1. Big Idea 3

2. Types of Reactions Synthesis 2Mg +O2 2MgO
Decomposition HgO + heat Hg + O2
Acid –base neutralization HCl + NaOH NaCl +H2O
acid base ionic salt water
Oxidation-reduction- rxn that results in a the change of the oxidation states of some of the participating molecule Cu e Cu(s)
Precipitation – 2 aqueous solutions are mixed and a solid or precipitate is created. This can also be expressed as a net ionic equation once you cross out spectator ions
K2CO3 (aq) + Mg(NO3)2 (aq) KNO3 (aq) + MgCO3 (s)
2 K+ + CO Mg NO K+ + 2NO3- + MgCO3 (s)
CO Mg MgCO3 (s)

3. Ionic salts dissolved in water

4. DH = H products –H reactants
Enthalpy
DH if positive = endothermic
DH if negative = exothermic

5. For reaction to proceed the reactants need to have enough energy to reach transition state
Transition state is where reactants are part of the activated complex
This is highest point on graph
The amount of energy to reach the activated complex is called activation energy Ea
At this point all reactant bonds are broken but no product bonds formed yet
Exothermic if products are at lower energy than reactants = -DH

6. Catalysts Catalyst speeds up reaction by lowering activation energy
No effect on DH

7. Day 2

8. Oxidation #s- rules Elemental form =0
Ions in equation= charge of ion Na+ = +1
Oxygen in compound -2 (except in peroxide polyatomic ion)
Hydrogen in compound +1 (except if metal hydride (CaH2 H=-1)
Compounds- all elements add up to zero
Polyatomic ion – all elements add up to charge of polyatomic ion
Last resort rule- If you still can’t calculate all the elements from previous rule this rule(ish) you use as last resort.
If in Group 1A oxidation # = +1
If in Group 2A oxidation # = +2
If in Group 7A oxidation # = -1

9. Figure out what is oxidized and what is reduced in equation
What’s the lions name? What does he say?
LEO
GER +2 -2
X =0
2Ca + O2 2CaO
Ca  Ca +2
Gets more + (lost e-) so oxidized
O  O -2
Gets more - (gain e-) so reduced

10. Voltaic cells
Definition of electricity: form of energy that is made by the movement and interaction of electrons
1. The RED CAT GETS FAT!
An Ox
Reduction occurs at Cathode
Oxidation occurs at Anode
2. Electrons always flow from the anode to the cathode!

11. A student sets up a voltaic cell.
One electrode is Cu
One electrode is Zn
Which metal will be reduced?
Which metal on the reduction potential table has a more
+ EO ?
Cu2+ (aq) + 2e-  Cu(s) reduced
Zn (s)  Zn2+(aq) + 2e- oxidized

12. Half Reactions Voltaic Cell Zn electrode gets smaller
Flow of e-  Cu
Cu electrode Zn
Zn electrode
Zn  Zn2+ + 2e-
2e-
Cu e- → Cu
Lose e-
(Table Reduction potential)
Gains e-
(Table Reduction potential)
Changes to
Zn2+ aqueous
Oxidized
Reduced
Cu+2
Anode (-)
Zn2+
Cu 2+
Changes to Cu solid
Cathode(+)
Zn+2
ZnSO4
CuSO4
Zn electrode gets smaller
Cu electrode gets bigger

13. Charge build up!!! Connected this way the reaction starts
Stops because charge builds up.
Charge build up!!! Ag Cu e- e- e-
Zn2+ Zn Zn e- e-
Ag+
Zn2+
Cu+2
Zn2+
Zn2+
Anode
Cathode

14. Voltaic (Galvanic) Cell
Now Electricity can travels in a complete circuit
Salt Bridge allows ions to migrate.
Cl-
Na+
Cl-
Na+
This keeps charge in solution neutral. Zn
Anode - Cu
Cathode + e-
+ soln
Neutral soln
Neutral soln
- soln

15. Which one is the cathode?
Pb2+ + 2e- Pb(s) = -.13V
Al3+ + 3e- Al(s) = -1.66V
E0 = Ecathode + Eanode
cell
Positive Ecell means it produces that much electricity
Postive Ecell is
spontaneous
Pb2+ + 2e- Pb(s) = -.13V cathode
Al(s)  Al3+ + 3e- = +1.66V anode
Negative Ecell means it requires that much electricity to proceed
Negative Ecell is
NONspontaneous
E0 = -.13V V
cell
= +1.53V

16. The half-cell reactions are reversible
The sign of E0 changes when the reaction is reversed
Changing the stoichiometric coefficients of a half-cell reaction does not change the value of E0

17. ELECTROLYTIC CELLS use electricity to drive a
non-spontaneous chemical reaction forward

18. Differences between Electrochemical Cells
Electrolytic Cells are different from Voltaic Cells because …
They need Electricity to force a redox rxn to occur
There is an external power source req’d
They don’t produce electricity
The polarities are reversed
The Anode is positive
The Cathode is negative
Electrolytic cell reactions are also different because they usually take place in one solution/one cell
“A POX on Electrolytic cells”
Anode
Positive
Oxidation
The Cathode is then just
the opposite
Cathode / Negative / Reduction

19. 1. Draw and completely analyze a molten NaBr electrolytic cell.
Electrons flow through the wire from anode to cathode.
Power Source e- e- Pt Pt _
reduction
cathode
2Na+ + 2e- → 2Na(l)
-2.71 v +
oxidation
anode
2Br- → Br2(g)+ 2e-
-1.09 v
Na+
Br-
2Na+ + 2Br- → Br2(g)+ 2Na(s)
E0 = v MTV = v

20. Electrolytic in a solution

21. Electrolysis of Aqueous Salts
In all electrolytic cells the most easily oxidized species is oxidized and the most easily reduced species is reduced.
Need to compare to water!
Water oxidizing
Water reducing

22. 2. Draw and completely analyze a 1.0 M KI electrolytic cell.
Power Source Pt +
Anode
oxidation -
Cathode
Reduction
H2 gas will form at the cathode
I2 solid will form at the anode K+ I-
H2O
O2 +4H+ +4e-  2H V
I2 + 2e-  2I V
2H20 +2e-  H2 + 2OH V
K+ + e  K V
Which is more +?
Which is more -?
H+ and OH-
Remember to write as oxidation
2H20 +2e-  H2 + 2OH V
2I-  I2 + 2e V
2H2O+ 2I- → 2H2+ I2(s) + 2OH- E0 = v

23. Calculate DGo of the reaction:
Zn (s) + CuSO4 (aq)  ZnSO4 (aq) + Cu(s)
DGo = -nFEo
DGo = -(2mol of e-) (96485 Coulomb/mol of e-) (1.10 V)
DGo = -(2mol of e-) (96485 Coulomb/mol of e-) (1.10J/Coulomb)
DGo = J (spontaneous)
From balanced eqn

24. How much Ca will be produced in an electrolytic cell of molten CaCl2 if a current of A is passed through the cell for 1.5 hours?
Anode:
2Cl- (l) Cl2 (g) + 2e-
2 mole e- = 1 mole Ca and I =q/t and F= Coulomb/mol of e-
Cathode:
Ca2+ (l) + 2e Ca (s)
Ca2+ (l) + 2Cl- (l) Ca (s) + Cl2 (g)
0.452C 1s
1.5 hr 3600 s hr
mol Ca =
96,500 C
1 mol e-
2 mol e-
1 mol Ca
= mol Ca
= 0.50 g Ca