1. Balancing Redox Reactions Chem 12

2. Application of oxidation numbers: Oxidation = an increase in oxidation number Reduction = a decrease in oxidation number

3. Using oxidation numbers to identify redox reactions: Example: Determine if the following reaction is a redox reaction: CH 4 (g) + Cl 2 (g)  CH 3 Cl(g) + HCl(g)  Oxidation number of C increases from -4 to -2  Oxidation number of Cl decreases from 0 to -1 Yes, redox rxn! 0+1-4 +1 +1 -2

4. Zn + Cu +  Zn 2+ + Cu Zn  Zn 2+ Cu +  Cu TO BALANCE A REDOX REACTION… BREAK IT INTO HALF-REACTIONS!!!

5. Zn + Cu +  Zn 2+ + Cu Zn  Zn 2+ + 2e - Cu +  Cu THEN BALANCE EACH REACTION… …FOR BOTH MASS AND CHARGE

6. Zn + Cu +  Zn 2+ + Cu Zn  Zn 2+ + 2e - 1e - + Cu +  Cu THEN BALANCE EACH REACTION… …FOR BOTH MASS AND CHARGE

7. Zn + Cu +  Zn 2+ + Cu Zn  Zn 2+ + 2e - 2(1e - + Cu +  Cu) EQUILIZE ELECTRON TRANSFER…

8. Zn + Cu +  Zn 2+ + Cu Zn  Zn 2+ + 2e - 2(1e - + Cu +  Cu) + Zn + 2Cu +  Zn 2+ + 2Cu AND ADD THE REACTIONS TOGETHER

9. C 2 O 4 2- + Fe 3+  CO 2 + Fe 2+ C 2 O 4 2-  CO 2 Fe 3+  Fe 2+ BREAK IT INTO HALF-REACTIONS!!!

10. C 2 O 4 2- + Fe 3+  CO 2 + Fe 2+ C 2 O 4 2-  2CO 2 Fe 3+  Fe 2+ THEN BALANCE EACH REACTION… …FOR BOTH MASS AND CHARGE

11. C 2 O 4 2- + Fe 3+  CO 2 + Fe 2+ C 2 O 4 2-  2CO 2 + 2e - Fe 3+  Fe 2+ …FOR BOTH MASS AND CHARGE THEN BALANCE EACH REACTION…

12. C 2 O 4 2- + Fe 3+  CO 2 + Fe 2+ C 2 O 4 2-  2CO 2 + 2e - 1e - + Fe 3+  Fe 2+ THEN BALANCE EACH REACTION… …FOR BOTH MASS AND CHARGE

13. C 2 O 4 2- + Fe 3+  CO 2 + Fe 2+ C 2 O 4 2-  2CO 2 + 2e - 2(1e - + Fe 3+  Fe 2+ ) EQUILIZE ELECTRON TRANSFER…

14. C 2 O 4 2- + Fe 3+  CO 2 + Fe 2+ C 2 O 4 2-  2CO 2 + 2e - 2(1e - + Fe 3+  Fe 2+ ) + C 2 O 4 2- + 2Fe 3+  2CO 2 + 2Fe 2+ AND ADD THE REACTIONS TOGETHER

15. BALANCING IN ACIDIC SOLUTION (the steps…..) 1. Assign oxidation charges to all atoms in equation 2. Break into half reactions 3. Balance each half reaction for all atoms (except O and H) 4. Add H 2 O to balance O (because in an acidic solution) 5. Add H + ions to balance H. 6. Add electrons to reduction reaction to balance charges 7. Equalize electron transfer (make e - lost = e - gained) 8. Add reactions!

16. BALANCING IN BASIC SOLUTION (the steps…..) Steps 1-5 same as acidic! 6. Adjust for basic conditions by adding to both sides the same number of OH - ions as H + already there. 7. Simplify each half reaction by combining H + and OH - to make water 8. Cancel any water present on both sides of the half reactions 9. Add electrons to reduction reaction to balance charges 10. Equalize electron transfer (make e - lost = e - gained) 11. Add reactions!

17. Examples: 1. Write a balanced half reaction that shows the reduction of permanganate ions to maganese (II) ions in acidic conditions. 2. Write a balanced half reaction that shows the oxidation of S 2 O 3 2- to SO 3 2- in basic conditions

18. Day 1: Try it: Pg 728 # 13 and section review # 2 Pg 732 # 19, 20 Pg 734 # 23, 24

19. Ni 2+ + Cr 2 O 7 2-  Cr 3+ + Ni 4+ BALANCING IN ACIDIC SOLUTION Ni 2+  Ni 4+ Cr 2 O 7 2-  Cr 3+ BREAK IT INTO HALF-REACTIONS!!!

20. Ni 2+ + Cr 2 O 7 2-  Cr 3+ + Ni 4+ BALANCING IN ACIDIC SOLUTION Ni 2+  Ni 4+ + 2e - Cr 2 O 7 2-  Cr 3+ THEN BALANCE EACH REACTION… …FOR BOTH MASS AND CHARGE

21. Ni 2+ + Cr 2 O 7 2-  Cr 3+ + Ni 4+ BALANCING IN ACIDIC SOLUTION Ni 2+  Ni 4+ + 2e - Cr 2 O 7 2-  2Cr 3+

22. Ni 2+ + Cr 2 O 7 2-  Cr 3+ + Ni 4+ BALANCING IN ACIDIC SOLUTION Ni 2+  Ni 4+ + 2e - Cr 2 O 7 2-  2Cr 3+ + 7H 2 O WE CAN BALANCE OXYGEN WITH WATER MOLECULES BECAUSE THE REACTION HAPPENS IN SOLUTION

