NAME PER DATE DUE ACTIVE LEARNING IN CHEMISTRY EDUCATION CHAPTER 23 OXIDATION AND REDUCTION. (Part 1) , A.J. Girondi

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1 NAME PER DATE DUE ACTIVE LEARNING IN CHEMISTRY EDUCATION CHAPTER 23 OXIDATION AND REDUCTION (Part 1) , A.J. Girondi

2 NOTICE OF RIGHTS All rights reserved. No part of this document may be reproduced or transmitted in any form by any means, electronic, mechanical, photocopying, or otherwise, without the prior written permission of the author. Copies of this document may be made free of charge for use in public or nonprofit private educational institutions provided that permission is obtained from the author. Please indicate the name and address of the institution where use is anticipated A.J. Girondi, Ph.D. 505 Latshmere Drive Harrisburg, PA Website: , A.J. Girondi

3 SECTION 23.1 Introduction To Oxidation Reduction Reactions In earlier chapters, we studied chemical reactions labeling them as combination, decomposition, single replacement, and double replacement. In this chapter, we will study yet another group called oxidation-reduction reactions. We can greatly simplify the classification of reactions by grouping them all into two broad classes. These two classes are: (1) reactions in which there is no electron transfer from one substance to another, and (2) reactions in which electrons transfer from one substance to another. All chemical reactions fit into one or the other of these two categories. In previous chapters we made use of oxidation numbers, but we did not define them. Now it is time to do so. An oxidation number is a signed number which is assigned to an atom or ion according to a set of rules. It represents the "oxidation state" of the atom or ion. The "oxidation state" of an atom or ion changes when it loses or gains electrons. You have previously used oxidation numbers to write the correct formulas for compounds and polyatomic ions. You may recall that there is a table of common oxidation numbers in the reference section of your ALICE materials. The rules used to assign oxidation numbers to atoms and ions are listed below. You should know them well enough to use them from memory during tests and quizzes. Rules for Assigning Oxidation Numbers 1. The oxidation number of an atom of a free element is zero. Elements are free if they are not combined with other elements. If atoms of an element are combined with themselves, they are still considered to be free. For example, a free atom of Ag has an oxidation number of zero. In addition, atoms in molecules like H2, Cl2, N2, O2, F2, Br2, I2, P4, S8, etc., have oxidation numbers of zero. 2. The oxidation number of a monatomic ion is equal to its charge. A monatomic ion is one that formed from only one atom. Ex: Ag The algebraic sum of the oxidation numbers of the atoms in the formula of a compound is zero. 4. In compounds, the oxidation number of hydrogen is +1. (There is one exception. In compounds known as hydrides, it can be -1. Sodium hydride is NaH.) 5. In compounds, the oxidation number of oxygen is -2. (Exceptions: oxygen is -1 in peroxide compounds like H2O2, and +2 when it combines with fluorine in OF2.) 6. In combinations of nonmetal atoms, the oxidation number of the less electronegative element is positive and of the more electronegative element is negative. For example, in NO2, N = +4, and O = -2. The element with the positive oxidation number is written first in the formula of a compound, such as in NO2. An exception is ammonia, NH3, in which nitrogen (-3) is written before hydrogen (+1). 7. The algebraic sum of the oxidation numbers of the atoms in the formula of a polyatomic ion is equal to the charge on the polyatomic ion. Example: In Cr2O7 2-, each chromium atom is +6, while each oxygen atom is -2. Note that (+6 X 2) + (-2 X 7) = -2. Because of your work in previous chapters, you are already somewhat familiar with rules 3 through 7. Now, rules 1 and 2 will also be needed in order to understand redox reactions. Reactions in which no electrons are transferred usually involve the separation and rejoining of atoms or ions. For example, silver nitrate (AgNO3) and sodium chloride (NaCl) exist as ions in solution. AgNO3(s) ----> Ag 1+ (aq) + NO3 1- (aq) , A.J. Girondi

