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Redox titrations

Redox titrations. HIGHER GRADE CHEMISTRY CALCULATIONS. Redox titrations are used to find out information about one reactant, using known information about the other. Worked example 1.

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Redox titrations

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  1. Redox titrations

  2. HIGHER GRADE CHEMISTRY CALCULATIONS Redox titrations are used to find out information about one reactant, using known information about the other. Worked example 1. Iron(II) sulphate can be oxidised using acidified potassium permanganate solution. Calculate the mass of iron(II) sulphate which will completely react with 200 cm3 of 0.25 mol l-1 acidified permanganate solution. Redox Titration. Write the redox equation:- 2MnO4- + 16H+ + 5 Fe2+ 2Mn2+ + 8H2O + 5Fe3+ Calculate the number of moles of the ‘known’ reactant No. of moles = C x V(litres) = 0.25 x 0.2 = 0.05 mol Use mole ratio in equation to calculate the number of moles of the ‘unkown’ reactant. 0.05 mol of MnO4- reacts with 5/2 x 0.05 = 0.125 mol of Fe2+ Use mass = no. of mole x gfm to calculate the mass of iron(II) sulphate Mass of FeSO4 = number of moles x gfm Mass of FeSO4 = 0.125 x 152 = 19 g

  3. Calculations for you to try. • Iron(II) ions react with acidified dichromate solution as shown below:- • 6Fe2+ + Cr2O72- + 14H+ 6Fe3+ + 2Cr3+ + 7H2O • Calculate the number of moles of iron(II) ions which will completely • react with 25cm3 of 0.4 mol l-1 dichromate solution. Number of moles of dichromate = C x V(litres) = 0.4 x 25/1000 = 0.01 From the mole ratio in the balanced equation number of moles of iron(II) = 6 x 0.01 = 0.06 Higher Grade Chemistry

  4. 2. Hydrogen peroxide reacts with acidified permanganate solution as shown below:- 5H2O2 + 2MnO4- + 6H+ 2Mn2+ + 5O2 + 8H2O 100 cm3 of hydrogen peroxide solution reacts with 12 cm3 of 0.2 mol l-1 permanganate solution. Calculate the concentration of the hydrogen peroxide solution. Number of moles of permanganate = C x V(litres) = 0.2 x 10/1000 = 0.002 From the mole ratio in the balanced equation number of moles of H2O2 = 5/2 x 0.002 = 0.005 Use Concentration = number of moles/Volume (litres) = 0.05 / 0.1 = 0.5 mol l-1 Higher Grade Chemistry

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