1. Redox Titrations A redox titration is the same as an acid-base titration except it involves a redox reaction and generally does not require an indicator. Reagents are chosen so that the reaction is spontaneous and one of the half reactions has a natural colour change.

2. Redox Titrations A redox titration is the same as an acid-base titration except it involves a redox reaction and generally does not require an indicator. Reagents are chosen so that the reaction is spontaneous and one of the half reactions has a natural colour change. Pick a suitable reagent for redox titration involving IO 3 - in acid solution.F - I - SO 4 2- Cl -

3. Pick the spontaneous reaction

4. 6.75 mL of 0.100 M KMnO 4 is required to titrate 25.0 mL of FeCl 2. Calculate the [Fe 2+ ]. MnO 4 - + 8H + +5Fe 2+ → Mn 2+ + 4H 2 O + 5Fe 3+

5. 6.75 mL of 0.100 M KMnO 4 is required to titrate 25.0 mL of FeCl 2. Calculate the [Fe 2+ ]. MnO 4 - + 8H + +5Fe 2+ → Mn 2+ + 4H 2 O + 5Fe 3+ 0.00675 L0.0250L 0.100 M? M

6. 6.75 mL of 0.100 M KMnO 4 is required to titrate 25.0 mL of FeCl 2. Calculate the [Fe 2+ ]. MnO 4 - + 8H + +5Fe 2+ → Mn 2+ + 4H 2 O + 5Fe 3+ 0.00675 L0.0250L 0.100 M? M [Fe 2+ ] =

7. 6.75 mL of 0.100 M KMnO 4 is required to titrate 25.0 mL of FeCl 2. Calculate the [Fe 2+ ]. MnO 4 - + 8H + +5Fe 2+ → Mn 2+ + 4H 2 O + 5Fe 3+ 0.00675 L0.0250L 0.100 M? M [Fe 2+ ] =

8. 6.75 mL of 0.100 M KMnO 4 is required to titrate 25.0 mL of FeCl 2. Calculate the [Fe 2+ ]. MnO 4 - + 8H + +5Fe 2+ → Mn 2+ + 4H 2 O + 5Fe 3+ 0.00675 L0.0250L 0.100 M? M 0.00675 L MnO 4 - x 0.100 mole [Fe 2+ ] = 1 L

9. 6.75 mL of 0.100 M KMnO 4 is required to titrate 25.0 mL of FeCl 2. Calculate the [Fe 2+ ]. 1MnO 4 - + 8H + +5Fe 2+ → Mn 2+ + 4H 2 O + 5Fe 3+ 0.00675 L0.0250L 0.100 M? M 0.00675 L MnO 4 - x 0.100 mole x 5 moles Fe 2+ [Fe 2+ ] = 1 L 1 mole MnO 4 -

10. 6.75 mL of 0.100 M KMnO 4 is required to titrate 25.0 mL of FeCl 2. Calculate the [Fe 2+ ]. MnO 4 - + 8H + +5Fe 2+ → Mn 2+ + 4H 2 O + 5Fe 3+ 0.00675 L0.0250L 0.100 M? M 0.00675 L MnO 4 - x 0.100 mole x 5 moles Fe 2+ [Fe 2+ ] = 1 L 1 mole MnO 4 - 0.0250 L

11. 6.75 mL of 0.100 M KMnO 4 is required to titrate 25.0 mL of FeCl 2. Calculate the [Fe 2+ ]. MnO 4 - + 8H + +5Fe 2+ → Mn 2+ + 4H 2 O + 5Fe 3+ 0.00675 L0.0250L 0.100 M? M 0.00675 L MnO 4 - x 0.100 mole x 5 moles Fe 2+ [Fe 2+ ] = 1 L 1 mole MnO 4 - 0.0250 L =0.135 M

12. Write the anode and cathode reactions. Pt H 2 O 2(aq) MnO 4 - in acid NaNO 3aq) Inert electrodes- look at the solution for the reactions voltmeter

13. Cathode Anode

14. Cathode: MnO 4 - + 8H + + 5e - → Mn 2+ + 4H 2 O Anode: H 2 O 2 → O 2 + 2H + + 2e - What happens to the mass of the cathode? Constant What happens to the mass of the anode? Constant What happens to the pH of the cathode? Increases What happens to the pH of the anode? Decreases

15. Non-Inert Electrodes A non-inert Anode might oxidize. The Cathode will stay inert. DC Power - + K 2 SO 4(aq) Cu 2K + SO 4 2- H 2 O

16. - Non-Inert Electrodes A non-inert Anode might oxidize. The Cathode will stay inert. DC Power - + K 2 SO 4(aq) Cu 2K + SO 4 2- H 2 O

17. - Cathode Reduction Non-Inert Electrodes A non-inert Anode might oxidize. The Cathode will stay inert. DC Power - + K 2 SO 4(aq) Cu 2K + SO 4 2- H 2 O

18. - Cathode Reduction 2H 2 O + 2e - → H 2 +2OH - -0.41 v Non-Inert Electrodes A non-inert Anode might oxidize. The Cathode will stay inert. DC Power - + K 2 SO 4(aq) Cu 2K + SO 4 2- H 2 O

19. - Cathode Reduction 2H 2 O + 2e - → H 2 +2OH - -0.41 v Non-Inert Electrodes A non-inert Anode might oxidize. The Cathode will stay inert. DC Power - + K 2 SO 4(aq) Cu Cu might oxidize 2K + SO 4 2- H 2 O

20. You must look at the possible oxidation of: SO 4 2- H 2 O Cu Strongest Reducing Agent

21. - Cathode Reduction 2H 2 O+2e - → H 2(g) + 2OH - -0.41 v + Anode Oxidation Cu (s) → Cu 2+ +2e - -0.34 v Non-Inert Electrodes A non-inert Anode might oxidize. The Cathode will stay inert. DC Power - + K 2 SO 4(aq) Cu 2K + SO 4 2- H 2 O

22. 2H 2 O + Cu (s) → Cu 2+ + H 2 +2OH - E 0 = -0.75 v MTV = +0.75 v Non-Inert Electrodes A non-inert Anode might oxidize. The Cathode will stay inert. DC Power - + K 2 SO 4(aq) Cu - Cathode Reduction 2H 2 O+2e - → H 2(g) + 2OH - -0.41 v + Anode Oxidation Cu (s) → Cu 2+ +2e - -0.34 v 2K + SO 4 2- H 2 O

23. Is Al a reactive or non-reactive metal? Look on page 8 Reactive as Al is a relatively strong reducing agent. Why is Al used for boats, patio furniture, swing sets, and trucks boxes? Al makes a clear transparent Al 2 O 3 paint like coating that prevents further oxidation.