CHEM 101 1st Edition Lecture 9 Outline of Last Lecture I. Aqueous Solutions continuedII. Solubility of Ionic CompoundsIII. Acids and BasesOutline of Current Lecture I. Anhydrides of Acids and BasesII. Gas- Forming Reactions of Ionic CompoundsIII. Oxidation-Reduction ReactionsIV. Determining Oxidation NumbersCurrent Lecture- Anhydrides of Acids and Baseso Oxides of metals such as Na2O and CaO are called basic oxides, since they react to bases with water:Na2O(s) + H2O(l) 2 NaOH(aq)CaO(s) + H2O(l) Ca(OH)2(s)(lime) (Slaked lime)o These oxides are also the anhydrides of the corresponding bases.o Application of slaked lime as lime mortar:Ca(OH)2(s) + CO2(g) CaCO3(s)o Calcination of lime stone (to produce lime):CaCO3(s) CaO(s) + CO2(g) (T> 898 °C)- Gas-Forming Reactions of Ionic Compoundso Metal carbonates, sulfides, and sulfites from gases in the presence of an acid. Metal carbonates form CO2:CaCo3(s) + 2 HCl(aq) CaCl2(aq) + H2CO3(aq) H2CO3(aq) H2O(l) + CO2(g)o Overall reaction:CaCO3(s) + 2 HCl(aq) CaCl2(aq) + H2O(l) + CO2(g)o Sulfites form H2S and sulfites form SO2:Na2S(aq) + 2HCl(aq) 2 NaCl(aq) + H2S(g)Na2SO3 (aq) + 2HCl(aq) 2NaCl(aq) + SO2(g) + H2O(l)These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.o Ammonium salts form gases in the presence of a strong acid base:NH4Cl(aq) + NaOH(aq) NaCl(aq) + NH3(g) + H2O(g)- Clicker Question: Sulfuric acid is the product of the reaction of _______ and H2O.A. SO3 (Correct)B. SO2C. H2SD. SO4^(2-)- Oxidation-Reduction Reactionso Common oxidation numbers:o Ex: Fe2^(3+)O3^(2-)(s)+ 3C^(2+)O^(2-) (g) 2Fe(s) + 3C^(4+)O2^(2-) (g) Since there are 3O^(2-) then = -6Which balances the 2 Fe^(3+) then = +6 Since there are 3C^(4+) then = +12Which balances the 6 O2^(2-) = -12 Half Reaction:1. Fe2^(3+)O3(s)(Oxidizing agent) + 6e^(-) 2Fe^(0+) (s) Reduction2. 3 C^(2+)O(g)(Reducing agent) 3 C^(4+)O2 + 6e ^(-) OxidationReduction + Oxidation = Reddox Reactiono Pure copper wire in dilute AgNO3 solution after several hours forms silver crystals. Blue color weeks later is due to Cu^(2+) ions formed in redox reaction. Ex: Aqueous solutuion of AgNO3 + Cu-wireIonic Equation 2Ag^(+)(aq) + 2NO3^(-)(aq) + Cu(s) 2Ag(s) + 2NO3^(-)(aq) + CU^(2+)(s)Half Reaction:1. 2Ag^(+)(aq) +2e^(-) 2Ag(s) Reduction2. Cu(s) Cu^(2+)(s) + 2e^(-) Oxidationreduction + oxidation = reddox reaction = net ionic equationo Oxidation-reduction reactions O2(g) + 2H2(g) 2H2O(l) (reddox reaction)- Determining Oxidation Numberso Oxidation numbers are the actual or apparent charge of the atom in the elemental state or combined in a compound. 1. The atoms of pure elements always have an oxidation bumber of zero:Ex: Mg(s), Hg(l), I2(s), O2(g) 2. If an atom is charged, then the charge is the same as the oxidation number:Ion Oxidation NumberMg^(2+) +2Cl^(-)(aq) -1Sn^(4+)(s) +4Hg2^(2+)(aq) +2/2 = +1 3. In a compound, fluorine always has an oxidation number of -1. 4. Oxygen most often has an oxidation number of -2Ex: H2O, MgO, NaOH- Exceptions: OF2: Oxidation number for O is +2Peroxides (H2O2, Na2O2): -1Superoxides (KO2): -1/2 5. In compounds, Cl, Br, an I have an oxidation number of -1Ex: NaCl, CCl4, HBrExceptions: ClO^(-), ClO2^(-), ClO3^(-), ClO4^(-) (compare with rule 4) 6. In compounds, the oxidation number of H is +1Exceptions: NaH, CaH2, so-called hydrides 7. For neutral compounds, the sum of the oxidation numbers equals zero. For a poly atomic ion, the sum equals the charge- Common Oxidizing AgentsOxidizing Agent Reaction ProductO2 O2^(-), H2OFe, Cl2, Br2, or I2 F^(-), Cl^(-), Br^(-) or I^(-) HNO3 NO (dilute HNO3)NO2 (concentrated HNO3)Cr2O7^(2-) Cr^(3+) (acidic solution)MnO4^(-) Mn^(2+) (acidic solution)- Common Reducing AgentsReducing Agent Reaction
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