Other Redox Systems

• You are required to perform calculations involving redox reactions of transition elements

• These include:

  • Constructing redox equations

  • Calculating oxidation states

  • Selecting suitable oxidising agents and reducing agents

  • Calculating cell potentials