1. Electrochemistry

2. Define oxidation and reduction.
Determine oxidation numbers for atoms.
Identify the oxidizing agent, the reducing agent.
Distinguish between redox and non-redox reactions.

3. Burning and corrosion needs oxygen – oxidation.
Oxidation-reduction reactions – (redox)
Chemical changes when electrons are transferred from one reactant to another.
Oxidation - an atom loses one or more electrons.
Reduction - an atom gains one or more electrons.
"LEO says GER” 
Losing Electrons is Oxidation, Gaining Electrons is Reduction

4. Magnesium is oxidized. Oxygen is reduced.
2 Mg (s) + O2 (g) → 2 MgO (s)
Mg – neutral Mg2+ ion
O – neutral O2– ion
Magnesium is oxidized. Oxygen is reduced.
Mg → Mg2+ 2e-
O + 2e- → O2-

5. Written difference between ion and oxidation:
Mg (s) + Cl2 (g) → MgCl2 (s)
Written difference between ion and oxidation:
Chlorine ion – Cl1- ion charge = 1-
 oxidation number = -1
Mg → Mg2+ 2e-
Cl + 1e- → Cl1-
Sometimes these numbers are the same (like above) sometimes they are very different – which is why we write them differently.

6. **You assign them to EACH atom**
Oxidation number represents the charge the atom would have if every bond were ionic.
**You assign them to EACH atom**
1. Assign known numbers first (below).
Then calculate the others.
All uncombined elements (diatomic) – zero.
Monatomic ions in ionic bond - equals ion charge.

7. In compounds:
Alkali metals - always +1.
Earth metals - always +2
Al: +3, F: -1, H: +1*, O: –2*
Neutral compound:
Sum of ox.numbers for each atom must be zero.
Polyatomic ion:
Sum of ox.numbers must be the charge of that ion.
Assign oxidation numbers to each atom in SO2.
S = +4 O = –2

8. Assign ox.numbers for each atom in K2Cr2O7
Step 1: Start with atoms which are known.
O: –2
 K: +1
Step 2: Solve for other atoms.
All O atoms: –2 × 7 = –14

All K atoms: +1 × 2 = +2
The total for the compound must be zero.
All Cr atoms: ?? = 0
Two Cr atoms - (+12) ÷ 2 = +6 each
K = +1 Cr = +6 O = –2.

9. Assign ox. numbers for each atom in Fe(NO3)3
Ionic bond between Fe3+ and NO3–
Fe: +3
Sum of ox.numbers for the compound must be 0.
All O atoms: × 9 = – (-18) + ?? = 0
All N atoms: ÷ 3 =
Fe = +3 N = +5 O = –2.

10. Ox.numbers do not change – no e- transferred – NOT a redox reaction.
Use ox.numbers to determine if reaction is a redox reaction. +4 -2 +1 -2 +1 +4 -2
SO H2O → H2SO3 +4 -4 +2 -2 +2 +4 -6
Ox.numbers do not change – no e- transferred – NOT a redox reaction.

11. Cu(s) + 2 AgNO3(aq) → CuNO3(aq) + 2 Ag(s)
Is the following reaction a redox reaction? +1 +5 -2 +1 +5 -2
Cu(s) + 2 AgNO3(aq) → CuNO3(aq) + 2 Ag(s) +1 +5 -6 +1 +5 -6
• Cu – oxidized (loss of electrons). 

• Ag – reduced (gain of electrons).
Oxidation cannot occur without reduction.

12. Cu(s) + 2 AgNO3(aq) → CuNO3(aq) + 2 Ag(s)
Oxidizing agent - causes the oxidation of another substance.
AgNO3 is the oxidizing agent
Reducing agent - causes the reduction of another substance.
Cu is the reducing agent
Oxidizing agent becomes reduced and the reducing agent becomes oxidized. +1
Cu(s) + 2 AgNO3(aq) → CuNO3(aq) + 2 Ag(s) +1

13. 2 HNO3(aq) + 3 H2S(g) → 2 NO(g) + 3 S(s) + 4 H2O(l)
Identify the substance oxidized, the substance reduced, the oxidizing agent and the reducing agent. +1 +5 -2 +1 -2 +2 -2 +1 -2
2 HNO3(aq) + 3 H2S(g) → 2 NO(g) + 3 S(s) + 4 H2O(l) +1 +5 -6 +2 -2 +2 -2 +2 -2
S – oxidized
N – reduced
H2S – reducing agent
HNO3 – oxidizing agent

14. How many electrons are transferred in the reaction below:
gains 3e- +1 +5 -2 +1 -2 +2 -2 +1 -2
HNO3(aq) + H2S(g) → 2 NO(g) + 3 S(s) + 4 H2O(l) 2 3
loses 2e-
S: (3 atoms) x (2e- lost) = 6 electrons lost
N: (2 atoms) x (3e- gained) = 6 electrons gained
Stoichiometry used to determined total electrons transferred.

15. Strong oxidizing agents: Are very reactive – will take from anything
Oxidizing Agents Reaction Products
O O2–, H2O, CO2
F2, Cl2, Br2, I2 F–, Cl–, Br–, I–
MnO4– Mn2+
Cr2O72– Cr3+
HNO NO, NO2
H2O O2, H2O
Strong reducing agents:
Are very reactive – will give to anything.
Metals, substances that burn easily – H2, CxHy