4 NaCl(s) ----> Na 1+ (aq) + Cl 1- (aq) When added together, the positive Ag 1+ ions combine with the negative Cl 1- ions to form a precipitate of AgCl. The Na 1+ and NO3 1- ions are spectator ions which will combine only if the water is removed from the system. Formula Equation: AgNO3(aq) + NaCl(aq) ----> AgCl(s) + NaNO3(aq) Ionic Equation: Ag 1+ (aq) + NO3 1- (aq) + Na 1+ (aq) + Cl 1- (aq) ----> AgCl(s) + NO3 1- (aq) + Na 1+ (aq) Net Ionic Equation: Ag 1+ (aq) + Cl 1- (aq) ----> AgCl(s) Ag has an oxidation number of {1} when combined with NO3 1- and with Cl 1-. The oxidation number remains the same before and after the reaction. When the oxidation number does not change, there has been no transfer of electrons. You can see from the equations above that there has been neither an increase nor a decrease in oxidation number. This is true for the silver ions and for the other ions involved in the reaction. In this chapter, however, we want to concentrate on the second type of reaction in which electrons are transferred from one particle to another. Such oxidation-reduction equations are often called "redox" equations for short. You have seen some redox reactions in previous chapters, although you probably did not realize it. For example, single replacement reactions are also redox redox reactions. We will now study some of the "whys" and "hows" of these reactions. Let's begin with some definitions. Oxidation is defined as the LOSS of electrons by a substance. Reduction is defined as the GAIN of electrons by a substance. An example of a typical redox equation is: 2 Mg(s) + O2(g) ----> 2 MgO(s) The reactants, Mg and O2, are neutral particles. The product, MgO, is made of equal numbers of Mg 2+ and O 2- ions. These oppositely charged ions attract each other, and the result in an ionically bonded compound. We can gain a better understanding of the reaction between Mg and O2 by breaking it into two separate parts, something that can be done to any redox equation. The separate parts are called halfequations or half-reactions. In the first half-equation below, magnesium is shown losing two electrons (e - ). Remember that each electron has a charge of {2}. Magnesium is undergoing oxidation. Oxygen, on the other hand, is shown gaining two electrons per atom (4 electrons total) and is, therefore, undergoing reduction. In oxidation half-equations, the electrons are always written on the product side (right side) to show that they are being lost: Oxidation half-equation: Mg ----> Mg e - In the half-equation above, each magnesium atom is losing {3} electrons. In reduction half-equations, the electrons are written on the reactant side to show that they are being gained. Reduction half-equation: O2 + 4 e > 2 O 2- In the half-equation above, a total of {4} electrons are being gained. Since two oxygen atoms are involved, each oxygen atom is gaining {5} electrons , A.J. Girondi

5 In the overall equation, note that the oxidation number of the magnesium atom is changing from 0 to {6}. Also notice that the oxidation number of each of the two oxygen atoms is changing from 0 to {7} Mg(s) + O2(g) ----> 2 MgO(s) These changes in the oxidation numbers of magnesium and oxygen can be explained by the loss and gain of electrons as shown in the half-equations. You must learn to easily identify whether half-equations represent oxidations or reductions. One way to remember the definitions is to think of electrons going Out in Oxidation, and electrons Returning in Reduction. Problem 1. Complete the exercises below by labeling each of the half-equations as oxidations or as reductions. Half-Equations Oxidation or Reduction a. Cu 1+ + e - ---> Cu b. F2 + 2e - ---> 2 F 1- c. Cr > Cr e - d. Bi > Bi e - e. N e - ---> N 3+ f. 2 Cl > Cl2 + 2e - SECTION 23.2 Balancing Redox Half-equations The total charge on each side of a redox equation must be equal. This is also true for halfequations. Notice that the sum of the oxidation numbers is zero on both sides of this equation taken from section 23.1: Mg(s) + O2(g) ----> 2 MgO(s) The total charge on each side of a redox half-equation must also be equal. In each of the halfequations shown previously, this is also true. For example: Cr > Cr e - You can see that the total charge on both sides of the half-reaction is {8} , A.J. Girondi

