1. More on Redox
UNIT III
REDOX

2. Reduction / Oxidation L.E.O. G.E.R.s (Leo the lion growls!)
Looses electrons; Oxidized
Gains electrons; Reduced

3. __Al + __Cu2+ --> __Cu + __Al3+
What do you do first?

4. __Al + __Cu2+ --> __Cu + __Al3+
1. Start by writing half reactions (Oxidation and reduction)

5. __Al + __Cu2+ --> __Cu + __Al3+
1. Start by writing half reactions (Oxidation and reduction)
(USE ELECTRONS TO BALANCE THE CHARGES!!!) 
Oxidation:    Al --> Al 3+
Reduction:    Cu2+ --> Cu

6. __Al + __Cu2+ --> __Cu + __Al3+
1. Start by writing half reactions (Oxidation and reduction)
(Electrons go on the more positive side) 
Oxidation:    Al --> Al e-
Reduction:    2e- + Cu2+ --> Cu

7. __Al + __Cu2+ --> __Cu + __Al3+
1. Start by writing half reactions (Oxidation and reduction)
(Electrons go on the more positive side) 
Oxidation:    Al --> Al e-
Reduction:    2e- + Cu2+ --> Cu
2. Balance the electrons by finding the common multiple and multiply the half reactions accordingly. 

8. __Al + __Cu2+ --> __Cu + __Al3+
1. Start by writing half reactions (Oxidation and reduction)
(Electrons go on the more positive side) 
Oxidation:    Al --> Al e-
Reduction:    2e- + Cu2+ --> Cu
2. Balance the electrons by finding the common multiple and multiply the half reactions accordingly. 
The common multiple of the electrons is 6 so…
Oxidation:    2 x (Al --> Al e-)
Reduction:    3 x ( 2e- + Cu2+ --> Cu)

9. __Al + __Cu2+ --> __Cu + __Al3+
Recombine the reactions:                                            
6e- + 2 Al  + 3 Cu2+--> 2 Al Cu + 6e-
The electrons must cancel.              
                                    
2 Al  + 3 Cu2+--> 2 Al Cu
Atoms and charges must be conserved. 
Which was OXIDIZED?
WHICH WAS REDUCED?

10. __Al + __Cu2+ --> __Cu + __Al3+
Recombine the reactions:                                            
6e- + 2 Al  + 3 Cu2+--> 2 Al Cu + 6e-
The electrons must cancel.              
                                    
2 Al  + 3 Cu2+--> 2 Al Cu
Atoms and charges must be conserved. 
Which was OXIDIZED? Aluminum
WHICH WAS REDUCED? Copper

11. Balancing Redox Reactions (Acidic Conditions)
Given:         
(acidic conditions – you have H+)
MnO4- + I- --> I2 + Mn2+

12. Balancing Redox Reactions (Acidic Conditions)
Given:         
(acidic conditions – you have H+)
MnO4- + I- --> I2 + Mn2+
Step 1 - Half Reactions

13. Balancing Redox Reactions (Acidic Conditions)
Given:         
(acidic conditions – you have H+)
MnO4- + I- --> I2 + Mn2+
Step 1 - Half Reactions
MnO4-  -->   Mn2+
I-   -->     I2

14. Balancing Redox Reactions (Acidic Conditions)
Given:         
(acidic conditions – you have H+)
MnO4- + I- --> I2 + Mn2+
Step 1 - Half Reactions
MnO4-  -->   Mn2+
I-   -->     I2
Let’s balance the reduction one first.

15. Balancing Redox Reactions (Acidic Conditions)
Given:         
(acidic conditions – you have H+)
MnO4- + I- --> I2 + Mn2+
Step 1 - Half Reactions
MnO4-  -->   Mn2+
I-   -->     I2
Let’s balance the reduction one first.
for every Oxygen add a water on the other side

16. Balancing Redox Reactions (Acidic Conditions)
Given:         
(acidic conditions – you have H+)
MnO4- + I- --> I2 + Mn2+
Step 1 - Half Reactions
MnO4-  -->   Mn2+
I-   -->     I2
Let’s balance the reduction one first.
for every Oxygen add a water on the other side
MnO4-  -->   Mn2+ + 4H2O

