Presentation on theme: "Figure 4.11 A summary of terminology for oxidation-reduction (redox) reactions. X Y e-e- transfer or shift of electrons X loses electron(s)Y gains electron(s)"— Presentation transcript:
Figure 4.11 A summary of terminology for oxidation-reduction (redox) reactions. X Y e-e- transfer or shift of electrons X loses electron(s)Y gains electron(s) X is oxidizedY is reduced X is the reducing agentY is the oxidizing agent X increases its oxidation number Y decreases its oxidation number
2 Oxidation-Reduction Reactions (electron transfer reactions) 2Mg 2Mg 2+ + 4e - O 2 + 4e - 2O 2- Oxidation half-reaction (lose e - ) Reduction half-reaction (gain e - ) 2Mg + O 2 + 4e - 2Mg 2+ + 2O 2- + 4e - 2Mg + O 2 2MgO
Table 4.3 Rules for Assigning an Oxidation Number (O.N.) 1. For an atom in its elemental form (Na, O 2, Cl 2, etc.): O.N. = 0 2. For a monoatomic ion: O.N. = ion charge 3. The sum of O.N. values for the atoms in a compound equals zero. The sum of O.N. values for the atoms in a polyatomic ion equals the ion’s charge. General rules Rules for specific atoms or periodic table groups 1. For Group 1A(1):O.N. = +1 in all compounds 2. For Group 2A(2):O.N. = +2 in all compounds 3. For hydrogen:O.N. = +1 in combination with nonmetals 4. For fluorine:O.N. = -1 in combination with metals and boron 6. For Group 7A(17):O.N. = -1 in combination with metals, nonmetals (except O), and other halogens lower in the group 5. For oxygen:O.N. = -1 in peroxides O.N. = -2 in all other compounds(except with F)
Sample Problem 4.6 Determining the Oxidation Number of an Element PROBLEM: Determine the oxidation number (O.N.) of each element in these compounds: (a) zinc chloride(b) sulfur trioxide(c) nitric acid PLAN: SOLUTION: The O.N.s of the ions in a polyatomic ion add up to the charge of the ion and the O.N.s of the ions in the compound add up to zero. (a) ZnCl 2. The O.N. for zinc is +2 and that for chloride is -1. (b) SO 3. Each oxygen is an oxide with an O.N. of -2. Therefore the O.N. of sulfur must be +6. (c) HNO 3. H has an O.N. of +1 and each oxygen is -2. Therefore the N must have an O.N. of +5.
Figure 4.10 Highest and lowest oxidation numbers of reactive main-group elements.
7 Zn (s) + CuSO 4 (aq) ZnSO 4 (aq) + Cu (s) Zn is oxidizedZn Zn 2+ + 2e - Cu 2+ is reducedCu 2+ + 2e - Cu Zn is the reducing agent Cu 2+ is the oxidizing agent Copper wire reacts with silver nitrate to form silver metal. What is the oxidizing agent in the reaction? Cu (s) + 2AgNO 3 (aq) Cu(NO 3 ) 2 (aq) + 2Ag (s) Cu Cu 2+ + 2e - Ag + + 1e - AgAg + is reducedAg + is the oxidizing agent
8 NaIO 3 Na = +1O = -2 3x(-2) + 1 + ? = 0 I = +5 IF 7 F = -1 7x(-1) + ? = 0 I = +7 K 2 Cr 2 O 7 O = -2K = +1 7x(-2) + 2x(+1) + 2x(?) = 0 Cr = +6 What are the oxidation numbers of all the elements in each of these compounds? NaIO 3 IF 7 K 2 Cr 2 O 7
9 Types of Oxidation-Reduction Reactions Combination Reaction A + B C 2Al + 3Br 2 2AlBr 3 Decomposition Reaction 2KClO 3 2KCl + 3O 2 C A + B 00 +3 +1+5-2+10
10 Types of Oxidation-Reduction Reactions Combustion Reaction A + O 2 B S + O 2 SO 2 00 +4-2 2Mg + O 2 2MgO 00 +2-2
11 Displacement Reaction A + BC AC + B Sr + 2H 2 O Sr(OH) 2 + H 2 TiCl 4 + 2Mg Ti + 2MgCl 2 Cl 2 + 2KBr 2KCl + Br 2 Hydrogen Displacement Metal Displacement Halogen Displacement Types of Oxidation-Reduction Reactions 0 +1+20 0+40+2 0 0
12 The Activity Series for Halogens Halogen Displacement Reaction Cl 2 + 2KBr 2KCl + Br 2 0 0 F 2 > Cl 2 > Br 2 > I 2 I 2 + 2KBr 2KI + Br 2
13 The same element is simultaneously oxidized and reduced. Example: Disproportionation Reaction Cl 2 + 2OH - ClO - + Cl - + H 2 O Types of Oxidation-Reduction Reactions 0 +1 oxidized reduced
14 Ca 2+ + CO 3 2- CaCO 3 NH 3 + H + NH 4 + Zn + 2HCl ZnCl 2 + H 2 Ca + F 2 CaF 2 Precipitation Acid-Base Redox (H 2 Displacement) Redox (Combination) Classify each of the following reactions.
15 Solution Stoichiometry The concentration of a solution is the amount of solute present in a given quantity of solvent or solution. M = molarity = moles of solute liters of solution What mass of KI is required to make 500. mL of a 2.80 M KI solution? volume of KI solutionmoles KIgrams KI M KI 500. mL= 232 g KI 166 g KI 1 mol KI x 2.80 mol KI 1 L soln x 1 L 1000 mL x
17 Dilution is the procedure for preparing a less concentrated solution from a more concentrated solution. Dilution Add Solvent Moles of solute before dilution (i) Moles of solute after dilution (f) = MiViMiVi MfVfMfVf =
18 How would you prepare 60.0 mL of 0.200 M HNO 3 from a stock solution of 4.00 M HNO 3 ? M i V i = M f V f M i = 4.00 M M f = 0.200 MV f = 0.0600 L V i = ? L V i = MfVfMfVf MiMi = 0.200 M x 0.0600 L 4.00 M = 0.00300 L = 3.00 mL Dilute 3.00 mL of acid with water to a total volume of 60.0 mL.