6 This is an important rule about all redox equations: In any redox equation or half-equation, the total charge of the reactants must equal the total charge of the products. This rule makes it possible for you to determine exactly how many electrons are gained or lost in any halfequation. If you were told that manganese changes from a charge of +7 to +4, you could write a halfequation for this change using the previous examples as your guide. For example, examine the halfequation below: Mn > Mn 4+ The charges are not balanced (equal) on the reactant and product sides. By adding electrons (negative charges) to the reactant (left) side, the charges can be balanced: Mn e > Mn 4+ Each side of the half-equation now has a total charge of {9}. You have already learned that a chemical equation must be balanced according to mass, which means that it must have the same number of atoms or ions on both sides. This is because of the Law of Conservation of Mass which states that matter cannot be created or destroyed in ordinary chemical reactions. Now you have also learned that charge cannot be created or destroyed, either. This is the Law of Conservation of Charge. A redox reaction or half-reaction must be balanced in two ways: first, according to {10}, and, second, according to {11}. Problem 2. Complete the exercises below by adding the correct number of electrons needed to balance the charges in each half-equation. Then, label each half-equation as an oxidation or reduction. Review the first two rules for assigning oxidation numbers before you begin. Half-Reactions Oxidation or Reduction a. Na ---> Na 1+ + e - b. Cl2 + e - ---> 2 Cl 1- c. 2 H 1+ + e - ---> H2 d. 2 Al ---> 2 Al 3+ + e - e. 3 F2 + e - ---> 6 F 1- f. Fe > Fe 3+ + e - g. Mn 7+ + e - ---> Mn 2+ h. 6 Cl > 3 Cl2 + e - i. Cr 6+ + e - ---> Cr 3+ j. Cl > Cl 5+ + e , A.J. Girondi

7 Problem 3. Write balanced half-equations for each situation presented below. You will need to decide on which side to place the electrons. Make sure your half- equations are balanced according to mass and charge. Then label each change as oxidation or reduction. Half-Equation Balanced Half-Equation Ox or Red? a. Cu > Cu 1+ b. I > I 1+ c. 3 Br2 ---> 6 Br 1- d. Sn > Sn 4+ e. P > P 3+ f. Cu > Cu g. 2 Br2 ---> 4 Br 1- h. 6 N > 3 N2 SECTION 23.3 Adding Redox Half-Equations Together Oxidation and reduction are processes that must occur together. This is true because it is impossible to create or destroy electrons in any chemical change. Electrons can only be transferred from one substance to another. The electrons that are lost in oxidation are gained in a reduction reaction. This is why all oxidation-reduction reactions are the sum of two half-reactions. Let's consider the case in which sodium reacts with fluorine: Sodium atoms tend to lose {12} electron(s) to become a stable ion: Na ----> Na 1+ + e - Fluorine atoms tend to gain one electron to become stable. Since fluorine is diatomic, F2 would gain a total of {13} electrons: F2 + 2e > 2 F 1- After the diatomic molecule gains electrons, is it still diatomic? {14} It has been changed into two independent stable ions. Since F2 needs two electrons and sodium gives up only one electron per atom, two sodium atoms will have to react with one F2 molecule, so we multiply through the oxidation halfequation by two: 2 (Na ----> Na 1+ + e - ) OR 2 Na ----> 2 Na e - and then we add the two half-equations together: 2 Na ----> 2 Na e - <-- oxidation half-equation F2 + 2e > 2 F 1- <-- reduction half-equation 2 Na + F2 + 2e > 2 Na F e - <-- overall redox equation Since there are two electrons on each side of the net equation, they can be dropped to yield the final redox equation: 2 Na(s) + F2(g) ----> 2 NaF(s) , A.J. Girondi

