Continuation of Chapter 4 Oxidation-Reduction Reactions

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1 Continuation of Chapter 4 Oxidation-Reduction Reactions Oxidation-reduction reactions involve the transfer of electrons from one species to another. Oxidation is defined as the loss of electrons. Reduction is defined as the gain of electrons. Oxidation and reduction always occur simultaneously. 1 Oxidation-Reduction Reactions The reaction of an iron nail with a solution of copper(ii) sulfate, CuSO 4, is an oxidation- reduction reaction. The molecular equation: Fe(s) CuSO4 FeSO4 Total net ionic: Fe (s) + Cu + + SO - 4 Cu(s) Fe + + SO Cu (s) Net ionic: Fe(s) Cu Fe Cu(s) 1

2 The net ionic equation shows the reaction of iron metal with Cu + to produce iron(ii) ion and copper metal. Fe(s) Loss of e - oxidation Cu Fe Cu(s) Gain of e - reduction 3 Oxidation Numbers The concept of oxidation numbers is a simple way of keeping track of electrons in a reaction. The oxidation number (or oxidation state) of an atom in a substance is the actual charge of the atom if it exists as a monatomic ion. Alternatively, it is hypothetical charge assigned to the atom in the substance by simple rules. 4

3 Oxidation Number Rules Rule 1 Applies to Elements Statement The oxidation number of an atom in an element is zero. 3 Monatomic ions Oxygen The oxidation number of an atom in a monatomic ion equals the charge of the ion. The oxidation number of oxygen is in most of its compounds. (An exception is O in H O and other peroxides, where the oxidation number is 1.) Oxidation Number Rules Rule Applies to Hydrogen Halogens Compound s and ions Statement The oxidation number of hydrogen is +1 in most of its compounds. Fluorine is 1 in all of its compounds. The other halogens are 1 unless the other element is another halogen or oxygen. The sum of the oxidation numbers of the atoms in a compound is zero. The sum in a polyatomic ion equals the charge on the ion. 6 3

4 Sample Question The oxidation state of nitrogen given for all the following species is correct EXCEPT 1) N H 4 ( ). ) NH OH ( 1). 3) N O (+1). 4) HN 3 ( 1). 5) HNO (+3). 7 Sample Problem Assign oxidation numbers to all atoms in the following: CO K Cr O 7 PCl 5 HNO 4

5 Describing Oxidation-Reduction Reactions A half-reaction is one of the two parts of an oxidation- reduction reaction. One involves the loss of electrons (oxidation) and the other involves the gain of electrons (reduction). Fe(s) Fe(s) Cu Fe Fe e Cu(s) We can write this reaction in terms of two half-reactions. oxidation half-reaction Cu e Cu(s) reduction half-reaction 9 Describing Oxidation-Reduction Reactions An oxidizing agent is a species that oxidizes another species; it is itself reduced. A reducing agent is a species that reduces another species; it is itself oxidized. reducing agent Fe(s) Loss of e - Cu oxidizing agent oxidation Fe Gain of e - reduction Cu(s) 10 5

6 Sample Question In the following reactions, label the oxidizing agent and the reducing agent. a) CO (g) + C (s) CO (g) b) 8 Fe (s) + S 8 (s) 8FeS (s) Sample Question Which of the following conversions requires an oxidizing agent? 1) Mn 3+ Mn + ) C H 4 C H 6 3) SiF 4 SiF 6 4) CrO 4 Cr O 7 5) SO SO 3 1 6

7 Sample Question Which of the following reactions is an oxidation reduction reaction? 1) CaCO 3 + HCl CaCl + H O + CO ) NH 4 NO 3 N O + H O 3) AgNO 3 + KI AgI + KNO 3 4) H SO 4 + NaOH Na SO 4 + H O 5) CaO + SO 3 CaSO 4 13 Some Common Oxidation-Reduction Reactions Most fall into one of the following simple categories: Combination Reactions Decomposition Reactions Displacement Reactions Combustion Reactions 14 7

