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Get Started Now- Intro Lesson: a6:11
- Intro Lesson: b4:21
- Intro Lesson: c6:51
- Lesson: 15:14
- Lesson: 24:30

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 (KMnO
_{4}) is an oxidizing agent that is purple in solution, but turns colorless when reduced to Mn^{2+}ions. - Potassium iodide (KI) in solution gives I
^{-}ions that get oxidized (lots of chemicals can be used for this part) into brown-colored I_{2}in solution. Then in a redox titration, I_{2}can be reduced back to colorless I^{-}ions. Starch can be added (it acts like an indicator for I_{2}) to this, which is blue-black when I_{2}is present, the color fading when I_{2}becomes I^{-}again.

- Potassium permanganate (KMnO
- WORKED EXAMPLE:

A solution containing Co^{2+}ions of unknown concentration is made. 25mL of this Co^{2+}solution was measured and was titrated by 0.2M MnO_{4}^{-}solution until equivalence point was reached. 19.40 mL of the MnO_{4}^{-}solution was required.

The first thing that needs doing is the finding out of the two half-reactions:- Manganese in MnO
_{4}^{-}will be reduced to Mn^{2+}ions as shown in the half-equation:MnO _{4}^{-}+ 8H^{+}+ 5e^{-}$\enspace$→$\enspace$Mn^{2+}+ 4H_{2}O - Co
^{2+}ions can be oxidized to Co^{3+}according to the half-equation:Co ^{2+}$\enspace$→$\enspace$Co^{3+}+ 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 equation1 x [ MnO _{4}^{-}+ 8H^{+}+ 5e^{-}$\enspace$→$\enspace$Mn^{2+}+ 4H_{2}O ]5 x [ Co ^{2+}$\enspace$→$\enspace$Co^{3+}+ e^{-}]

These balance for electrons and give the overall equation:MnO _{4}^{-}+ 8H^{+}+ 5Co^{2+}$\enspace$→$\enspace$Mn^{2+}+ 4H_{2}O + 5Co^{3+}

This reaction has the cobalt solution as the unknown, so MnO_{4}^{-}with known concentration is being added by burette. MnO_{4}^{-}is purple and as it is added to the cobalt solution, the purple color will disappear as Co^{2+}reacts it away.__When equivalence point is reached the purple color will no longer be removed__as there will be no more Co^{2+}to remove the MnO_{4}^{-}and the purple color that it causes. Therefore the equivalence point is shown by the appearance of the purple color of the MnO_{4}^{-}that’s now in excess.

The number of moles of MnO_{4}^{-}can be calculated using the information in the question:Mol MnO _{4}^{-}= 19.40 mL * $\frac{1\; L}{1000 \; m L} \;* \; \frac{0.2\; mol \; MnO_{4}^{-}}{1 \; L}$ = 3.88 * 10^{-3}mol MnO_{4}^{-}

Looking at the equation, we can see a 1:5 MnO_{4}^{-}to Co^{2+}ratio. The equivalence point will have five times as many moles of cobalt as manganese, then.Mol Co ^{2+}= 3.88 * 10^{-3}MnO_{4}^{-}$\,$* $\frac{5\; mol \; Co^{2+}}{1 \; mol \; MnO_{4}^{-}}$ = 0.0194 mol Co^{2+}

With the moles of Co^{2+}ions now found in 25 mL volume of the sample used, we can calculate the concentration.[Co ^{2+}] = $\frac{0.0194\; mol \; Co^{2+}}{0.025 \; L }$ = 0.776 M Co^{2+} - Manganese in MnO

- Introduction
__Using titration for redox reactions.__a)Recap of titration.b)Titration for redox reactions.c)Redox titration: Worked example - 1.
**Find the full equation and concentration of substances in a redox titration.**

A solution containing Co^{2+}ions of unknown concentration is made. 25mL of this Co^{2+}solution was measured and was titrated by 0.25M MnO_{4}^{-}solution until equivalence point was reached. An average titre of 16.20 mL MnO_{4}^{-}solution was required. The reaction produces Mn^{2+}and Co^{3+}ions.

Write the full redox equation for this reaction and find the concentration of the aqueous Co^{2+}solution. - 2.
**Find the full equation and concentration of substances in a redox titration.**

A solution containing I^{-}ions of unknown concentration is made. 25mL of this solution is measured precisely and is titrated by 0.18M MnO_{4}^{-}solution until the equivalence point is reached. This is repeated, to find an average titre of 19.55 mL MnO_{4}^{-}solution being needed to completely react the I^{-}ions.

Write the full redox equation for this reaction and find the concentration of the aqueous I^{-}solution.