8 It's fairly simple to write redox equations from two half-equations. Let's try to write the redox equation for the reaction between Cu atoms and N 3+ ions. The two half-reactions are: Cu ---> Cu 2+ and 2 N > N2 The 2 is needed in front of N 3+ in the second half-reaction because nitrogen is diatomic as a free element, and the equation must be balanced by mass. While the two half-equations are balanced according to mass, you can see that they are not balanced according to {15}. To balance the charge, we add electrons to the proper side of each half-reaction: Cu ---> Cu e - <---- oxidation 2 N e - ---> N2 <---- reduction Now, since the number of electrons lost in the oxidation must equal the number of electrons gained in the reduction, we multiply through the oxidation half-equation by {16} : 3 Cu ----> 3 Cu e - (When trying to balance some redox equations, you may have to multiply through the reduction halfequation, or you may have to multiply through both half-equations.) Then, we add the two half-reactions together: 3 Cu ----> 3 Cu e - <-- oxidation half-equation 2 N e > N2 <-- reduction half-equation 2 N Cu + 6e > N2 + 3 Cu e - <-- overall redox reaction Since there are {17} electrons on both sides of the equation, they should be eliminated to give the overall redox equation: 2 N Cu ----> N2 + 3 Cu 2+ How can you tell that the equation above is balanced according to mass? {18} Notice that in the equation above, {19} electrons are lost in the oxidation and {20} electrons are gained in the reduction. Electrons must always be balanced in this way in a redox equation. Using the same procedure as above, write the overall redox equation for each of the half-equation combinations given below. Identify each half-equation as oxidation or reduction. (Some elements in these equations may have oxidation numbers that do not appear on your list of common oxidation numbers.) Problem 4. Li ----> Li 1+ and O2 ----> 2 O , A.J. Girondi

9 Problem 5. Fe ----> Fe 2+ and Hg > Hg Problem 6. Cr > Cr 2+ and Al ----> Al 3+ Problem 7. F2 ----> 2 F 1- and Zn ----> Zn 2+ SECTION 23.4 Changes In Oxidation Numbers So far, the equations we have examined have been pretty simple. All of them involve either single elements or single ions. In such cases, it is pretty easy to determine which elements or ions are losing electrons and which are gaining electrons. Many redox equations are more complicated than this. Some involve polyatomic ions like PO4 3- or NO3 1-. Before we can attempt to understand redox equations that include such ions, we must know how to determine the oxidation number of each atom in a polyatomic ion. This is not new to you. You worked with this concept back in Chapter 14. For example, what is the oxidation number of the Mn atom in the ion MnO4 1-? The total of the oxidation numbers of the atoms in this ion must must equal the charge on the ion, which in this case is -1. Rule 5 in section 23.1 tells us that the oxidation number of oxygen is -2. Since there are four of them, the total charge from oxygen is -8. If the total is to be -1, the oxidation number of the one manganese atom present must be +7. Let's try another one. What is the oxidation number of each chromium atom in the Cr2O7 2- ion? The seven oxygen atoms have a total charge of -14. The total charge must add up to -2, since that is the charge on the whole ion. The total positive charge from the chromium atoms must, therefore, be +12. Since there are two Cr atoms present, each one must have an oxidation number of {21}. You may want to reread the rules in section 23.1 and review the work you did with oxidation numbers in Chapter 14 before completing problem , A.J. Girondi

10 Problem 8. Oxidation Numbers of Elements in Compounds or Polyatomic Ions Compound or Ion Determine Ox. No. of: Oxidation No. is: a. HAsO2 As b. HBr Br c. KI I d. MnO2 Mn e. MnO4 1- Mn f. H3PO4 P g. HCO3 1- C h. CO3 2- C i. ClO4 1- Cl j. ClO2 1- Cl k. SeO4 2- Se l. Cr2O7 2- Cr We can now use oxidation numbers to determine which atoms are being oxidized (losing electrons) and which are being reduced (gaining electrons). In addition to losing and gaining electrons, another definition of oxidation and reduction can be given in terms of changes in oxidation number: When an atom or ion undergoes oxidation, its oxidation number goes up (becomes more positive or less negative). When an atom or ion undergoes reduction, its oxidation number goes down (becomes more negative or less positive.) Oxidation involves an increase in oxidation number, while {22} involves a decrease in oxidation number. According to our other definition, the half-equation below would be labeled as an oxidation because electrons are being lost. Note that the oxidation number of Fe increases from 0 to +3, so it also fits our new definition of oxidation: Fe ----> Fe e - oxidation A similar relationship occurs when reduction occurs. In the half-equation shown below, Br2 gains 2 electrons to become 2 Br 1-. Electrons are gained, so this is reduction. The oxidation number of bromine goes from 0 to -1. A decrease in oxidation number also indicates that a reduction has occurred: Br2 + 2 e > 2 Br 1- reduction Use these additional definitions of oxidation and reduction to determine whether iron, Fe, is being oxidized or reduced in each of the following processes , A.J. Girondi

