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-→Mn2+ + 4H2O
- Co2+ ions can be oxidized to Co3+ according to the half-equation:
Co2+→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-→Mn2+ + 4H2O ]
5 x [ Co2+→Co3+ + e- ]
These balance for electrons and give the overall equation:
MnO4- + 8H+ + 5Co2+→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 * = 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- * = 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+] = = 0.776 M Co2+