1. Redox Titrations

2. Oxidation-reduction reactions involve a transfer of electrons. The oxidising agent accepts electrons and the reducing agent gives electrons. In working out the equation for a redox reaction we can use the half- equation method. In this method we use the half-equation for the oxidising agent and the half-equation for the reducing agent then add them together.

3. Examples of half-reaction equations Iron(III) salts are reduced to iron(II) salts. The half equation is Fe 3+  Fe 2+ For the equation to balance the charge on the RHS must equal the charge on the LHS. This can be accomplished by inserting an electron on the LHS Fe 3+ + e -  Fe 2+

4. When chlorine acts as an oxidising agent it is reduced to Cl - ions Cl 2  2Cl - To obtain a balanced equation 2 electrons muct be inserted on the LHS Cl 2 + 2e -  2Cl -

5. Potassium mangante(VII) is an oxidising agent. In acid solutions it is reduced to a manganese(II) salt MnO 4 - + 8H + + 5e -  Mn 2+ + 4H 2 O Potassium dichromate(VI) is also an oxidising agent. It is reduced to a chromium(III) salt Cr 2 O 7 - + 14H + + 6e -  2Cr 3+ + 7H 2 O

6. Using half equations to obtain the equation for the reaction Find the equation for the reaction of iron(III) with chloride ions The 2 half reactions needed are 1 Fe 3+ + e -  Fe 2+ 2 Cl 2 + 2e -  2Cl - We need iron(III) and chloride as reactants so equation 2 must be reversed 3 2Cl -  Cl 2 + 2e -

7. Now we must balance the electrons, then add the half equations together to obtain the full equation. To balance the electrons we multiply equation 1 by 2 1Fe 3+ + e -  Fe 2+ x2 4 2Fe 3+ + 2e -  2Fe 2+ 3 2Cl -  Cl 2 + 2e - 2Fe 3+ + 2Cl -  2Fe 2+ + Cl 2 There must be the same no of electrons on each side which then cancel.

8. Construct an equation for the reaction of potassium manganate(VII) with ethanedioate ion Ethanedioate ion is C 2 O 4 2- and the half equation is C 2 O 4 2-  2CO 2 + 2e - 2MnO 4 - + 16H + + 5C 2 O 4 2-  2Mn 2+ + 8H 2 O + 10CO 2

9. Now we can do a titration calculation using this equation! A 25.0ml portion of sodium ethanedioate solution of concentration 0.200 mol/L is warmed and titrated against a solution of potassium manganate(VII). If 17.2ml of potassium manganate(VII) is required what is its concentration?

10. No of moles ethanedioate in 25.0ml of solution of conc 0.200 mol/L = 0.200 x 25 = 0.005000 mols 1000 Ratio of ethanedioate to manganate(VII) = 2:5  no of mols of manganate(VII) in 17.2ml = 0.005000 x 2 = 0.002000 mols 5  conc of manganate(VII) = 0.002000 x 1000 17.2 = 0.116 mol/L