23. Ni 2+ + Cr 2 O 7 2-  Cr 3+ + Ni 4+ BALANCING IN ACIDIC SOLUTION Ni 2+  Ni 4+ + 2e - 14H + + Cr 2 O 7 2-  2Cr 3+ + 7H 2 O WE CAN BALANCE HYDROGEN WITH H + IONS SINCE IT IS AN ACIDIC SOLUTION

24. Ni 2+ + Cr 2 O 7 2-  Cr 3+ + Ni 4+ BALANCING IN ACIDIC SOLUTION Ni 2+  Ni 4+ + 2e - 6e - + 14H + + Cr 2 O 7 2-  2Cr 3+ + 7H 2 O

25. Ni 2+ + Cr 2 O 7 2-  Cr 3+ + Ni 4+ BALANCING IN ACIDIC SOLUTION 3(Ni 2+  Ni 4+ + 2e - ) 6e - + 14H + + Cr 2 O 7 2-  2Cr 3+ + 7H 2 O EQUILIZE ELECTRON TRANSFER…

26. Ni 2+ + Cr 2 O 7 2-  Cr 3+ + Ni 4+ BALANCING IN ACIDIC SOLUTION 3(Ni 2+  Ni 4+ + 2e - ) 6e - + 14H + + Cr 2 O 7 2-  2Cr 3+ + 7H 2 O + 14H + + Cr 2 O 7 2- + 3Ni 2+  3Ni 4+ + 2Cr 3+ + 7H 2 O AND ADD THE REACTIONS TOGETHER

27. Ag + NO 3 -  Ag + + NO BALANCING IN ACIDIC SOLUTION Ag  Ag + NO 3 -  NO SEE IF YOU CAN PREDICT EACH STEP BEFORE ADVANCING THE SLIDE

28. Ag + NO 3 -  Ag + + NO BALANCING IN ACIDIC SOLUTION Ag  Ag + + 1e - NO 3 -  NO

29. Ag + NO 3 -  Ag + + NO BALANCING IN ACIDIC SOLUTION Ag  Ag + + 1e - NO 3 -  NO + 2H 2 O

30. Ag + NO 3 -  Ag + + NO BALANCING IN ACIDIC SOLUTION Ag  Ag + + 1e - 4H + + NO 3 -  NO + 2H 2 O

31. Ag + NO 3 -  Ag + + NO BALANCING IN ACIDIC SOLUTION Ag  Ag + + 1e - 3e - + 4H + + NO 3 -  NO + 2H 2 O

32. Ag + NO 3 -  Ag + + NO BALANCING IN ACIDIC SOLUTION 3(Ag  Ag + + 1e - ) 3e - + 4H + + NO 3 -  NO + 2H 2 O

33. Ag + NO 3 -  Ag + + NO BALANCING IN ACIDIC SOLUTION 3(Ag  Ag + + 1e - ) 3e - + 4H + + NO 3 -  NO + 2H 2 O + 4H + + NO 3 - + 3Ag  3Ag + + NO + 2H 2 O

34. BALANCING IN BASIC SOLUTION To balance a redox reaction in a basic solution, where hydroxide (OH-) ions outnumber hydrogen (H+) ions, FIRST BALANCE IT AS IF IT WERE IN ACIDIC SOLUTION

35. BALANCING IN BASIC SOLUTION In the previous example… 4H + + NO 3 - + 3Ag  3Ag + + NO + 2H 2 O … is balanced in acidic solution

36. BALANCING IN BASIC SOLUTION To balance in basic solution… 4H + + NO 3 - + 3Ag  3Ag + + NO + 2H 2 O … add an amount of hydroxide equal to the amount of H + to each side of the reaction 4OH -

37. BALANCING IN BASIC SOLUTION To balance in basic solution… 4H + + NO 3 - + 3Ag  3Ag + + NO + 2H 2 O …Combine the H + and OH - to make water 4OH -

38. BALANCING IN BASIC SOLUTION To balance in basic solution… 4H 2 O + NO 3 - + 3Ag  3Ag + + NO + 2H 2 O …Combine the H + and OH - to make water 4OH -

39. BALANCING IN BASIC SOLUTION To balance in basic solution… 4H 2 O + NO 3 - + 3Ag  3Ag + + NO + 2H 2 O …since there is water on both sides, we can cancel some out 4OH - 2

40. BALANCING IN BASIC SOLUTION To balance in basic solution… 2H 2 O + NO 3 - + 3Ag  3Ag + + NO + 4OH - … and rewrite the final balanced equation

41. Zn + Cu +  Zn 2+ + Cu 0+1+20 The oxidation number of zinc increases from zero to positive two by losing two electrons. ZINC IS OXIDIZED The oxidation number of copper decreases from positive one to zero by gaining one electron. COPPER IS REDUCED ASSIGN OXIDATION NUMBERS TO EACH ELEMENT IN THE REACTION

42. Try it: Page 739 # 27(a, b, c), 28

43. H + + MnO 4 - + Fe 2+  Mn 2+ + Fe 3+ + H 2 O The oxidation number of Mn increases from +7 to +2 by gaining five electrons. MANGANESE IS REDUCED The oxidation number of Fe increases from +2 to +3 by losing one electron. IRON IS OXIDIZED +1+7-2+2 +3+1-2 ASSIGN OXIDATION NUMBERS TO EACH ELEMENT IN THE REACTION

44. H + + MnO 4 - + Fe 2+  Mn 2+ + Fe 3+ + H 2 O +1+7-2+2 +3+1-2 MANGANESE IS REDUCED IRON IS OXIDIZED The compound that contains the element being reduced is called the OXIDIZING AGENT The compound that contains the element being oxidized is called the REDUCING AGENT