11 Problem 9. Determine the oxidation number of iron in each case. Write the oxidation number of iron above each Fe shown, whether it is alone or combined with another element. Process Ox or Red a. Fe2O3 becomes Fe b. FeO becomes Fe2O3 c. FeF2 becomes Fe d. FeO becomes FeCl3 SECTION 23.5 Oxidizing Agents And Reducing Agents When discussing oxidation-reduction reactions, the terms oxidizing agent and reducing agent are often used. The substance that undergoes a decrease in its oxidation number is called the oxidizing agent. The substance whose oxidation number increases is called the reducing agent. The oxidizing agent does the oxidizing, and, as a result, gets reduced. The reducing agent does the reducing, and, as a result, gets oxidized. This can be a little confusing at first, so let's organize the data in another way. Oxidizing Agent Reducing Agent 1. its oxidation number decreases 1. its oxidation number increases 2. it does the oxidizing 2. it does the reducing 3. it gets reduced (gains e - ) 3. it gets oxidized (loses e - ) It makes sense if you think about it. The oxidizing agent does the oxidizing, meaning that it causes something else to lose electrons. In the process the oxidizing agent gains those electrons; thus, the oxidizing agent gets reduced. On the other hand, the reducing agent has to give away electrons in order to do the reducing. Therefore, the reducing agent ends up losing electrons and being oxidized. Complete problem 10 by identifying the reducing agent and oxidizing agent in each equation. As you complete the table, be sure to indicate whether you are answering with the atom of an element or the ion (the first two are done for you). Be sure to include the charge when answering with ions. Problem 10. Identifying Oxidizing and Reducing Agents Reducing Agent Oxidizing Agent a. 2 Na + Ni > 2 Na 1+ + Ni Na Ni 2+ b. Hg Fe > Hg + 2 Fe 3+ Fe 2+ Hg 2+ c. 2 Al + 3 Pb > 2 Al Pb d. Mg Li ----> Mg + 2 Li 1+ e. O2 + 4 K ----> 2 K2O f. Fe + Ni > Fe 2+ + Ni g. Cu + 2 Ag > Cu Ag h. Cd + 2 H > Cd 2+ + H2 i. 3 Co Al ----> 3 Co + 2 Al , A.J. Girondi

12 ACTIVITY 23.6 Observing Redox Reactions So far you have worked with redox equations. Now it is time to actually carry out several redox reactions in the laboratory. You will be able to see changes in substances being oxidized and reduced, and you will be attempting to identify the substances being oxidized and reduced. Be sure to wear safety glasses and an apron. Procedure - Part A: 1. Place 20 ml of 0.2 M CuSO4 solution into a clean 125 ml Erlenmeyer flask. This solution contains Cu 2+ ions. We can ignore the SO4 2- ions, because they do not get involved in the reaction. 2. Put a small amount of powdered zinc metal into the Cu 2+ solution. Allow the reaction to proceed for at least five minutes before recording any observations. Swirl the mixture in the flask every minute or so. What happens to the zinc metal after 5 minutes? {23} What happens to the solution's color? {24} 3. Atoms of zinc metal, Zn, have been converted into colorless Zn 2+ ions which go into solution. The blue Cu 2+ ions present in the CuSO4 solution have been converted into solid copper atoms, Cu, which appear on the bottom of the flask. Write a redox equation that describes the reaction between Zn and Cu 2+ : {25} 4. Complete the items below. Be sure to indicate whether your answer refers to an atom or an ion of an element. substance oxidized: {26} substance reduced: {28} oxidizing agent: {27} reducing agent: {29} Procedure - Part B: 1. Obtain a solution of 3.0 M HCl. Place 20 ml of the HCl into a 125 ml Erlenmeyer flask. This solution contains H 1+ ions. Place the beaker under the fume hood and add a small (2 cm) piece of magnesium metal to the solution. Observe what happens. What evidence was there that a chemical reaction occurred? {30} 2. Magnesium atoms, Mg(s), react with the hydrogen ions, H 1+, in the HCl solution to form hydrogen gas, H2(g), and magnesium ions, Mg 2+ (aq), which go into solution. Write a redox equation for the reaction that occurred. {31} 3. Complete the items below. Be sure to indicate whether your answer refers to an atom or an ion of an element. substance oxidized: {32} substance reduced: {34} oxidizing agent: {33} reducing agent: {35} , A.J. Girondi