17. Balancing Redox Reactions (Acidic Conditions)
for every Oxygen add a water on the other side:
MnO4-  -->   Mn2+ + 4H2O
For every hydrogen add a H+ to the other side:

18. Balancing Redox Reactions (Acidic Conditions)
for every Oxygen add a water on the other side:
MnO4-  -->   Mn2+ + 4H2O
For every hydrogen add a H+ to the other side:
8H+ + MnO4-  -->   Mn2+ + 4H2O

19. Balancing Redox Reactions (Acidic Conditions)
for every Oxygen add a water on the other side:
MnO4-  -->   Mn2+ + 4H2O
For every hydrogen add a H+ to the other side:
8H+ + MnO4-  -->   Mn2+ + 4H2O
Balance the imbalance of charge with electrons (+7 vs. +2)

20. Balancing Redox Reactions (Acidic Conditions)
for every Oxygen add a water on the other side:
MnO4-  -->   Mn2+ + 4H2O
For every hydrogen add a H+ to the other side:
8H+ + MnO4-  -->   Mn2+ + 4H2O
Balance the imbalance of charge with electrons (+7 vs. +2)
5e- + 8H+ + MnO4-  -->   Mn2+ + 4H2O

21. Balancing Redox Reactions (Acidic Conditions)
for every Oxygen add a water on the other side:
MnO4-  -->   Mn2+ + 4H2O
For every hydrogen add a H+ to the other side:
8H+ + MnO4-  -->   Mn2+ + 4H2O
Balance the imbalance of charge with electrons (+7 vs. +2)
5e- + 8H+ + MnO4-  -->   Mn2+ + 4H2O
Reduction side is done.

22. Balancing Redox Reactions (Acidic Conditions)
NOW let’s balance the oxidation.
I-   -->     I2

23. Balancing Redox Reactions (Acidic Conditions)
I-   -->     I2
Balance the atoms:

24. Balancing Redox Reactions (Acidic Conditions)
I-   -->     I2
Balance the atoms:
2I-   -->     I2

25. Balancing Redox Reactions (Acidic Conditions)
I-   -->     I2
Balance the atoms:
2I-   -->     I2
Balance the imbalance of charge with electrons (-2 vs. 0)

26. Balancing Redox Reactions (Acidic Conditions)
I-   -->     I2
Balance the atoms:
2I-   -->     I2
Balance the imbalance of charge with electrons (-2 vs. 0)
2I-   -->     I2 + 2e-
Step 1 is done.

27. Balancing Redox Reactions (Acidic Conditions)
Given:         
(acidic conditions – you have H+)
MnO4- + I- --> I2 + Mn2+
Step 1 - Half Reactions
5e- + 8H+ + MnO4-  -->   Mn2+ + 4H2O
2I-   -->     I2 + 2e-
Step 2 - Balance the electrons by finding the common multiple and multiply the half reactions accordingly. 

28. Balancing Redox Reactions (Acidic Conditions)
Given:         
(acidic conditions – you have H+)
MnO4- + I- --> I2 + Mn2+
Step 1 - Half Reactions
5e- + 8H+ + MnO4-  -->   Mn2+ + 4H2O
2I-   -->     I2 + 2e-
Step 2 - Balance the electrons by finding the common multiple and multiply the half reactions accordingly. 
What is the Common Multiple here?

29. Balancing Redox Reactions (Acidic Conditions)
Given:         
(acidic conditions – you have H+)
MnO4- + I- --> I2 + Mn2+
Step 1 - Half Reactions
5e- + 8H+ + MnO4-  -->   Mn2+ + 4H2O
2I-   -->     I2 + 2e-
Step 2 - Balance the electrons by finding the common multiple and multiply the half reactions accordingly. 
What is the Common Multiple here?
Common Multiple here is 10.

