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Redox Titrations

Key Concepts

  • A redox titration is a volumetric method that relies on the oxidation of the analyte (substance to be analysed).

  • The titrant (solution of known concentration) is often an oxidising agent.

  • At the equivalence point E(forward) = E(reverse), or, ΔE(cell) = 0

  • If the redox reaction does not produce a well-defined colour change at the equivalence point, an indicator should be used in the titration.

  • The redox titration curve is a plot of Electrode Potential (volts) vs volume of titrant or analyte.


  1. Write a balanced half equation for the oxidation reaction.

  2. Write a balanced half equation for the reduction reaction.

  3. Add the oxidation and reduction half equations together to give a balanced redox reaction equation.

  4. Extract all the relevant information from the question.

  5. Check the data for consistency, for example, concentrations are often given in M or mol L-1 but volumes are often given in mL. You will need to convert the mL to L for consistency. The easiest way to do this is to multiply the volume in mL x 10-3 (enter the volume in mL into your calculator, then click "EXP", then enter -3 and the conversion is done!).

  6. Calculate the moles of reactant (titrant) (n) for which you have both volume (V) and concentration (molarity, c): n = c x V

  7. From the redox reaction equation find the mole ratio of known reactant (titrant) : unkown reactant (analyte)

  8. Calculate moles of unknown reactant (analyte) using this mole ratio.

  9. From the volume (V) of the unknown reactant (analyte) and its calculated moles (n), calculate its concentration ( molarity, c): c = n ÷ V


A 0.0484 mol L-1 standard solution of potassium permanganate was titrated against 25.00 mL of an iron(II) sulfate solution.

The equivalence point, as indicated by a faint pink colour, was reached when 15.50 mL of potassium permanganate solution had been added.

Calculate the concentration of the iron(II) sulfate solution.

  1. reduction half equation:
    MnO4- + 8H+ + 5e → Mn2+ + 4H2O

  2. oxidation half equation:
    Fe2+ → Fe3+ + e

  3. redox equation:
    MnO4- + 8H+ + 5e Mn2+ + 4H2O
    5Fe2+ 5Fe3+ + 5e

    MnO4- + 8H+ + 5Fe2+ Mn2+ + 4H2O + 5Fe3+

  4. Extract all the relevant information:
    MnO4- Fe2+
    [MnO4-] = 0.0484 mol/L [Fe2+] = ? mol/L
    V (MnO4-) = 15.50 mL V (Fe2+) = 25.00 mL

  5. Check the data for consistency: convert volumes in mL to L
    MnO4- Fe2+
    [MnO4-] = 0.0484 mol/L [Fe2+] = ? mol/L
    V (MnO4-) = 15.50 x 10-3 L V (Fe2+) = 25.00 x 10-3 L

  6. Calculate moles of known reactant (titrant):
    [MnO4-] = 0.0484 mol/L  
    V (MnO4-) = 15.50 x 10-3 L  
    n (MnO4-) = c x V = 0.0484 x 15.50 x 10-3 = 7.502 x 10-4 mol  

  7. Find the mole ratio from balanced redox equation
    MnO4- : Fe2+
    1 : 5

  8. Calculate moles (n) of Fe2+
    n (Fe2+) = 5 x n (MnO4-) = 5 x 7.502 x 10-4 = 0.003751 mol

  9. Calculate the concentration of Fe2+
    [Fe2+] = n ÷ V = 0.003751 ÷ 25.00 x 10-3 = 0.1500 mol/L

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