13 Procedure - Part C: 1. Add a small piece of copper metal, Cu, to 20 ml of 0.1 M HCl. Do you observe any reaction? 2. Based on your observation, is the following statement true or false {36}? "H 1+ ions will oxidize Mg metal much more readily than they oxidize Cu metal." From these results you see that oxidizing and reducing agents are not always strong enough for a redox reaction to occur spontaneously. H 1+ was able to oxidize Mg, but it was not strong enough to oxidize the copper. Cu + HCl ----> no reaction OR Cu + H > no reaction If this reaction had occurred, it would have fallen into the category of single replacement. Single replacement reactions are also redox reactions. Remember when you used the activity series in chapter 6? (A copy of the activity series can also be found in your reference notebook.) If you check it, you will find that copper is below hydrogen on the activity series, meaning no reaction will occur. You will also note in the series that Mg is located above both copper and hydrogen, which explains why the reactions in parts A and B of this activity did occur. SECTION 23.7 Balancing Redox Equations In Chapters 6 and 7 you learned how to balance chemical equations. The equations you worked with were classified into four groups - combination, decomposition, single replacement, and double replacement. You learned how to balance them by inspection. This means that you looked from side to side in each equation changing the coefficients, in your attempt to make the number of atoms of each element on the left side equal to the number of atoms of each element on the right side. Many redox equations are difficult to balance by inspection. For example:?? H2S +?? HNO3 ---->?? S +?? NO +?? H2O Balancing such equations can be made easier by using oxidation numbers. This fact is the basis for a method of balancing redox equations known as the electron-transfer method. It is a much more systematic method of balancing equations than is balancing by inspection. The steps involved in the electron-transfer method will be illustrated using the equation above. (For simplicity's sake, sulfur is represented as S in this example, not S8) Step 1: Assign oxidation numbers to each element in the equation. (Follow the rules in section 23.1) \ / \ / \ / \ / H 2 S + HNO > S + NO + H 2 O , A.J. Girondi

14 Step 2: Identify the elements that have been oxidized and reduced, and then write the half-equations (balanced according to mass and charge). Sulfur's oxidation number increases from -2 to 0, so it has been oxidized. Nitrogen's oxidation number decreases from +5 to +2, so it has been reduced. Make sure both half-equations are balanced according to mass and according to charge. oxidation: S > S + 2e - reduction: N e > N 2+ Step 3: Conserve electrons by multiplying each half-equation by a coefficient that makes the number of electrons lost in the oxidation equal to the number of electrons gained in the reduction. 3 (S > S + 2e - ) = 3 S > 3 S + 6e - 2 (N e > N 2+ ) = 2 N e > 2 N 2+ Step 4: Add the two half-reactions. (The electrons should cancel out.) oxidation: 3 S > 3 S + 6e - reduction: 2 N e > 2 N 2+ overall: 3 S N > 3 S + 2 N 2+ Step 5: Place the coefficients in front the proper substances in the original equation. 3 H2S + 2 HNO3 ----> 3 S + 2 NO +?? H2O Step 6: Add any other coefficients or make any changes needed to balance the equation. (In our example, you need a coefficient in front of the H2O. A 4 will balance it.) 3 H2S + 2 HNO3 ----> 3 S + 2 NO + 4 H2O DONE!! Here's another example. Let's follow the six steps to balance this equation: I2 + HNO3 ----> HIO3 + NO2 + H2O Step 1: \ / \ / \ / \ / I 2 + HNO > HIO 3 + NO 2 + H 2 O , A.J. Girondi