8 A combination reaction is a reaction in which two substances combine to form a third substance. Na (s) Cl (g) NaCl (s) A decomposition reaction is a reaction in which a single compound reacts to give two or more substances. HgO (s) Hg (l) O (g) A combustion reaction is a reaction in which a substance reacts with oxygen, usually with the rapid release of heat to produce a flame. 4 Fe (s) + 3 O (g) Fe O 3 (s) 15 Figure 4.16: Combustion reaction Photo courtesy of James Scherer. 4 Fe (s) + 3 O (g) Fe O 3 (s) 16 8

9 Figure 4. 14: Decomposition Reaction Photo courtesy of James Scherer. HgO (s) Hg (l) O (g) 17 A displacement reaction (also called a single- replacement reaction) is a reaction in which an element reacts with a compound, displacing an element from it. Zn(s) HCl ZnCl H(g) here, Zn has displaced H + from solution. Net ionic: Zn (s) + H + Zn + + H NOTE: any element higher in the activity series (Table 4.6) will reduce (displace) the ion of any element lower in the activity series Activity series looks at the relative reactivity of a neutral metal (M ) with an aqueous cation (Y n+ ) Net ionic: M (s) + Y n+ Mn n+ + Y (s) 18 9

10 Table 4.6 Sample Problem Using the activity series (Table 4.6), write balanced chemical and net ionic equations for the following reactions. If no reaction occurs write down NR. (a) Al(s) + ZnCl (b) Ni (s) + MnBr (c ) Zn (s) + HCl (d) Pt(s) + HBr (e) Ca (s) + H O (l) 0 10

11 Types of Chemical Reactions Oxidation-Reduction Reactions Balancing Simple Oxidation-Reduction Reactions At first glance, the equation representing the reaction of zinc metal with silver(i) ions might appear to be balanced. Zn(s) Ag Zn Ag(s) However, a balanced equation must have a charge balance as well as a mass balance. 1 Balancing Simple Oxidation-Reduction Reactions Since the number of electrons lost in the oxidation half-reaction must equal the number gained in the reduction half-reaction, Zn(s) Zn e oxidation half-reaction Ag e Ag(s) reduction half-reaction we must double the reaction involving the reduction of the silver. 11

12 Balancing Simple Oxidation-Reduction Reactions Adding the two half-reactions together, the electrons cancel, Zn(s) Zn e Ag e Ag(s) Zn(s) Ag Zn oxidation half-reaction reduction half-reaction Ag(s) which yields the balanced oxidation-reduction reaction. 3 Sample Problem Balance the following oxidation-reduction reactions by the half-reaction method: Cr 3+ + Zn (s) Cr (s) + Zn + 1

13 Working with Solutions The majority of chemical reactions discussed here occur in aqueous solution. When you run reactions in liquid solutions, it is convenient to dispense the amounts of reactants by measuring out volumes of reactant solutions When we dissolve a substance in a liquid, we call the substance the solute and the liquid the solvent. The general term concentration refers to the quantity of solute in a standard quantity of solution.. 5 Molar concentration,, or molarity (M),, is defined as the moles of solute dissolved in one liter (cubic decimeter) of solution. Molarity (M) moles of solute liters of solution Let s try an example. A sample of mol iron(iii) chloride, FeCl 3, was dissolved in water to give 5.0 ml of solution. What is the molarity of the solution? moles of FeCl3 Since molarity liters of solution mole of FeCl3 then M 1.36 M FeCl liter of solution 6 13

14 Sample Problem What is the molarity of 100 ml solution containing 15.6g of NaOH? SAMPLE PROBLEM What is the molarity of Br - ions in a 0.5 M aqueous solution of FeBr 3 assuming complete dissociation? 14

15 Working with Solutions Molar Concentration The molarity of a solution and its volume are inversely proportional. Therefore, adding water makes the solution less concentrated. This inverse relationship takes the form of: M i V i M f V So, as water is added, increasing the final volume, V f, the final molarity, M f, decreases. f 9 Sample Problem How many moles of HCl are contained in 50 ml of a molar solution? 15