30. Balancing Redox Reactions (Acidic Conditions)
Given:         
(acidic conditions – you have H+)
MnO4- + I- --> I2 + Mn2+
Step 1 - Half Reactions
5e- + 8H+ + MnO4-  -->   Mn2+ + 4H2O
2I-   -->     I2 + 2e-
Step 2 - Balance the electrons by finding the common multiple and multiply the half reactions accordingly. 
What is the Common Multiple here?
Common Multiple here is 10. (Look at the e-; 5 x 2 = 10)

31. Balancing Redox Reactions (Acidic Conditions)
5e- + 8H+ + MnO4-  -->   Mn2+ + 4H2O
2I-   -->     I2 + 2e-
Step 2 - Balance the electrons by finding the common multiple and multiply the half reactions accordingly. 
What is the Common Multiple here?
Common Multiple here is 10. (Look at the e-; 5 x 2 = 10)
2(5e- + 8H+ + MnO4-  -->   Mn2+ + 4H2O)
5(2I-   -->     I2 + 2e- )

32. Balancing Redox Reactions (Acidic Conditions)
5e- + 8H+ + MnO4-  -->   Mn2+ + 4H2O
2I-   -->     I2 + 2e-
Step 2 - Balance the electrons by finding the common multiple and multiply the half reactions accordingly. 
What is the Common Multiple here?
Common Multiple here is 10. (Look at the e-; 5 x 2 = 10)
2(5e- + 8H+ + MnO4-  -->   Mn2+ + 4H2O)
5(2I-   -->     I2 + 2e- )
Step 3 Check electrons, atoms and charge.

33. Balancing Redox Reactions (Acidic Conditions)
5e- + 8H+ + MnO4-  -->   Mn2+ + 4H2O
2I-   -->     I2 + 2e-
Step 2 - Balance the electrons by finding the common multiple and multiply the half reactions accordingly. 
What is the Common Multiple here?
Common Multiple here is 10. (Look at the e-; 5 x 2 = 10)
2(5e- + 8H+ + MnO4-  -->   Mn2+ + 4H2O)
5(2I-   -->     I2 + 2e- )
Step 3 Check electrons, atoms and charge.
10e- + 16H+ + 2MnO4-  + 10I--->5I2 + 2Mn2+ + 8H2O + 10e- 

34. Balancing Redox Reactions (Acidic Conditions)
Step 3 Check electrons, atoms and charge.
10e- + 16H+ + 2MnO4-  + 10I--->5I2 + 2Mn2+ + 8H2O + 10e- 

35. Balancing Redox Reactions (Acidic Conditions)
Step 3 Check electrons, atoms and charge.
10e- + 16H+ + 2MnO4-  + 10I--->5I2 + 2Mn2+ + 8H2O + 10e- 
Get rid of the “extra stuff”:

36. Balancing Redox Reactions (Acidic Conditions)
Step 3 Check electrons, atoms and charge.
10e- + 16H+ + 2MnO4-  + 10I--->5I2 + 2Mn2+ + 8H2O + 10e- 
Get rid of the “extra stuff”:
16H+ + 2MnO4-  + 10I--->5I2 + 2Mn2+ + 8H2O 

37. Balancing Redox Reactions (Acidic Conditions)
Step 3 Check electrons, atoms and charge.
10e- + 16H+ + 2MnO4-  + 10I--->5I2 + 2Mn2+ + 8H2O + 10e- 
Get rid of the “extra stuff”:
16H+ + 2MnO4-  + 10I--->5I2 + 2Mn2+ + 8H2O 
All done!

38. Balancing Redox Reactions (Basic Conditions)
Given:         
(basic conditions – you have OH-)
Cr(OH)3 + ClO > CrO42- + Cl-

39. Balancing Redox Reactions (Basic Conditions)
Given:         
(basic conditions – you have OH-)
Cr(OH)3 + ClO > CrO42- + Cl-
Step 1 - Half Reactions

40. Balancing Redox Reactions (Basic Conditions)
Given:         
(basic conditions – you have OH-)
Cr(OH)3 + ClO > CrO42- + Cl-
Step 1 - Half Reactions
ClO3- --> Cl- 
Cr(OH)3 --> CrO42-
Let’s balance the reduction one first.

41. Balancing Redox Reactions (Basic Conditions)
ClO3- --> Cl- 
For every Oxygen add a water on the other side:

42. Balancing Redox Reactions (Basic Conditions)
ClO3- --> Cl- 
For every Oxygen add a water on the other side:
ClO3- --> Cl- + 3H2O 

43. Balancing Redox Reactions (Basic Conditions)
ClO3- --> Cl- 
For every Oxygen add a water on the other side:
ClO3- --> Cl- + 3H2O 
For every hydrogen add a H+ to the other side:

44. Balancing Redox Reactions (Basic Conditions)
ClO3- --> Cl- 
For every Oxygen add a water on the other side:
ClO3- --> Cl- + 3H2O 
For every hydrogen add a H+ to the other side:
6H+ + ClO3- --> Cl- + 3H2O

45. Balancing Redox Reactions (Basic Conditions)
ClO3- --> Cl- 
For every Oxygen add a water on the other side:
ClO3- --> Cl- + 3H2O 
For every hydrogen add a H+ to the other side:
6H+ + ClO3- --> Cl- + 3H2O
HERE’S WHERE IT GETS TRICKY!

