Redox titrations - Redox and Electrochemistry

Redox titrations



In this lesson, we will learn:

  • To recall the practical use of titration experiments
  • How titration applies to redox reactions.
  • How to calculate chemical quantities required in redox reactions.


  • We learned the basics of a titration with its use in acid-base chemistry in Acid-base-titration.
    Just like acid-base titrations are used to find the concentration of acids and bases, a redox titration can be done to find the unknown concentration of a chemical in a redox process. The working out and calculations are detailed in Acid-base-titration and is summarized in the image below. Chemical A and chemical B in a redox titration would simply be the two chemicals in the redox (the reducing and oxidizing agent):

  • Redox titrations will involve a reducing and oxidizing agent reacting together, but indicator is normally not used like it is in acid-base titrations. This means that one of the reactants used has to be one with a color difference between its reduced and oxidized form. There are two good options:
    • Potassium permanganate (KMnO4) is an oxidizing agent that is purple in solution, but turns colorless when reduced to Mn2+ ions.
    • Potassium iodide (KI) in solution gives I- ions that get oxidized (lots of chemicals can be used for this part) into brown-colored I2 in solution. Then in a redox titration, I2 can be reduced back to colorless I- ions. Starch can be added (it acts like an indicator for I2) to this, which is blue-black when I2 is present, the color fading when I2 becomes I- again.

    A solution containing Co2+ ions of unknown concentration is made. 25mL of this Co2+ solution was measured and was titrated by 0.2M MnO4- solution until equivalence point was reached. 19.40 mL of the MnO4- solution was required.
    The first thing that needs doing is the finding out of the two half-reactions:
    • Manganese in MnO4- will be reduced to Mn2+ ions as shown in the half-equation:

      MnO4- + 8H+ + 5e-\enspace \enspace Mn2+ + 4H2O

    • Co2+ ions can be oxidized to Co3+ according to the half-equation:

      Co2+\enspace \enspace Co3+ + e-

      The method for working out half-equations in redox was covered in Half equations.

    Next, the combining of the two half-reactions will give us the overall equation

    1 x [ MnO4- + 8H+ + 5e-\enspace \enspace Mn2+ + 4H2O ]

    5 x [ Co2+\enspace \enspace Co3+ + e- ]

    These balance for electrons and give the overall equation:

    MnO4- + 8H+ + 5Co2+\enspace \enspace Mn2+ + 4H2O + 5Co3+

    This reaction has the cobalt solution as the unknown, so MnO4- with known concentration is being added by burette. MnO4- is purple and as it is added to the cobalt solution, the purple color will disappear as Co2+ reacts it away. When equivalence point is reached the purple color will no longer be removed as there will be no more Co2+ to remove the MnO4- and the purple color that it causes. Therefore the equivalence point is shown by the appearance of the purple color of the MnO4- that’s now in excess.

    The number of moles of MnO4- can be calculated using the information in the question:

    Mol MnO4- = 19.40 mL * 1L1000mL0.2molMnO41L\frac{1\; L}{1000 \; m L} \;* \; \frac{0.2\; mol \; MnO_{4}^{-}}{1 \; L} = 3.88 * 10-3 mol MnO4-

    Looking at the equation, we can see a 1:5 MnO4- to Co2+ ratio. The equivalence point will have five times as many moles of cobalt as manganese, then.

    Mol Co2+ = 3.88 * 10-3 MnO4- \,* 5molCo2+1molMnO4\frac{5\; mol \; Co^{2+}}{1 \; mol \; MnO_{4}^{-}} = 0.0194 mol Co2+

    With the moles of Co2+ ions now found in 25 mL volume of the sample used, we can calculate the concentration.

    [Co2+] = 0.0194molCo2+0.025L\frac{0.0194\; mol \; Co^{2+}}{0.025 \; L } = 0.776 M Co2+
  • Intro Lesson
    Using titration for redox reactions.
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Redox titrations

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