1. Oxidation-Reduction
Chapter 20

2. Oxidation states & balancing redox equations [20.1-20.2]
Cu Al →
Cu Al+3

3. Redox reaction: transfer of electrons
Reduction: process of gaining electrons; becoming more negative
LEO the lion
Oxidation: process of losing electrons; becoming more positive
goes GER

4. Oxidation and Reduction
Reactant reduced is the oxidizing agent.
H+ oxidizes Zn by taking electrons from it. (e- acceptor)
Reactant oxidized is the reducing agent.
Zn reduces H+ by giving it electrons. (e- donor)

5. Oxidation Numbers
In order to keep track of what loses electrons and what gains them, we assign oxidation numbers.

6. Oxidation numbers = charge of an atom Ca+2 F-1 Mn+7-2
-2 x 4 = -8
K2SO4
CO3-2
___ + 3(-2) = -2
Fe3O4
3(__) + 4(-2) = 0
Rules:
Elements = 0
Hydrogen +1; except in metal hydrides (NaH), then it is Placement is the key!
Oxygen = -2; except in peroxides (H2O2; O = -1)
In a compound, oxidation numbers must balance out to net charge of zero.
2(+1) + ___ + (-8) = 0

7. BALANCING SIMPLE REDOX REACTIONS
Cu+2 + Al → Cu + Al+3
1. Cu+2 → Cu Al → Al+3
2. no changes
e- + Cu+2 → Cu
Al → Al+3 + 3e-
4. (2e- + Cu+2 → Cu)3
(Al → Al+3 + 3e-)2
5. 6e- + 3Cu+2 → 3Cu
2Al → 2Al+3 + 6e-
2Al + 3Cu+2 → 3Cu + 2Al+3
2(0) + 3(+2) = 3(0) + 2(+3)
Half reaction method:
Write ½ reactions.
Balance atoms, if necessary (except H & O)
Balance charges by adding electrons.
Multiply (if needed) to cancel electrons.
Check work: charges & atoms
BALANCING SIMPLE REDOX REACTIONS

8. MnO4−(aq) + C2O42−(aq)  Mn2+(aq) + CO2(aq)
Half-Reaction Method
Consider the reaction between MnO4− and C2O42− :
MnO4−(aq) + C2O42−(aq)  Mn2+(aq) + CO2(aq)

9. MnO4− + C2O42-  Mn2+ + CO2 Half-Reaction Method
First, we assign oxidation numbers.
MnO4− + C2O42-  Mn2+ + CO2 +7 +3 +4 +2
Since the manganese goes from +7 to +2, it is reduced.
Since the carbon goes from +3 to +4, it is oxidized.

10. BALANCING REDOX REACTIONS IN ACIDIC SOLUTIONS
MnO4−(aq) + C2O42−(aq)  Mn2+(aq) + CO2(aq)
1. C2O42−  CO2
MnO4−  Mn2+
2. C2O42−  2 CO2
3. MnO4−  Mn H2O
4. 8 H+ + MnO4−  Mn H2O
5. C2O42−  2 CO2 + 2 e−
5 e− + 8 H+ + MnO4−  Mn2+ +4 H2O
6. 5 C2O42−  10 CO e−
10 e− + 16 H+ + 2 MnO4−  2 Mn H2O
16 H+ + 2 MnO4− + 5 C2O42− 
2 Mn H2O + 10 CO2
Write ½ reactions
Balance all atoms except H & O, if needed.
Balance oxygen atoms by adding H2O to one side since aqueous.
Balance hydrogen atoms by adding H+
Balance electric charge by adding electrons
Make electrons equal, to cancel
Check atoms & charge.
BALANCING REDOX REACTIONS IN ACIDIC SOLUTIONS

11. Balancing redox reactions in basic solutions
Zn + VO3-1  VO Zn+2
Zn  Zn+2
VO3-1  VO+2
VO3-1  VO+2 + 2H2O
4 H+ + VO3-1  VO+2 + 2H2O
2 (e- + 4 H+ + VO3-1  VO+2 + 2H2O)
Zn  Zn e-
6. 2e- + 8 H+ + 2VO3-1 2VO+2 +4H2O
Zn + 2 VO H+  2VO+2 + 4H2O + Zn+2
OH OH-
8 H2O
Zn + 2 VO H2O  2VO+2 + 4H2O + Zn OH-
Zn + 2 VO H2O  2VO+2 + Zn OH-
Write ½ reactions
Balance all atoms except H & O, if needed.
Balance oxygen atoms by adding H2O to one side since aqueous.
Balance hydrogen atoms by adding H+
Balance electric charge by adding electrons
Make electrons equal, to cancel
Check atoms & charge.
Now add OH- to both sides to eliminate H+
Balancing redox reactions in basic solutions