CHM1045 Lecture 13Outline of Last LectureI. Properties of AcidsII. Properties of BasesIII. AcidsIV. Neutralization ReactionV. Neutralization Reaction Involving a Weak ElectrolyteVI. Neutralization Reaction Producing a GasVII. Oxidation-Reduction ReactionsOutline of Current LectureI. Oxidation numberII. Types of Oxidation-Reduction ReactionsIII. The Activity Series for HalogensIV. Chemistry in Action: Breath AnalyzerV. Solution StoichiometryVI. Gravimetric AnalysisVII. TitrationsCurrent LectureOxidation number1. Free elements (uncombined state) have an oxidation number of zero.Na, Be, K, Pb, H2, O2, P4 = 02. In monatomic ions, the oxidation number is equal to the charge on the ion.Li+, Li = +1; Fe3+, Fe = +3; O2-, O = -23. The oxidation number of oxygen is usually –2. In H2O2 and O22- it is –1. 4. The oxidation number of hydrogen is +1 except when it is bonded to metals in binary compounds. In these cases, its oxidation number is –1.5. Group IA metals are +1, IIA metals are +2 and fluorine is always –1.6. The sum of the oxidation numbers of all the atoms in a molecule or ion is equal to the charge on the molecule or ion.The charge the atom would have in a molecule (or an ionic compound) if electrons were completely transferred7. Oxidation numbers do not have to be integers. The oxidation number of oxygen in the superoxide ion, O2-, is –½.Example 1 : Assign oxidation numbers to all the elements in the following compounds and ion:(a) Li2O(b) HNO3 (c)Answer: a.Lithium has an oxidation of +1 (Li+) and oxygens oxidation number is -2 (O2-)b. H has an oxidation number +1 . The other group (nitrate ion) must have a net oxidationof -1. Oxygen has an oxidation number of -2 and of we use x to represent the oxidation# of nitrogen, then the nitrate ion can be written: (N(x)O(2-)3)-x + 3(-2) = -1x= + 5c.The sum of oxidation numbers in the dichromate ion Cr2O2-7 must be -2. We know that the oxidation number of O is -2 so all the remains to determine the oxidation number of Cr which we call y. The dichromate ion is written: (Cr(y)zO7(2-))2-2(y) + 7(-2) = -2y=+6Types of Oxidation-Reduction ReactionsCombination ReactionA + B C2Al + 3Br2 2AlBr3Decomposition ReactionC A + B2KClO3 2KCl + 3O2Combustion ReactionA + O2 BS + O2 SO22Mg + O2 2MgODisplacement ReactionA + BC AC + BSr + 2H2O Sr(OH)2 + H2Hydrogen DisplacementTiCl4 + 2Mg Ti + 2MgCl2Metal DisplacementCl2 + 2KBr 2KCl + Br2Halogen DisplacementThe Activity Series for HalogensF2 > Cl2 > Br2 > I2Halogen Displacement ReactionCl2 + 2KBr 2KCl + Br2I2 + 2KBr 2KI + Br2Types of Oxidation-Reduction ReactionsDisproportionation ReactionThe same element is simultaneously oxidized and reduced.Example: ReducedCl2 + 2OH- ClO- + Cl- + H2O OxidizedExample 2: Classify the following redox reactions and indicate changes in the oxidation numbers of the elements:a)b)c)d)Answer:a) This is a decomposition reaction because one reactant is converted to 2 different products. The oxidation # of N changes from +1 to 0 , while that of O changes from -2 to 0.b) Combination reaction. Oxidation # of Li changes from 0 to +1 while that of N changes from 0 to -3c) Metal displacement reaction. Ni metals replaces the Pb2+ ion. The oxidation # of Ni increases from 0 to +2 while that of Pb decreases from +2 to 0d) Oxidation number of N is +4 in NO2 and it is +3 in HNO2 and +5 in HNO3. Because the oxidation # of the same element both decreases and increases, this is a disproportionation reaction.Chemistry in Action: Breath Analyzer3CH3CH2OH + 2K2Cr2O7 + 8H2SO4 3CH3COOH + 2Cr2(SO4)3 + 2K2SO4 + 11H2OSolution Stoichiometry- The concentration of a solution is the amount of solute present in a given quantity of solvent or solution.- M = Molarity = moles of soluteLiters of solutionExample 3:How many grams of potassium dichromate (K2Cr2O7) are required to prepare a 250-mL solution whose concentration is 2.16 M?Answer: moles of solute = molarity + L solnMoles of K2Cr2O7 = 2.16 mol K2Cr2O7 * .250 L soln1 L soln= .540 mol K2Cr2O7The molar mass of K2Cr2O7 is 294.2 gGrams of K2Cr2O7 needed= .540 mol K2Cr2O7 * 294.2 K2Cr2O7 1 mol K2Cr2O7= 159 g K2Cr2O7Example 4:In a biochemical assay, a chemist needs to add 3.81 g of glucose to a reaction mixture. Calculate the volume in milliliters of a 2.53 M glucose solution she should use for the addition. Answer: 3.81 g C6H12O6 * 1 mol C6H12O6 = 2.114 * 10^-2 mol C6H12O6180.2 g C6H12O6V=n 2.114 * 10^-2 mol C6H12O6 * 1000mL solnM 2.53 mol C6H12O6/ L soln 1 L soln = 8.36 mL soln- Dilution is the procedure for preparing a less concentrated solution from a more concentrated solution.Example 5: Describe how you would prepare 5.00 × 102 mL of a 1.75 M H2SO4 solution, starting withan 8.61 M stock solution of H2SO4.(8.61 M) (V1) = (1.75 M) (5.00 *102 mL)V1= (1.75 M) (5.00 *10 2 mL)8.61 M= 102 mLGravimetric Analysis1. Dissolve unknown substance in water2. React unknown with known substance to form a precipitate3. Filter and dry precipitate4. Weigh precipitate5. Use chemical formula and mass of precipitate to determine amount of unknown ionExample 6:A 0.5662-g sample of an ionic compound containing chloride ions and an unknown metal is dissolved in water and treated with an excess of AgNO3. If 1.0882 g of AgCl precipitate forms, what is the percent by mass of Cl in the original compound?% Cl = 35.45 g Cl * 100% 143.4 g AgCl= 24.72% mass of Cl = .2472 * 1.0882 g= .2690 g% Cl = .2690 g * 100% .5662 = 47.51%Titrations- In a titration, a solution of accurately known concentration is added gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete.- Equivalence point – the point at which the reaction is complete- Indicator – substance that changes color at (or near) the equivalence point- Titrations can be used in the analysis of Acid-base reactionsH2SO4 + 2NaOH 2H2O + Na2SO4Redox reactions5Fe2+ + MnO4- + 8H+ Mn2+ + 5Fe3+ + 4H2OExample 7A 16.42-mL volume of 0.1327 M KMnO4 solution is needed to oxidize 25.00 mL of a FeSO4 solution in an acidic medium. What is the concentration of the FeSO4 solution in molarity? The net ionic equation isAnswer: moles of KMnO4= .1327 mol KMnO4 * 16.42 mL
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