Lecture 7 & 8. 2Mg (s) + O 2 (g) 7.1 Oxidation-Reduction (Redox) Reactions. 8.1 Introduction to Coordination Chemistry. Back to the first lecture
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1 Lecture 7 & OxidationReduction (Redox) Reactions 8.1 Introduction to Coordination Chemistry Back to the first lecture Initial step of this reaction: BaO 2 + Mg BaO + MgO 2Mg (s) + O 2 (g) 2MgO (s) 1
2 The redox process in compound formation OxidationReduction Reactions (electron transfer reactions) 2Mg (s) + O 2 (g) 2MgO (s) OIL RIG 2Mg 2Mg e O 2 + 4e 2O 2 Oxidation halfreaction (lose e ) Reduction halfreaction (gain e ) 2Mg + O 2 + 4e 2Mg O 2 + 4e 2Mg + O 2 2MgO 2
3 Oxidation number (O.N.) The charge the atom would have in a molecule (or an ionic compound) if electrons were completely transferred. 1. Free elements (uncombined state) have an oxidation number of zero. Na, Be, K, Pb, H 2, O 2, P 4, S 8 = 0 2. In monatomic ions, the oxidation number is equal to the charge on the ion. Li +, Li = +1; Fe 3+, Fe = +3; O 2, O = 2 3. The oxidation number of oxygen is usually 2. In H 2 O 2 and O 2 2 it is 1. Rules for Assigning an Oxidation Number (O.N.) General rules 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. 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 5. For oxygen: O.N. = 1 in peroxides O.N. = 2 in all other compounds(except with F) 6. For Group 7A(17): O.N. = 1 in combination with metals, nonmetals (except O), and other halogens lower in the group 3
4 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. e.g. NaH 5. Group IA metals are +1, IIA metals are +2 and fluorine is always 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. HCO 3 Oxidation numbers of all the elements in HCO 3? O = 2 H = +1 3x(2) + 1 +? = 1 C = +4 Oxidation numbers of all the elements in the following? IF 7 F = 1 7x(1) +? = 0 I = +7 NaIO 3 Na = +1 O = 2 3x(2) + 1 +? = 0 I = +5 K 2 Cr 2 O 7 O = 2 K = +1 7x(2) + 2x(+1) + 2x(?) = 0 Cr = +6 4
5 Oxidation Numbers.. in salts = ionic charge Examples: Procedure to determine O.N. in covalent molecules Move bonding electrons to the electron negative partner and count the fictive charge: borane: H Methane: H C H H H C + I IV Sample Problem 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: 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. SOLUTION: (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. 5
6 Elements: N.O. = 0 Highest and lowest oxidation numbers of reactive maingroup elements Oxidation Numbers Periodic Table Max. Oxidation number: V.. max. +V Cr.. max. +VI Mn. max. +VII CrO 4 2 Cr 2 O 7 2 (NH 4 )VO
7 A summary of terminology for oxidationreduction (redox) reactions e X transfer or shift of electrons Y X loses electron(s) X is oxidized X is the reducing agent X increases its oxidation number Y gains electron(s) Y is reduced Y is the oxidizing agent Y decreases its oxidation number Spontaneous Redox Reactions Zn rod in CuSO 4 solution Observation: Elemental copper deposits on the Zn rod Zn (s) + CuSO 4 (aq) ZnSO 4 (aq) + Cu (s) Tip: The nobler metal likes to get reduced, wants to be in the elemental form. The unnoble metal gets oxidized. Cu 2+ is reduced Cu e Cu Zn is oxidized Zn Zn e Cu 2+ is the oxidizing agent Zn is the reducing agent 7
8 Redox Reaction and electrical conductivity electrical current Anodic Oxidation Anode (+) e volt meter e porous wall Cathode ( ) Cathodic Reduction Zinc sulfate solution Copper sulfate solution Cell 1 Cell 2 Zn Zn e Cu e Cu 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) silver nobler than copper HalfEquations: Cu Cu e Ag + + 1e Ag Ag + is reduced Ag + is the oxidizing agent 8
9 Sample Problem Recognizing Oxidizing and Reducing Agents (b) PbO(s) + CO(g) Pb(s) + CO 2 (g) The O.N. of C increases; it is oxidized; it is the reducing agent. The O.N. of Pb decreases; it is reduced; it is the oxidizing agent (c) 2H 2 (g) + O 2 (g) 2H 2 O(g) The O.N. of H increases; it is oxidized; it is the reducing agent. The O.N. of O decreases; it is reduced; it is the oxidizing agent. Balancing Redox Equations Oxalic acid H 2 C 2 O 4 is oxidised by the permanganate ion 4 in acidic solution. During the reaction Mn 2+ and CO 2 is formed. 4 (aq) + H 2 C 2 O 4 (aq) Mn 2+ (aq) + CO 2 (g) Calculate the oxidation numbers: 4 Mn + 4(2) = 1 Mn = +7 O H 2 C 2 O C C 4 C = +3 HO CO 2 C + 2(2) = 0 C = +4 O OH O C O 9