15 Step 2: oxidation: I2 ----> 2 I e - reduction: N 5+ + e > N 4+ Notice that if gases are diatomic in the equation to be balanced, they are also written in diatomic form in the half-reactions. This is why the oxidation half-reaction above is I2 ---> 2 I e - rather than I ---> I e -. A coefficient 2 is needed in front of I 5+ to balance mass, and then 10e - are needed to balance charge. Step 3: 1 (I2 ----> 2 I e - ) = I2 ----> 2 I e - 10 (N 5+ + e > N 4+ ) = 10 N e > 10 N 4+ Step 4: oxidation: I2 ----> 2 I e - reduction: 10 N e > 10 N 4+ overall: I N > 2 I N 4+ Step 5: I HNO3 ----> 2 HIO NO2 +?? H2O Step 6: I HNO3 ----> 2 HIO NO2 + 4 H2O DONE!! Problem 11. Use the six steps of the electron-transfer method to balance the redox equations below. Again, for simplicity, sulfur is represented as S instead of S8. (Some elements in these equations may have oxidation numbers that do not appear on your list of common oxidation numbers.) a. HNO3 + H2S -----> H2O + NO + S b. S + HNO > SO2 + NO + H2O c. HIO3 + NO2 + H2O ----> HNO3 + I , A.J. Girondi

16 d. Al + H 1+ + SO > Al 3+ + SO2 + H2O (In order to balance more complex equations, you may need to change one or more coefficients even after you have completed the 6 steps. This will be necessary in equation e below. e. HCl + KMnO4 ----> KCl + MnCl2 + Cl2 + H2O The equations below are optional. You may balance them on a separate page if you feel that you need more practice. f. HNO3 + KI ----> KNO3 + I2 + NO + H2O g. Fe 2+ + MnO H > Mn 2+ + Fe 3+ + H2O h. Sn 2+ + Ce > Sn 4+ + Ce 3+ j. Sn + H 1+ + NO > SnO2 + NO2 + H2O j. NaI + H2SO4 ----> H2S + I2 + Na2SO4 + H2O (This one can be fun!) , A.J. Girondi

17 SECTION 23.8 Learning Outcomes This is the end of Chapter 23. The subject of oxidation-reduction is continued in Chapter 24. Review the learning outcomes below. Arrange to take any quizzes or exams on Chapter 23, and then move on to Chapter Distinguish between oxidation and reduction. 2. Identify the oxidizing and reducing agents in a redox equation. 3. Assign oxidation numbers to all elements in any compound or polyatomic ion in a redox equation using the rules for assigning oxidation numbers. 4. Determine the number of electrons transferred in redox half- equations and overall equations. 5. Write balanced net redox equations given two half-equations. 6. Balance redox equations by the electron-transfer method , A.J. Girondi

