1. Chapter 4B: Balancing Redox Reactions
LEO SAYS GER
West Valley High School
AP Chemistry
Mr. Mata

2. Oxidation and Reduction (Redox)
Electrons are transferred
Spontaneous redox rxns can transfer energy
Electrons (electricity)
Heat
Non-spontaneous redox rxns can be made to happen with electricity

3. Oxidation Reduction Reactions (Redox)
Each sodium atom loses one electron:
Each chlorine atom gains one electron:

4. LEO says GER :
Lose Electrons = Oxidation
Sodium is oxidized
Gain Electrons = Reduction
Chlorine is reduced

5. Not All Reactions are Redox Reactions
Reactions in which there has been no change in oxidation number are not redox rxns.
Examples:

6. Rules for Assigning Oxidation Numbers Rules 1 & 2
The oxidation number of any uncombined element is zero
2. The oxidation number of a monatomic ion
equals its charge

7. Rules for Assigning Oxidation Numbers Rules 3 & 4
3. The oxidation number of oxygen in
compounds is -2
4. The oxidation number of hydrogen in compounds is +1

8. Rules for Assigning Oxidation Number Rule 5
5. The sum of the oxidation numbers in the formula of a compound is 0
2(+1) + (-2) = 0
H O
(+2) + 2(-2) + 2(+1) = 0
Ca O H

9. Rules for Assigning Oxidation Numbers Rule 6
6. The sum of the oxidation numbers in the
formula of a polyatomic ion is equal to
its charge
X + 4(-2) = -2
S O
X + 3(-2) = -1
N O
 X = +5
 X = +6

10. The Oxidation Number Rules - SIMPLIFIED
1. The sum of the oxidation numbers in
ANYTHING is equal to its charge
2. Hydrogen in compounds is +1
3. Oxygen in compounds is -2

11. Reducing Agents and Oxidizing Agents
The substance reduced is the oxidizing agent
The substance oxidized is the reducing agent
Sodium is oxidized – it is the reducing agent
Chlorine is reduced – it is the oxidizing agent

12. Trends in Oxidation and Reduction
Active metals:
Lose electrons easily
Are easily oxidized
Are strong reducing agents
Active nonmetals:
Gain electrons easily
Are easily reduced
Are strong oxidizing agents

13. Redox Reaction Prediction #1
Important Oxidizers
Formed in reaction
MnO4- (acid solution)
MnO4- (basic solution)
MnO2 (acid solution)
Cr2O72- (acid)
CrO42-
HNO3, concentrated
HNO3, dilute
H2SO4, hot conc
Metallic Ions
Free Halogens
HClO4
Na2O2
H2O2
Mn(II)
MnO2
Cr(III)
NO2 NO
SO2
Metallous Ions
Halide ions
Cl-
OH- O2

14. Redox Reaction Prediction #2
Important Reducers
Formed in reaction
Halide Ions
Free Metals
Metalous Ions
Nitrite Ions
Sulfite Ions
Free Halogens (dil, basic sol)
Free Halogens (conc, basic sol)
C2O42-
Halogens
Metal Ions
Metallic ions
Nitrate Ions
SO42-
Hypohalite ions
Halate ions
CO2

15. Oxidation Reduction
Oxidation means an increase in oxidation state - lose electrons.
Reduction means a decrease in oxidation state - gain electrons.
The substance that is oxidized is called the reducing agent.
The substance that is reduced is called the oxidizing agent.

16. Agents Oxidizing agent gets reduced. Gains electrons.
More negative oxidation state.
Reducing agent gets oxidized.
Loses electrons.
More positive oxidation state.

17. Half-Reactions
All redox reactions can be thought of as happening in two halves.
One produces e-’s = Oxidation half.
One requires e-’s = Reduction half.

18. Balancing Redox Reactions
In aqueous solutions the key is the number of electrons produced must be the same as those required.
For reactions in acidic solution an 8 step procedure.
Write separate half reactions
For each half reaction balance all reactants except H and O
Balance O using H2O

19. Acidic Solution Balance H using H+ Balance charge using e-
Multiply equations to make electrons equal
Add equations and cancel identical species
Check that charges and elements are balanced.

20. Balancing Redox Reactions
__ Al + __ Cu2+ --> __ Cu + __ Al3+
Start by writing half reactions (Oxidation and reduction)
(Electrons go on the more positive side)
Oxidation: Al --> Al e-
Reduction: 2e- + Cu2+ --> Cu

21. Balancing Redox Reactions
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)

22. Balancing Redox Reactions
Oxidation: 2 x (Al --> Al e-)
Reduction: 3 x ( 2e- + Cu2+ --> Cu)
_________________________________
Recombine: 6e-+2 Al + 3Cu2+-->2Al 3++ 3Cu + 6e-
The electrons must cancel.
2 Al + 3 Cu2+--> 2 Al Cu
Atoms and charges must be conserved.