16 Sample Problem How many ml of 15.0 M nitric acid solution should be dilute to make 50. ml of a 0.50 M solution? Quantitative Analysis Analytical chemistry deals with the determination of composition of materials that is, the analysis of materials. Quantitative analysis involves the determination of the amount of a substance or species present in a material. Gravimetric analysis is a type of quantitative analysis in which the amount of a species in a material is determined by converting the species into a solid product that can be isolated and weighed. Precipitation reactions are often used in gravimetric analysis. 3 16

17 Gravimetric Analysis Consider the problem of determining the amount of lead in a sample of drinking water. Adding sodium sulfate (Na SO 4 ) to the sample will precipitate lead(ii) sulfate. NaSO4 Pb Na PbSO4(s) The PbSO 4 can then be filtered, dried, and weighed. Suppose a 1.00 L sample of polluted water was analyzed for lead(ii) ion, Pb +, by adding an excess of sodium sulfate to it. The mass of lead(ii) sulfate that precipitated was 9.8 mg. What is the mass of lead in a liter of the water? Express the answer as mg of lead per liter of solution. 33 First we must obtain the mass percentage of lead in lead(ii) sulfate, by dividing the molar mass of lead by the molar mass of PbSO 4, then multiplying by 100. %Pb 07. g/mol g/mol % Then, calculate the amount of lead in the PbSO 4 precipitated. Amount Pb in sample = 9.8 mg PbS0 4 x = mg Pb Concentration (mass/v) of Pb in the water = mg/l 34 17

18 Sample Problem How much AgCl will be formed from mixing 1.50 L of M AgNO 3 with 1.75 L of M NaCl? Quantitative Analysis Volumetric Analysis An important method for determining the amount of a particular substance is based on measuring the volume of the reactant solution. Titration is a procedure for determining the amount of substance A by adding a carefully measured volume of a solution with known concentration of B until the reaction of A and B is just complete. (See Figure 4.0) Volumetric analysis is a method of analysis based on titration

19 Figure 4.0: Titration of an unknown amount of HCl with NaOH (reaction completion is indicated by an indictor colour change (i.e., clear to pink) 37 Quantitative Analysis Volumetric Analysis Consider the reaction of sulfuric acid, H SO 4, with sodium hydroxide, NaOH: HSO4 NaOH HO(l) NaSO4 Suppose a beaker contains 35.0 ml of M H SO 4. How many milliliters of 0.50 M NaOH must be added to completely react with the sulfuric acid? 38 19

20 Quantitative Analysis Volumetric Analysis First we must convert the L (35.0 ml) to moles of H SO 4 (using the molarity of the H SO 4 ). Then, convert to moles of NaOH (from the balanced chemical equation). Finally, convert to volume of NaOH solution (using the molarity of NaOH). (0.0350L) mole HSO4 1 L H SO solution L NaOH solution 4 mol NaOH 1mol H SO 4 or 1 L NaOH soln mol NaOH (49.0 ml of NaOH solution) 39 Sample Problem What volume of 0.00 M HCl solution is need to neutralize 5.0 ml of a M KOH solution? The balanced molecular equation is: HCl(aq aq) ) + KOH(aq aq) KCl + H O(l) 0

21 SAMPLE PROBLEM The concentration of aqueous I 3- solution can be determined by titration with aqueous sodium thiosulfate, Na S O 3. What is the molarity of I - 3 if 4.55 ml of 0.10 M Na S O 3 is needed for complete reaction with ml of the I - 3 solution? The net ionic equation is: S O I 3- S 4 O I - Operational Skills Using solubility rules. Writing net ionic equations. Deciding whether precipitation occurs. Classifying acids and bases as weak or strong. Writing an equation for a neutralization. Writing an equation for a reaction with gas formation. Assigning oxidation numbers. Balancing simple oxidation-reduction reactions. 4 1

22 Operational Skills Calculating molarity from mass and volume. Using molarity as a conversion factor. Diluting a solution. Determining the amount of a substance by gravimetric analysis. Calculating the volume of reactant solution needed. Calculating the quantity of a substance by titration. 43

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