18 SECTION 23.9 Answers to Questions and Problems Questions: {1} +1; {2} -1; {3} 2; {4} 4; {5} 2; {6} +2; {7} -2; {8} +3; {9} +4; {10} mass; {11} charge; {12} one; {13} 2; {14} no; {15} charge; {16} three; {17} six; {18} It has the same number of atoms of each element on both sides of the equation; {19} six; {20} six; {21} +6; {22} reduction; {23} It gets oxidized (Zn(s) ---> Zn 2+ (aq) + 2e - ; {24} It fades as the blue Cu 2+ ions are reduced to Cu atoms; {25} Zn(s) + Cu 2+ (aq) ---> Zn 2+ (aq) + Cu(s) {26} Zn; {27} Cu 2+ ; {28} Cu 2+ ; {29} Zn; {30} A gas is evolved (given off); {31} Mg(s) + 2 H 1+ (aq) ---> Mg 2+ (aq) + H2(g); {32} Mg; {33} H 1+ ; {34} H 1+ ; {35} Mg; {36} true Problems: 1. a. red; b. red; c. ox; d. ox; e. red; f. ox 2. a. Na ---> Na e - ox b. Cl2 + 2 e - ---> 2 Cl 1- red c. 2 H e - ---> H2 red d. 2 Al ---> 2 Al e - ox e. 3 F2 + 6 e - ---> 6 F 1- red f. Fe > Fe e - ox g. Mn e - ---> Mn 2+ red h. 6 Cl > 3 Cl2 + 6 e - ox i. Cr e - ---> Cr 3+ red j. Cl > Cl e - ox 3. a. Cu > Cu 1+ Cu e - ---> Cu 1+ red b. I > I 1+ I e - ---> I 1+ red c. 3 Br2 ---> 6 Br 1-3 Br2 + 6e - ---> 6 Br 1- red d. Sn > Sn 4+ Sn > Sn e - ox e. P > P 3+ P e - ---> P 3+ red f. Cu > Cu Cu e - ---> Cu red g. 2 Br2 ---> 4 Br 1-2 Br2 + 4e - ---> 4 Br 1- red h. 6 N > 3 N2 6 N > 3 N e - ox 4. ox: 4(Li ---> Li 1+ + e - ) 5. ox: Fe ---> Fe e - red: O2 + 4e - ---> 2 O 2- red: Hg e - ---> Hg overall: 4 Li + O2 ---> 4 Li O 2- overall: Fe + Hg > Fe 2+ + Hg 6. ox: Al ---> Al e - 7. ox: Zn ---> Zn e - red: 3(Cr 3+ + e - ---> Cr 2+ ) red: F2 + 2e - ---> 2 F 1- overall: 3 Cr 3+ + Al ---> 3 Cr 2+ + Al 3+ overall: F2 + Zn ---> 2 F 1- + Zn a. 3+; b. 1-; c. 1-; d. 4+; e. 7+; f. 5+; g. 4+; h. 4+; i. 7+; j. 3+; k. 6+; l a. red; b. ox; c. red; d. ox 10. c. Al, Pb 2+ ; d. Li, Mg 2+ ; e. K, O2; f. Fe, Ni 2+ ; g. Cu, Ag 1+ ; h. Cd, H 1+ ; i. Al, Co , A.J. Girondi

19 11. a. ox: 3 (S > S + 2e - ) red: 2 (3e - + N > N 2+ ) overall: 3 S N > 3 S + 2 N 2+ final: 2 HNO3 + 3 H2S ---> 4 H2O + 2 NO + 3 S 11. b. ox: 3 (S ---> S e - ) red: 4(3e - + N > N 2+ ) overall: 3 S + 4 N > 3 S N 2+ final: 3 S + 4 HNO3 ---> 3 SO2 + 4 NO + 2 H2O 11. c. ox: 10(N > N 5+ + e - ) red: 1(10e I > I2) overall: 10 N I > 10 N 5+ + I2 final: 2 HIO NO2 + 4 H2O ---> 10 HNO3 + I2 11. d. ox: 2(Al ---> Al e - ) red: 3(2e - + S > S 4+) overall: 2 Al + 3 S > 2 Al S 4+ final: 2 Al + 12 H SO > 2 Al SO2 + 6 H2O 11. e. ox: 5(2 Cl > Cl2 + 2e-) red: 2(5e - + Mn > Mn 2+ ) overall: 10 Cl Mn > 5 Cl2 + 2 Mn 2+ final: 16 HCl + 2 KMnO4 ---> 2 KCl + 2 MnCl2 + 5 Cl2 + 8 H2O 11. "f" through "j" - answers not given to the optional ones. You figure them out! , A.J. Girondi

20 SECTION Student Notes , A.J. Girondi