Manganate Redox Titration The redox titration between Fe2+ with MnO4 – (purple) is a very common exercise. This titration is self indicating because of the significant colour change from reactant to product. MnO4 -(aq) + 8H+ (aq) + 5Fe2+ (aq) Mn2+ (aq) + 4H2O (l) + 5Fe3+ (aq) Purple colourless Choosing correct acid for manganate titrations. The acid is needed to supply the 8H+ ions. Some acids are not suitable as they set up alternative redox reactions and hence make the titration readings inaccurate. Only use dilute sulphuric acid for manganate titrations. Insufficient volumes of sulphuric acid will mean the solution is not acidic enough and MnO2 will be produced instead of Mn2+ . MnO4 -(aq) + 4H+ (aq) + 3e- MnO2 (s) + 2H2O The brown MnO2 will mask the colour change and lead to a greater (inaccurate) volume of Manganate being used in the titration. Using a weak acid like ethanoic acid would have the same effect as it cannot supply the large amount of hydrogen ions needed (8H+ ). It cannot be conc HCl as the Clions would be oxidised to Cl2 by MnO4 – as the Eo MnO4 -/Mn2+ > Eo Cl2 /Cl- MnO4 -(aq) + 8H+ (aq) + 5e– Mn2+ (aq) + 4H2O(l) E+1.51V Cl2 (aq) +2e– 2Cl– (aq) E +1.36V This would lead to a greater volume of manganate being used and poisonous Cl2 being produced. It cannot be nitric acid as it is an oxidising agent. It oxidises Fe2+ to Fe3+ as Eo NO3 -/HNO2> Eo Fe3+/Fe2+ NO3 – (aq) + 3H+ (aq) + 2e– HNO2 (aq) + H2O(l) Eo +0.94V Fe3+ (aq)+e– Fe2+ (aq) Eo +0.77 V This would lead to a smaller volume of manganate being used. The purple colour of manganate can make it difficult to see the bottom of meniscus in the burette. If the manganate is in the burette then the end point of the titration will be the first permanent pink colour. Colourless purple
Other useful manganate titrations With hydrogen peroxide Ox H2O2 O2 + 2H+ + 2e- Red MnO4 -(aq) + 8H+ (aq) + 5e- Mn2+ (aq) + 4H2O Overall 2MnO4 -(aq) + 6H+ (aq) + 5H2O2 5O2 + 2Mn2+ (aq) + 8H2O Ox C2O4 2- 2CO2 + 2e- Red MnO4 -(aq) + 8H+ (aq) + 5e- Mn2+ (aq) + 4H2O Overall 2MnO4 -(aq) + 16H+ (aq) + 5C2O4 2-(aq) 10CO2 (g) + 2Mn2+(aq) + 8H2O(l) With ethanedioate With Iron (II) ethanedioate both the Fe2+ and the C2O4 2- react with the MnO4 – 1MnO4 – reacts with 5Fe2+ and 2 MnO4 – reacts with 5C2O4 2- MnO4 -(aq) + 8H+ (aq) + 5Fe2+ Mn2+ (aq) + 4H2O + 5Fe3+ 2MnO4 -(aq) + 16H+ (aq) + 5C2O4 2- 10CO2 + 2Mn2+ (aq) + 8H2O So overall 3MnO4 -(aq) + 24H+ (aq) + 5FeC2O4 10CO2 + 3Mn2+ (aq) + 5Fe3+ + 12H2O So overall the ratio is 3 MnO4 – to 5 FeC2O4 The reaction between MnO4 – and C2O4 2- is slow to begin with (as the reaction is between two negative ions). To do as a titration the conical flask can be heated to 60o C to speed up the initial reaction. Be able to perform calculations for these titrations and for others when the reductant and its oxidation product are given. A 2.41g nail made from an alloy containing iron is dissolved in 100cm3 acid. The solution formed contains Fe(II) ions. 10cm3 portions of this solution are titrated with potassium manganate (VII) solution of 0.02M. 9.80cm3 of KMnO4 were needed to react with the solution containing the iron. What is the percentage of Iron by mass in the nail? Manganate titration example MnO4 – (aq) + 8H+ (aq) + 5Fe2+ Mn2+ (aq) + 4H2O + 5Fe3+ Step1 : find moles of KMnO4 moles = conc x vol 0.02 x 9.8/1000 = 1.96×10-4 mol Step 2 : using balanced equation find moles Fe2+ in 10cm3 = moles of KMnO4 x 5 = 9.8×10-4 mol Step 3 : find moles Fe2+ in 100cm3 = 9.8×10-4 mol x 10 = 9.8×10-3 mol Step 4 : find mass of Fe in 9.8×10-3 mol mass= moles x RAM = 9.8×10-3 x 55.8 = 0.547g Step 5 : find % mass %mass = 0.547/2.41 x100 = 22.6% N Goalby chemrevise.org A 1.412 g sample of impure FeC2O4 .2H2O was dissolved in an excess of dilute sulphuric acid and made up to 250 cm3 of solution. 25.0 cm3 of this solution decolourised 23.45 cm3 of a 0.0189 mol dm–3 solution of potassium manganate(VII). What is the percentage by mass of FeC2O4 .2H2O in the original sample? Step1 : find moles of KMnO4 moles = conc x vol 0.0189 x 23.45/1000 = 4.43×10-4 mol Step 2 : using balanced equation find moles FeC2O4 .2H2O in 25cm3 = moles of KMnO4 x 5/3 (see above for ratio) = 7.39×10-4 mol Step 3 : find moles FeC2O4 .2H2O in 250 cm3 = 7.39×10-4 mol x 10 = 7.39×10-3 mol Step 4 : find mass of FeC2O4 .2H2O in 7.39×10-3 mol mass= moles x Mr = 7.39×10-3 x 179.8 = 1.33g Step 5 ; find % mass %mass = 1.33/1.412 x100 =
188.8.131.52 Variable oxidation states (A-level only)
The redox titrations of Fe2+ and C2O4 2– with MnO4 –
Students should be able to perform calculations for these titrations and similar redox reactions.