1. Write and balance the equation and identify the reaction type for each of the following reactions: Potassium metal reacts with chlorine gas Zn metal reacts with aqueous copper(II) sulfate Aqueous silver nitrate reacts with aqueous potassium chromate Mercury(II) oxide reacts when heated Hexane reacts with oxygen

2. 2 major categories of inorganic chemical reactions: Double ReplacementOxidation - Reduction Precipitation Neutralization Synthesis Decomposition Single Replacement Combustion Precipitation – Driving Force is formation of Precipitate (Use solubility rules to determine) Ex: AgNO 3 + K 2 CrO 4 - Molecular -Complete ionic -Net ionic

5. You try: Write the molecular, complete ionic and net ionic equations for the reaction of aqueous aluminum sulfate with aqueous sodium hydroxide. Neutralization – Arrhenius acid with Arrhenius base (driving force is the formation of H 2 O) Ex: HCl + NaOH -Molecular -Complete ionic -Net ionic Calculate the mass of aluminum hydroxide that could be produced if 30.0 mL of 0.150M NaOH(aq) is combined with 25.0 mL of 0.250M Al 2 (SO 4 ) 3 (aq)

6. If any of the following are produced in an acid-base neutraliztion, decomposition will occur to produce a gas: H2S (it itself is a gas) H 2 SO 3 (aq)  SO 2 (g) + H 2 O(l) H 2 CO 3 (aq)  CO 2 (g) + H 2 O(l) NH 4 + (aq) + OH - (aq)  NH 3 (g) + H 2 O(l) Ex: NaHSO 3 (aq) + HCl(aq) HSO 3 - + H + (aq)  SO 2 (g) + H 2 O(l)

7. Oxidation – Reduction (Redox) The following is the rxn. for the formation of rust: Fe(s)+O 2 (g)  Fe 2 O 3 (s) What is the oxidation number of iron? What is the oxidation number of iron in Fe 2 O 3 ? What happened? Because it involved reacting with O 2, corrosion of a metal was originally called oxidation. It was later found that similar reactions occurred without O 2. Now oxidation simply means that an element has LOST electrons.

8.  Look back at the reaction, what happened to the O 2 ?  When an element GAINS electrons, we say that it has been reduced.  The oxidation number become less/lower.  A way to remember REDOX, is to think of a lion.  LEO goes GER In the following reaction, determine the oxidation state for each element and then determine what is reduced, oxidized; the oxidizing agent and the reducing agent: Zn + CuSO 4 → ZnSO 4 + Cu

9. ¬ Uncombined elements = 0 ­ Monatomic ions have an oxidation state equal to their charge Including ions within compounds e.g. Na = +1 in Na 2 CO 3 ® O = -2 as oxide, O = -1 as peroxide (O 2 -2 ) ¯ H = +1; except in MH x, H = -1 (where M = metal)

10. ° Most electronegative nonmetal in a compound has a negative oxidation state assigned; base it on position on the Periodic Table ± For a neutral compound, the sum of the oxidation states = 0 ² For a polyatomic ion, the sum of the oxidation states = the charge on the ion

11. Assign oxidation states to each element in the following reactions: 1. CH 4 + 2O 2  CO 2 + 2H 2 O 2. Al + 3I 2  2AlI 3 3. 2PbS + 3O 2  2PbO + 2SO 2 Determine which in each is oxidized (Reducing agent) or reduced (Oxidizing agent)

12. ¬ Write the oxidation and reduction half-reactions ­ For Each Half-Reaction Balance all elements except hydrogen and oxygen Balance O using H 2 O Balance H using H + Balance the charge using electrons ® If necessary, multiply the half-reactions to equalize the number of electrons transferred ¯ Add the half-reactions and cancel identical species ° Check to ensure elements and charges are balanced

13. Balance the Following: 1.MnO 4 -1 (aq) + Fe +2 (aq)  Fe +3 (aq) + Mn +2 (aq) 2. Cr 2 O 7 2- (aq) + C 2 H 5 OH(aq)  Cr 3+ (aq) + CO 2 (g)

14. Analysis by titration can also be done using redox reactions that go rapidly to completion in aqueous solution. The method is the same but equivalence refers to moles of electrons transferred rather than H + or OH -. A 1.026g sample of iron-containing ore requires 24.35 mL of 0.0195 M potassium permanganate solution to reach the equivalence point. What is the mass % of iron in the ore? Answer:12.9% Fe N (normality) = # of equivalents per L, so # equivalents = NV. At the equivalence point # equivalents = # equivalents so N1V1 = N2V2.