10 Balancing Redox Equations Half Equations (aq) Mn 2+ (aq) H 2 C 2 O 4 (aq) 2CO 2 (g) Reduction Oxidation 1 st step: Identify the oxidation and the reduction and balance then the total oxidation numbers with electrons (Red: electrons on the left side, Ox: electrons on the right side) e Mn 2+ 2(+3) 2(+4) H 2 C 2 O 4 2CO 2 + 2e 2 nd step: Here the reaction is performed under acidic condition. So balance the charge of the half equations with protons. (If a reaction occurs under basic conditions OH ions are used to balance the equation) e + 8H + Mn 2+ charge: 6 charge: +2 2(+3) +4 H 2 C 2 O 4 2CO 2 + 2e charge: 2 + 2H + 3 rd step: Balance with water, so that you obtain proper half equations e + 8H + Mn H 2 O H 2 C 2 O 4 2 CO 2 + 2e + 2H + 10
11 4 th step: Multiply the equations to have the same number of electrons on each side. Simplify and add the equations e + 8H + Mn H 2 O x 2 H 2 C 2 O 4 2CO 2 + 2e + 2H + x e + 16H + 2 Mn H 2 O 5H 2 C 2 O 4 10CO e + 10H + Redox: H 2 C 2 O 4 + 6H + 2Mn CO 2 + 8H 2 O Remember: in alkaline solutions you have to balance with OH. A redox titration known concentration unknown concentration all the oxalic acid used for the reduction 11
12 Sample Problem Balancing Redox Equations by the Oxidation Number Method PROBLEM: Use the oxidation number method to balance the following equations: (a) Cu(s) + HNO 3 (aq) Cu(NO 3 ) 2 (aq) + NO 2 (g) + H 2 O(l) SOLUTION: (2) (a) Cu(s) + HNO 3 (aq) Cu(NO 3 ) 2 (aq) + NO 2 (g) + H 2 O(l) O.N. of Cu increases because it loses 2e ; it is oxidized and is the reducing agent. O.N. of N decreases because it gains1e ; it is reduced and is the oxidizing agent. loses 2e Cu(s) + HNO 3 (aq) balance other ions Cu(NO 3 ) 2 (aq) + NO 2 (g) + H 2 O(l) gains 1e x2 to balance e balance unchanged polyatomic ions Sample Problem continued Balancing Redox Equations by the Oxidation Number Method Cu 0 Cu e HNO 3 + 1e NO 2 Cu 0 Cu e HNO 3 + 1e + H + NO 2 Cu 0 Cu e HNO 3 + 1e + H + NO 2 + H 2 O Cu 0 Cu e 2HNO 3 + 2e + 2H + 2NO 2 + 2H 2 O x2 Cu 0 + 2HNO 3 + 2e + 2H + Cu NO 2 + 2H 2 O HNO 3 Cu(s) + 4HNO 3 (aq) Cu(NO 3 ) 2 (aq) + 2NO 2 (g) + 2 H 2 O(l) 12
13 Exercise Permanganate ions react with bromine ions in basic solution to form 2 and BrO 3. Write the balanced equation for the reaction. O.N. Br Br = 1 BrO 3 Br = +5 O.N. 4 Mn = +7 2 Mn = +4 Half reactions 4 +3e 2 Reduction Br BrO 3 +6e Oxidation Balance the equations and add the two half reaction together. continued 4 +3e 2 Br BrO 3 +6e Charge: 4 0 Charge: e 2 + 4OH Br + 6OH BrO 3 +6e 4 +3e + 2H 2 O 2 + 4OH Br + 6OH BrO 3 +6e + 3H 2 O 13
14 continued 4 +3e + 2H 2 O 2 + 4OH x2 Br + 6OH BrO 3 +6e + 3H 2 O e + 4H OH 2 O 2 Br + 6OH BrO 3 +6e + 3H 2 O 2 4 (aq) + Br (aq) + H 2 O(l) 2 2 (s) + BrO 3 (aq) + 2OH (aq) Displacing one metal with another 14
15 The activity series of the metals strength as reducing agents Li K Ba Ca Na Mg Al Mn An Cr Fe Cd Co Ni Sn Pb H 2 Cu Hg Ag Au can displace H 2 from water can displace H 2 from steam can displace H 2 from acid cannot displace H 2 from any source 15
16 The activity series of the metals Reduced Form = oxidized form + ze Standard Reduction Potential E o (V) less noble metals noble metals, nonmetals and their compounds Sample Problem PROBLEM: Finding an Unknown Concentration by a Redox Titration Calcium ions (Ca 2+ ) are required for blood to clot and for many other cell processes. An abnormal Ca 2+ concentration is indicative of disease. To measure the Ca 2+ concentration, 1.00mL of human blood was treated with Na 2 C 2 O 4 solution. The resulting CaC 2 O 4 precipitate was filtered and dissolved in dilute H 2 SO 4. This solution required 2.05mL of 4.88x10 4 M K 4 to reach the end point. The unbalanced equation is: K 4 (aq) + CaC 2 O 4 (s) + H 2 SO 4 (aq) MnSO 4 (aq) + K 2 SO 4 (aq) + CaSO 4 (s) + CO 2 (g) + H 2 O(l) (a) Calculate the amount (mol) of Ca 2+. (b) Calculate the Ca 2+ ion concentration expressed in units of mg Ca 2+ /100mL blood. 16
17 (a) Calculate the amount (mol) of Ca 2+. volume of K 4 soln mol of K 4 c=n/v n= c V x10 L 4 mol 10 3 L = 1.00x10 6 mol K 4 mol of K 4 molar ratio mol of CaC 2 O 4 see previous equations: 2:5 2 mol K 4 correspond to 5 mol CaC 2 O x10 6 mol 5 2 = 2.50x10 6 mol CaC 2 O 4 = 2.50x10 6 mol Ca 2+ (b) Calculate the Ca 2+ ion concentration expressed in units of mg Ca 2+ /100mL blood. mol Ca 2+ /1mL blood multiply by 100 mol Ca 2+ /100mL blood 2.50x10 6 mol Ca 2+ 1mL blood x100 = 2.50x10 4 mol Ca mL blood multiply by M g Ca 2+ /100mL blood n=m/m m=nm in 100mL blood: 2.50x10 4 mol x 40.08g/mol = 100.2x10 4 g =10.0mg Ca 2+ /100mL blood 17
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