23. Balancing Redox Reactions (Acidic Conditions)
MnO4- + I- --> I2 + Mn2+ (acidic)
Step 1 Half Reactions:
MnO4- --> Mn2+
I- --> I2

24. Lets balance the reduction one first:
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

25. Now for the oxidation I- --> I2 Balance the atoms: 2I- --> I2
Balance the imbalance of charge with electrons (-2 vs. 0):
2I- --> I2 + 2e-

26. Balancing Redox Reactions
Balance the electrons by finding the common multiple and multiply the half reactions accordingly.
Common Multiple here is 10.

27. Balancing Redox Reactions
2( 5e- + 8H+ + MnO4- --> Mn2+ + 4H2O )
5( 2I- --> I2 + 2e- )
Step 3 Check electrons, atoms and charge. Clean it up.
10e- + 16H+ + 2MnO I--->5I2 + 2Mn2+ + 8H2O + 10e-
16H+ +2MnO4- +10I-->5I2 +2Mn2+ + 8H2O

28. Basic Solution
Do everything you would with acid, but add one more step.
Add enough OH- to both sides to neutralize the H+
Makes water

29. Basic Solution Cr(OH)3 +ClO3- ->CrO42- + Cl- (basic)
Step 1 Half Reactions:
Lets balance the reduction one first:
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

30. Basic Solution Each H+ will react with an OH- on both sides:
6 OH- + 6H++ClO3- -> Cl- +3H2O + 6OH-
H+ and OH- make water:
6H2O + ClO3- --> Cl- + 3H2O + 6OH-
cancel the waters:
3H2O + ClO3- --> Cl- + 6OH-
Balance the imbalance of charge with e-’s (-1 vs. -7):
6e- + 3H2O + ClO3- --> Cl- + 6OH-

31. Basic Solution Now for the oxidation: Cr(OH)3 --> CrO42-
for every O, add a H2O on other side:
H2O + Cr(OH)3 --> CrO42-
For every H, add a H+ to the other side:
H2O + Cr(OH)3 --> CrO H+
Each H+ will react with OH- on both sides:
5 OH-+H2O+Cr(OH)3 ->CrO42-+5H++5OH-

32. Basic Solution 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-

33. Basic Solution
Balance the electrons by finding the common multiple and multiply the half reactions accordingly.
Common Multiple here is 6:
1(6e- + 3H2O + ClO3- -> Cl- + 6OH-)
2(5 OH-+Cr(OH)3 -> CrO42-+4H2O + 3e-)

34. Basic Solution
Step 3: Check electrons, atoms and charge then clean it up:
6e- + 3H2O + ClO OH- + 2Cr(OH)3 -->Cl- + 6OH- + 2CrO H2O + 6e-
ClO OH- + 2Cr(OH)3 -->Cl- + 2CrO H2O

35. Practice Redox Reaction #1
Mn BiO3 - => MnO Bi 3+
2 (4 H2O + Mn 2+ => MnO H e)
5 ( 2 e H BiO3 - => Bi H2O)
14 H Mn BiO3 - => 2 MnO Bi H2O

36. Practice Redox Reaction #2
ClO Cl - => Cl2 + ClO2
2 (e H ClO3 - => ClO2 + H2O)
Cl - => Cl e)
4 H ClO Cl - => 2 ClO H2O + Cl2

37. Practice Redox Reaction #3
P + Cu 2+ => Cu + H2PO4 -
2 (4 H2O + P => H2PO H e)
5 (2 e + Cu 2+ => Cu)
8 H2O P Cu 2+ => 2 H2PO H Cu)

38. Practice Redox Reaction #4
MnO C2O4 2- => MnO2 + CO2
2 (3 e H MnO4 - => MnO H2O)
3 (C2O => 2 CO e)
4 H2O MnO C2O4 2- => 2 MnO OH CO2

39. Practice Redox Reaction #5
ClO2 => ClO2 - + ClO3 - e + ClO2 => ClO2 - H2O + ClO2 => ClO H + + e 2 OH ClO2 => ClO2 - + ClO3 - + H2O

40. Practice Redox Reaction #6
H2O + Zn + NO3 - => Zn(OH) NH3
8 e H NO3 - => NH H2O
4 (4 H2O + Zn => Zn(OH) H e
6 H2O + NO OH Zn => NH Zn(OH)4 2-