Big Idea 3

Types of Reactions Synthesis 2Mg +O2 2MgO Decomposition 2HgO + heat 2Hg + 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 2+ + 2e- 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) 2KNO3 (aq) + MgCO3 (s) 2 K+ + CO3 2- + Mg 2+ + 2NO3- 2 K+ + 2NO3- + MgCO3 (s) CO3 2- + Mg 2+ MgCO3 (s)

Ionic salts dissolved in water

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

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

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

Day 2

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

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 + -2 =0 2Ca + O2 2CaO Ca0  Ca +2 Gets more + (lost e-) so oxidized O20  O -2 Gets more - (gain e-) so reduced

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!

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

Half Reactions Voltaic Cell Zn electrode gets smaller Flow of e-  Cu Cu electrode Zn Zn electrode Zn  Zn2+ + 2e- 2e- Cu 2+ + 2e- → 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

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

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

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 + +1.66V cell = +1.53V

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

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

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

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 = -3.80 v MTV = +3.80 v

Electrolytic in a solution

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

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-  2H20 +1.23V I2 + 2e-  2I- +0.54V 2H20 +2e-  H2 + 2OH- -0.83V K+ + e-  K -2.92V Which is more +? Which is more -? H+ and OH- Remember to write as oxidation 2H20 +2e-  H2 + 2OH- -0.83V 2I-  I2 + 2e- -0.54V 2H2O+ 2I- → 2H2+ I2(s) + 2OH- E0 = -1.37 v

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 = -21200 J (spontaneous) From balanced eqn

How much Ca will be produced in an electrolytic cell of molten CaCl2 if a current of 0.452 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= 96485 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 = 0.0126 mol Ca = 0.50 g Ca