46. Balancing Redox Reactions (Basic Conditions)
ClO3- --> Cl- 
For every Oxygen add a water on the other side:
ClO3- --> Cl- + 3H2O 
For every hydrogen add a H+ to the other side:
6H+ + ClO3- --> Cl- + 3H2O
HERE’S WHERE IT GETS TRICKY!
Each H+  will react with an  OH- on both sides:

47. Balancing Redox Reactions (Basic Conditions)
ClO3- --> Cl- 
For every Oxygen add a water on the other side:
ClO3- --> Cl- + 3H2O 
For every hydrogen add a H+ to the other side:
6H+ + ClO3- --> Cl- + 3H2O
HERE’S WHERE IT GETS TRICKY!
Each H+  will react with an  OH- on both sides:
6 OH-  + 6H+ + ClO3- --> Cl- + 3H2O + 6 OH- 

48. Balancing Redox Reactions (Basic Conditions)
6 OH-  + 6H+ + ClO3- --> Cl- + 3H2O + 6 OH-  
H+ and OH-  make water:

49. Balancing Redox Reactions (Basic Conditions)
6 OH-  + 6H+ + ClO3- --> Cl- + 3H2O + 6 OH-  
H+ and OH-  make water:
6H2O + ClO3- --> Cl- + 3H2O + 6 OH-

50. Balancing Redox Reactions (Basic Conditions)
6 OH-  + 6H+ + ClO3- --> Cl- + 3H2O + 6 OH-  
H+ and OH-  make water:
6H2O + ClO3- --> Cl- + 3H2O + 6 OH-
cancel the waters

51. Balancing Redox Reactions (Basic Conditions)
6 OH-  + 6H+ + ClO3- --> Cl- + 3H2O + 6 OH-  
H+ and OH-  make water:
6H2O + ClO3- --> Cl- + 3H2O + 6 OH-
cancel the waters
3H2O + ClO3- --> Cl- + 6 OH-

52. Balancing Redox Reactions (Basic Conditions)
6 OH-  + 6H+ + ClO3- --> Cl- + 3H2O + 6 OH-  
H+ and OH-  make water:
6H2O + ClO3- --> Cl- + 3H2O + 6 OH-
cancel the waters
3H2O + ClO3- --> Cl- + 6 OH-
Balance the imbalance of charge with electrons (-1 vs. -7)

53. Balancing Redox Reactions (Basic Conditions)
6 OH-  + 6H+ + ClO3- --> Cl- + 3H2O + 6 OH-  
H+ and OH-  make water:
6H2O + ClO3- --> Cl- + 3H2O + 6 OH-
cancel the waters
3H2O + ClO3- --> Cl- + 6 OH-
Balance the imbalance of charge with electrons (-1 vs. -7)
6e- + 3H2O + ClO3- --> Cl-  + 6 OH-

54. Balancing Redox Reactions (Basic Conditions)
Now for the oxidation
Cr(OH)3 --> CrO42-

55. Balancing Redox Reactions (Basic Conditions)
Cr(OH)3 --> CrO42-
For every Oxygen add a water on the other side:

56. Balancing Redox Reactions (Basic Conditions)
Cr(OH)3 --> CrO42-
For every Oxygen add a water on the other side:
H2O + Cr(OH)3 --> CrO42-

57. Balancing Redox Reactions (Basic Conditions)
Cr(OH)3 --> CrO42-
For every Oxygen add a water on the other side:
H2O + Cr(OH)3 --> CrO42-
For every hydrogen add a H+ to the other side:

58. Balancing Redox Reactions (Basic Conditions)
Cr(OH)3 --> CrO42-
For every Oxygen add a water on the other side:
H2O + Cr(OH)3 --> CrO42- +  
For every hydrogen add a H+ to the other side:
H2O + Cr(OH)3 --> CrO H+

