1. Single & Double Displacement Reactions

2. The Activity Series K Na Li Ca Mg Al Zn Fe Ni Sn Pb H Cu Hg Ag Au
The Activity Series is used to predict the outcome of single displacement reactions
The higher a metal is on the Activity Series, the more likely it will react with those below it
A metal will not displace a metal higher on the activity series than itself

3. The Activity Series Cont.
The metals near the top are more reactive because their valence electrons are more easily removed
H is the only nonmetal in the series, H2 may be displaced from acids or can be given off when a metal reacts with H2O producing H2 + metal hydroxide

4. The Activity Series PracticeK Na Li Ca Mg Al Zn Fe Ni Sn Pb H Cu Hg Ag Au
The Activity Series Practice
Which of these will react?
K + NaCl  ???
K + NaCl  Na + KCl
Al + AgNO3  ???
Al + AgNO3  Ag + Al(NO3)3
Cu + CaCO3  ???
Cu + CaCO3  No Rxn
Pb + ZnO  ???
Pb + ZnO  No Rxn

5. The Solubility Table
The Solubility Table is used to predict the outcome of double replacement reactions

6. The Solubility Table Cont.
If both products of a DR reaction are soluble then no reaction occurs
If 1+ product of a DR reaction is solid then a reaction occurs

7. The Solubility Table Practice
Which of these will react?
NaCl (aq) + Ba3(PO4)2 (s)  ???
NaCl (aq) + Ba3(PO4)2 (s)  No Rxn
FeCO3 (s) + CaCl2 (aq) ???
FeCO3 (s) + CaCl2 (aq)  FeCl2 (aq) + CaCO3 (s)
LiNO3 (aq) + MgBr2 (aq)  ???
LiNO3 (aq) + MgBr2 (aq)  No Rxn
Zn(OH)2 (s) + NiSO4 (aq)  ???
Zn(OH)2 (s) + NiSO4 (aq)  ZnSO4 (aq) + Ni(OH)2 (s)

8. Oxidation-Reduction Reactions

9. OXIDATION-REDUCTION REACTIONS
Oxidation and Reduction reactions always take place simultaneously resulting from the gain or loss of electrons.
Loss of electrons – oxidation (Increase in Oxidation Number)
Na > Na+1 + e-1
Gain of electrons - reduction (Decrease in Oxidation Number)
Cl2 + 2 e > 2 Cl-1
LEO says GER

10. Which substance is oxidized?
Oxidation occurs when a molecule
does any of the following:   
Loses electrons
Loses hydrogen
Gains oxygen
Example: 2Fe2O3 + 3C  4Fe + 3CO2
Which substance is oxidized?
If a molecule undergoes oxidation, it has been oxidized and it is the reducing agent .
Metals are usually good reducing agents the higher it is on the activity series the better reducing agent it is.

11. Which substance is reduced?
Reduction occurs when a molecule does any of the
following:
Gains electrons
Gains hydrogen
Loses oxygen
Example: 2Fe2O3 + 3C  4Fe + 3CO2
Which substance is reduced?
If a molecule undergoes reduction, it has been reduced and it is the oxidizing agent

12. Zinc is being oxidized while the copper is being reduced. Why?

13. Oxidation Numbers number assigned to an atom to indicate its degree of oxidation (+) or reduction (-)
Rules for Assigning Oxidation Numbers
Ions are given the same oxidation number as their charge
Oxygen is usually -2
Hydrogen is usually +1
Elemental or diatomic molecules are 0
For compounds the sum of oxidation numbers must equal 0

14. Steps for Balancing a Redox Reaction: Half Reaction Method
S + HNO3  SO2 + NO + H2O
Write the unbalanced equation in ionic form
Write separate half reactions for the oxidation and reduction processes
Balance the atoms in the half reactions
Add electrons to one side of each half reaction to balance charge
Multiply each half reaction by an appropriate number to make the electrons equal on both sides.
S + H+ + NO3-  SO2 + NO + H2O
Ox: S  SO2 Red: NO3-  NO
Ox: S + 2H2O SO2 + 4H+ Red: 4H+ + NO3-  NO + 2H2O
Ox: S + 2H2O SO2 + 4H++ 4e- Red: 4H+ + NO3- + 3e- NO + 2H2O
Ox: 3S + 6H2O 3SO2 + 12H++ 12e- Red: 16H+ + 4NO e- 4NO + 8H2O

15. Balancing Redox continued
6. Add balanced ½ equations 7. Add “Spectator Ions” and balance equation
Ox: 3S + 6H2O 3SO2 + 12H++ 12e-
Red: 16H+ + 4NO e- 4NO + 8H2O
Total: 3S + 4H+ + 4NO3-  3SO2 + 4NO + 2H2O
3S + 4HNO3 3SO2 + 4NO + 2H2O

16. Practice Half-Reactions
Don’t forget to determine the charge of each species first!
4 Li + O2  2 Li2O
Oxidation Half-Reaction:
Reduction Half-Reaction:
Zn + Na2SO4  ZnSO4 + 2 Na

17. Percent Error
𝐸𝑟𝑟𝑜𝑟=𝑒𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡𝑎𝑙 𝑣𝑎𝑙𝑢𝑒 −𝑎𝑐𝑐𝑒𝑝𝑡𝑒𝑑 𝑣𝑎𝑙𝑢𝑒 𝑃𝑒𝑟𝑐𝑒𝑛𝑡 𝐸𝑟𝑟𝑜𝑟= 𝑒𝑟𝑟𝑜𝑟 𝑎𝑐𝑐𝑒𝑝𝑡𝑒𝑑 𝑣𝑎𝑙𝑢𝑒 ×100%