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Reduction-Oxidation Reactions (1) 213 PHC 4th lecture (1) Gary D. Christian, Analytical Chemistry,6 th edition.

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Presentation on theme: "Reduction-Oxidation Reactions (1) 213 PHC 4th lecture (1) Gary D. Christian, Analytical Chemistry,6 th edition."— Presentation transcript:

1 Reduction-Oxidation Reactions (1) 213 PHC 4th lecture (1) Gary D. Christian, Analytical Chemistry,6 th edition.

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

3 Half-Reaction Method I - + MnO 4 - → I 2 + Mn 2+ Separate the reaction into two half reactions: I - → I 2 MnO 4 - → Mn 2+ 3 Balancing Redox Reactions

4 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. 4

5 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 5

6 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) 6

7 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 7

8 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 8

9 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 9

10 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. 10

11 Self-Indication Highly colored e.g. potassium permanganate purple) MnO 4 - + 5e = Mn 2+ (purple) (colorless) 11 Detection Of The End Point

12 Starch Indicator Used with iodine Strach-I 2 complex (dark blue) 12

13 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 13

14 SUMMARY Balancing redox reactions. Indicators used to detect the end point in redox reaction. Thank You 14


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