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.
  
  
  
• WORKED EXAMPLE:
  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 * $\frac{1\; L}{1000 \; m L} \;* \; \frac{0.2\; mol \; MnO_{4}^{-}}{1 \; L}$ = 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^- $\,$* $\frac{5\; mol \; Co^{2+}}{1 \; mol \; MnO_{4}^{-}}$ = 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⁺] = $\frac{0.0194\; mol \; Co^{2+}}{0.025 \; L }$ = 0.776 M Co²⁺