59. Balancing Redox Reactions (Basic Conditions)
Cr(OH)3 --> CrO42-
For every Oxygen add a water on the other side:
H2O + Cr(OH)3 --> CrO42- +  
For every hydrogen add a H+ to the other side:
H2O + Cr(OH)3 --> CrO H+
REMEMBER: HERE’S WHERE IT GETS TRICKY!
Each H+  will react with an  OH- on both sides:

60. Balancing Redox Reactions (Basic Conditions)
Cr(OH)3 --> CrO42-
For every Oxygen add a water on the other side:
H2O + Cr(OH)3 --> CrO42- +  
For every hydrogen add a H+ to the other side:
H2O + Cr(OH)3 --> CrO H+
REMEMBER: HERE’S WHERE IT GETS TRICKY!
Each H+  will react with an  OH- on both sides:
5 OH-  + H2O + Cr(OH)3 --> CrO H+ + 5 OH- 

61. Balancing Redox Reactions (Basic Conditions)
5 OH-  + H2O + Cr(OH)3 --> CrO H+ + 5 OH   
H+ and OH-  make water:

62. Balancing Redox Reactions (Basic Conditions)
5 OH-  + H2O + Cr(OH)3 --> CrO H+ + 5 OH   
H+ and OH-  make water:
5 OH-  + H2O + Cr(OH)3 --> CrO H2O

63. Balancing Redox Reactions (Basic Conditions)
5 OH-  + H2O + Cr(OH)3 --> CrO H+ + 5 OH   
H+ and OH-  make water:
5 OH-  + H2O + Cr(OH)3 --> CrO H2O
cancel the waters

64. Balancing Redox Reactions (Basic Conditions)
5 OH-  + H2O + Cr(OH)3 --> CrO H+ + 5 OH   
H+ and OH-  make water:
5 OH-  + H2O + Cr(OH)3 --> CrO H2O
cancel the waters
5 OH-  + Cr(OH)3 --> CrO H2O

65. Balancing Redox Reactions (Basic Conditions)
5 OH-  + H2O + Cr(OH)3 --> CrO H+ + 5 OH   
H+ and OH-  make water:
5 OH-  + H2O + Cr(OH)3 --> CrO H2O
cancel the waters
5 OH-  + Cr(OH)3 --> CrO H2O
Balance the imbalance of charge with electrons (-2 vs. 0)

66. Balancing Redox Reactions (Basic Conditions)
5 OH-  + H2O + Cr(OH)3 --> CrO H+ + 5 OH   
H+ and OH-  make water:
5 OH-  + H2O + Cr(OH)3 --> CrO H2O
cancel the waters
5 OH-  + Cr(OH)3 --> CrO H2O
Balance the imbalance of charge with electrons (-5 vs. -2)
5 OH-  + Cr(OH)3 --> CrO H2O + 3e-

67. Balancing Redox Reactions (Basic Conditions)
Step 2 - Balance the electrons by finding the common multiple and multiply the half reactions accordingly. 
What is the Common Multiple here?
Common Multiple here is 6. (Look at the e-; 1 x 6 = 6 & 2 X 3=6)
1(6e- + 3H2O + ClO3- --> Cl-  + 6 OH-)
2(5 OH-  + Cr(OH)3 --> CrO H2O + 3e- )

68. Balancing Redox Reactions (Basic Conditions)
Step 3 Check electrons, atoms and charge.
6e- + 3H2O + ClO OH-  + 2Cr(OH)3 -->Cl-  + 6OH- + 2CrO H2O + 6e- 

69. Balancing Redox Reactions (Basic Conditions)
Step 3 Check electrons, atoms and charge.
6e- + 3H2O + ClO OH-  + 2Cr(OH)3 -->Cl-  + 6OH- + 2CrO H2O + 6e- 
Get rid of the “extra stuff”:

70. Balancing Redox Reactions (Basic Conditions)
Step 3 Check electrons, atoms and charge.
6e- + 3H2O + ClO OH-  + 2Cr(OH)3 -->Cl-  + 6OH- + 2CrO H2O + 6e- 
Get rid of the “extra stuff”:
ClO OH-  + 2Cr(OH)3 -->Cl-  + 2CrO H2O 
All done!