1. Reduction-Oxidation Reactions (1) 213 PHC 11th lecture Dr. Mona Alshehri (1) Gary D. Christian, Analytical Chemistry,6 th edition.

2. By the end of the lecture the student should be able to: Detect the end point in redox reaction. Balance a redox reaction. 2

3. Detection Of The End Point 3

4. Self-Indication Highly colored e.g. potassium permanganate (purple) MnO 4 - + 5e = Mn 2+ (purple) (colorless) 4

5. Starch Indicator Used with iodine Strach-I 2 complex (dark blue) 5

6. Redox Indicators Highly colored dyes Weak reducing or oxidizing agents The color of the reduced and oxidized forms are different The color will change with E change during titration E o of indicator must be near the E of eq. point The redox indicator reaction must be rapid E.g. Diphenylamine 6

7. Balancing Redox Reactions 7

8. Half-Reaction Method I - + MnO 4 - → I 2 + Mn 2+ Separate the reaction into two half reactions: I - → I 2 MnO 4 - → Mn 2+ 8

9. Balance the atoms of each half-reaction: first balance all of the atoms except H and O. For an acidic solution add H 2 O to balance the O atoms and H + to balance the H atoms. In a basic solution use OH - and H 2 O to balance the O and H. Balance the iodine atoms: 2 I - → I 2 The Mn in the permanganate reaction is already balanced, so let's balance the oxygen: MnO 4 - → Mn 2+ + 4 H 2 O Add H + to balance the 4 waters molecules: MnO 4 - + 8 H + → Mn 2+ + 4 H 2 O The two half-reactions are now balanced for atoms. 9

10. Balance the charges in each half- reaction: The reduction half-reaction should accept the same number of electrons as the oxidation half- reaction loss. This is accomplished by adding electrons to the reactions: 2 I - → I 2 + 2e - 5 e - + 8 H + + MnO 4 - → Mn 2+ + 4 H 2 O 10

11. Multiply the electrons by a number: The two half-reactions should have the same number of electrons so they can cancel each other: 5 (2I - → I 2 +2e - ) 2 (5e - + 8H + + MnO 4 - → Mn 2+ + 4H 2 O) 11

12. Add the two half-reactions: 10 I - → 5 I 2 + 10 e - 16 H + + 2 MnO 4 - + 10 e - → 2 Mn 2+ + 8 H 2 O This gives the final equation: 10 I - + 10 e - + 16 H + + 2 MnO 4 - → 5 I 2 + 2 Mn 2+ + 10 e - + 8 H 2 O 12

13. Get the overall equation: Cancel out the electrons and H 2 O, H +, and OH - that may appear on both sides of the equation: 10 I - + 16 H + + 2 MnO 4 - → 5 I 2 + 2 Mn 2+ + 8 H 2 O 13

14. Check the numbers: Make certain that the mass and charge are balanced. In this example, the atoms are balanced with a +4 net charge on each side of the reaction. 10 I - + 16 H + + 2 MnO 4 - → 5 I 2 + 2 Mn 2+ + 8 H 2 O 14

15. Follow the link: http://www.youtube.com/watch?v=EtOrcK w6wso&feature=endscreen&NR=1 How to balance a redox reaction in acidic solution. http://www.youtube.com/watch?v=imqYliG 5BCU&feature=fvwrel How to balance a redox reaction in basic solution. 15

16. Questions? 16

17. SUMMARY Indicators used to detect the end point in redox reaction. Balancing redox reactions. Thank You 17