1. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 14-1 Chapter 14: Oxidation- Reduction Reactions

2. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 14-2 Questions for Consideration 1. What occurs in an oxidation-reduction reaction? 2. How do we account for the loss and gain of electrons in an oxidation-reduction reaction? 3. How do chemical reactions provide electricity in batteries? 4. How are simple oxidation-reduction reactions balanced? 5. How are complex oxidation-reduction reactions balanced? 6. How do oxidation-reduction reactions generate electricity in spontaneous reactions? How is electricity used to drive oxidation-reduction reactions that do not normally occur? 7. How can corrosion be prevented?

3. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 14-3 Chapter 14 Topics: 1. What Is an Oxidation-Reduction Reaction? 2. Oxidation Numbers 3. Batteries 4. Balancing Simple Oxidation-Reduction Reactions 5. Balancing Complex Oxidation-Reduction Reactions 6. Electrochemistry 7. Corrosion Prevention

4. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 14-4 14.1 What is an Oxidation-Reduction Reaction? A reaction in which electrons are transferred A reaction in which electrons are transferred Also called redox reactions Also called redox reactions An example: An example: Zn(s) + Cu 2+ (aq) → Zn 2+ (aq) + Cu(s) In this reaction, electrons are transferred from zinc to copper. Figure 14.5

5. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 14-5 Oxidation-Reduction Reaction Figure 14.6

6. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 14-6 Oxidation and Reduction Oxidation Oxidation The process of losing one or more electrons The process of losing one or more electrons Reduction Reduction The process of gaining one or more electrons The process of gaining one or more electrons Oxidation and reduction are coupled reactions. Oxidation and reduction are coupled reactions. Oxidation does not occur without reduction. Oxidation does not occur without reduction.

7. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 14-7 0 2+ 1  2+ 1  0 0 2+ 1  2+ 1  0 Zn(s) + CuCl 2 (aq) → ZnCl 2 (aq) + Cu(s) Only the reactants that change charge are involved in the redox reaction. Only the reactants that change charge are involved in the redox reaction. Zinc is oxidized because it changed charge from no charge (0) to +2, and thus lost two electrons. Zinc is oxidized because it changed charge from no charge (0) to +2, and thus lost two electrons. Copper is reduced because it changed charge from a +2 to no charge (0), and thus gained two electrons. Copper is reduced because it changed charge from a +2 to no charge (0), and thus gained two electrons. Chlorine is not involved in the reaction because it did not change its charge. Chlorine is not involved in the reaction because it did not change its charge. Oxidation and Reduction Reduction Oxidation

8. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 14-8 Oxidizing and Reducing Agents Oxidizing agent Oxidizing agent A reactant that gains electrons and is reduced is the oxidizing agent because it accepts the electrons that are lost by the reactant that is oxidized. A reactant that gains electrons and is reduced is the oxidizing agent because it accepts the electrons that are lost by the reactant that is oxidized. Reducing agent Reducing agent A reactant that loses electrons and is oxidized is the reducing agent because it provides electrons to the reactant that gets reduced. A reactant that loses electrons and is oxidized is the reducing agent because it provides electrons to the reactant that gets reduced.

9. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 14-9 Oxidizing and Reducing Agents 0 2+ 1  2+ 1  0 0 2+ 1  2+ 1  0 Zn(s) + CuCl 2 (aq) → ZnCl 2 (aq) + Cu(s) Zinc is oxidized and loses two electrons to copper, making zinc the reducing agent. Zinc is oxidized and loses two electrons to copper, making zinc the reducing agent. Copper is reduced and gains two electrons from zinc, making copper the oxidizing agent. Copper is reduced and gains two electrons from zinc, making copper the oxidizing agent. Reduced & oxidizing agent Oxidized & reducing agent

10. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 14-10 Activity: Oxidation-Reduction Reactions As shown in the figure, when iron metal is placed in a solution of copper(II) nitrate, Cu(NO 3 ) 2, a single- displacement reaction occurs in which copper deposits on the surface of the iron and aqueous iron(III) nitrate forms. As shown in the figure, when iron metal is placed in a solution of copper(II) nitrate, Cu(NO 3 ) 2, a single- displacement reaction occurs in which copper deposits on the surface of the iron and aqueous iron(III) nitrate forms. a) Write a balanced chemical equation for this reaction. b) Identify the element oxidized and the element reduced. c) Identify the oxidizing and reducing agents. d) On a molecular level, what’s happening at the surface of the iron metal? Figure from p. 576

11. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 14-11 Activity Solutions: Oxidation- Reduction Reactions a) Write a balanced chemical equation for this reaction. 2Fe(s) + 3Cu(NO 3 ) 2 (aq) → 2Fe(NO 3 ) 3 (aq) + 3Cu(s) b) Identify the element oxidized and the element reduced. 0 2+ 5+ 2  3+ 5+ 2- 0 0 2+ 5+ 2  3+ 5+ 2- 0 2Fe(s) + 3Cu(NO 3 ) 2 (aq) → 2Fe(NO 3 ) 3 (aq) + 3Cu(s) 2Fe(s) + 3Cu(NO 3 ) 2 (aq) → 2Fe(NO 3 ) 3 (aq) + 3Cu(s) Iron is oxidized Copper is reduced Reduction – gains electrons, decreases charge Oxidation – loses electrons, increases charge

12. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 14-12 Activity Solutions: Oxidation- Reduction Reactions c) Identify the oxidizing and reducing agents. 0 2+ 5+ 2  3+ 5+ 2  0 0 2+ 5+ 2  3+ 5+ 2  0 2Fe(s) + 3Cu(NO 3 ) 2 (aq) → 2Fe(NO 3 ) 3 (aq) + 3Cu(s) 2Fe(s) + 3Cu(NO 3 ) 2 (aq) → 2Fe(NO 3 ) 3 (aq) + 3Cu(s) Iron is oxidized, and therefore acts as the reducing agent. Copper is reduced, and therefore acts as the oxidizing agent. Reduced and oxidizing agent Oxidized and reducing agent

13. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 14-13 d) On a molecular level, what’s happening at the surface of the iron metal? On the surface of the iron metal, iron atoms act according to the half-reaction: act according to the half-reaction: Fe(s) → Fe 3+ + 3e  and move from the nail as a solid into and move from the nail as a solid into solution as ions. solution as ions. The copper atoms act according to the half-reaction: Cu 2+ + 2e  → Cu(s) and move from the solution as ions onto the and move from the solution as ions onto the nail as a solid. nail as a solid. Activity Solutions: Oxidation- Reduction Reactions

14. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 14-14 Oxidation Numbers A charge assigned to the atoms in any compound A charge assigned to the atoms in any compound Also called oxidation states Also called oxidation states For ionic compounds, the oxidation number is the same as the ion’s charge. For ionic compounds, the oxidation number is the same as the ion’s charge. We treat covalent compounds as if they’re ionic compounds, assigning oxidation numbers to elements using oxidation number rules. We treat covalent compounds as if they’re ionic compounds, assigning oxidation numbers to elements using oxidation number rules. We use rules to assign oxidation numbers (Table 14.1). We use rules to assign oxidation numbers (Table 14.1). The rules are a hierarchy. The first rule that applies takes precedence over any subsequent rules that may apply. The rules are a hierarchy. The first rule that applies takes precedence over any subsequent rules that may apply.

15. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 14-15 Rules to Assign Oxidation Numbers

16. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 14-16 Some Notes on the Rules for Oxidation Numbers For an isolated atom or a molecule that contains only one element, only rule 1 need be applied. An uncombined element, whether occurring as an atom such as He or a molecule such as H 2, has an oxidation number of 0. Thus, the oxidation number of sulfur in S, S 2, and S 8 is 0. For an isolated atom or a molecule that contains only one element, only rule 1 need be applied. An uncombined element, whether occurring as an atom such as He or a molecule such as H 2, has an oxidation number of 0. Thus, the oxidation number of sulfur in S, S 2, and S 8 is 0. A monoatomic ion has an oxidation number equal to its ionic charge, according to rule 2. A monoatomic ion has an oxidation number equal to its ionic charge, according to rule 2.

17. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 14-17 Some Notes on the Rules for Oxidation Numbers Rule 2 can be applied whenever oxidation numbers have been assigned to all but one element. Rule 2 can be applied whenever oxidation numbers have been assigned to all but one element. Rule 5 uses position in the periodic table to cover situations that are not handled by one of the other rules. Rule 5 uses position in the periodic table to cover situations that are not handled by one of the other rules. To assign oxidation numbers to binary compounds, handle one of the elements with an appropriate rule and the other with rule 2. To assign oxidation numbers to binary compounds, handle one of the elements with an appropriate rule and the other with rule 2. Total positive +Total negative = net charge Total positive +Total negative = net charge oxidation numbers oxidation numbers

18. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 14-18 Activity: Assigning Oxidation Numbers Assign oxidation numbers to each element in the following formulas: Assign oxidation numbers to each element in the following formulas: 1. Mg 2. Mn 2 O 3 3. Na 2 SO 4 4. Cr 2 O 7 2 

19. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 14-19 Activity Solutions: Assigning Oxidation Numbers 1. Mg – uncombined element, oxidation number is 0 2. Mn 2 O 3 – the charge on the oxygen is 2  (according to Rule 6); thus, the oxidation number on Mn must be 3+ 2  Mn 2 O 3 According to Rule 2, the sum of the oxidation numbers must equal the total charge, in this case 0. Using y to designate the unknown charge on the Mn: 2y + (3 ×  2) = 0 y = +3

20. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 14-20 Activity Solutions: Assigning Oxidation Numbers 3. Na 2 SO 4 - the charge on the oxygen is 2  (according to Rule 6) and the charge on the Na is 1+ (according to Rule 5); thus, the oxidation number on S must be 6+ 1+ y 2  1+ y 2  Na 2 SO 4 According to Rule 2, the sum of the oxidation numbers must equal the total charge, in this case 0. Using y to designate the unknown charge on S: (2 × +1) + y + (4 × –2) = 0 y = +6

21. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 14-21 Activity Solutions: Assigning Oxidation Numbers 4. Cr 2 O 7 2  - the charge on the oxygen is 2- (according to Rule 6) and the total charge for the polyatomic ion is 2-; thus, the oxidation number on Cr must be 6+ y 2  y 2  Cr 2 O 7 2  According to Rule 2, the sum of the oxidation numbers must equal the total charge, in this case 2  (the charge on the polyatomic ion). Using y to designate the unknown charge on the Cr: 2y + (7 ×  2) =  2 2y + (7 ×  2) =  2 y = +6

22. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 14-22 Identifying Oxidation-Reduction Reactions An oxidation-reduction reaction occurs if one or more elements changes its oxidation number. An oxidation-reduction reaction occurs if one or more elements changes its oxidation number. When the oxidation number increases, electrons have been lost and oxidation has occurred. When the oxidation number increases, electrons have been lost and oxidation has occurred. When the oxidation number decreases, electrons have been gained and reduction has occurred. When the oxidation number decreases, electrons have been gained and reduction has occurred. If a reaction has an uncombined element on one side that is combined on the opposite side, the reaction is an oxidation-reduction reaction. If a reaction has an uncombined element on one side that is combined on the opposite side, the reaction is an oxidation-reduction reaction.

23. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 14-23 Activity: Identifying Oxidation- Reduction Reactions Determine which of the following represent oxidation-reduction reactions. Determine which of the following represent oxidation-reduction reactions. For reactions that involve oxidation-reduction, identify the oxidizing and reducing agents. For reactions that involve oxidation-reduction, identify the oxidizing and reducing agents. 1. 2H 2 O(l) → 2H 2 (g) + O 2 (g) 2. HCl(aq) + NaOH(aq) → NaCl(aq) + H 2 O(l) 3. 4Al(s) + 3O 2 (g) → 2Al 2 O 3 (s) 4. Mg(s) + Zn(NO 3 ) 2 (aq) → Mg(NO 3 ) 2 (aq) + Zn(s)

24. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 14-24 Activity Solutions: Identifying Oxidation-Reduction Reactions 1. 2H 2 O(l) → 2H 2 (g) + O 2 (g) This is an oxidation-reduction reaction because the oxidation numbers change in the reaction. The oxidation numbers for the reaction are: 1 2  0 0 1 2  0 0 2H 2 O(l) → 2H 2 (g) + O 2 (g) Reduction and Oxidizing agent Oxidation and Reducing agent

25. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 14-25 Activity Solutions: Identifying Oxidation-Reduction Reactions 1+ 1  1+ 2  1+ 1+ 1  1+ 2  1+ 1  1+ 2  1+ 1+ 1  1+ 2  2. HCl(aq) + NaOH(aq) → NaCl(aq) + H 2 O(l) This is not an oxidation-reduction reaction. It is an acid-base neutralization. Notice how none of the elements change charge in the reaction.

26. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 14-26 Activity Solutions: Identifying Oxidation-Reduction Reactions 3. 4 Al(s) + 3O 2 (g) → 2Al 2 O 3 (s) This is an oxidation-reduction reaction because the oxidation numbers change in the reaction. The oxidation numbers for the reaction are: 0 0 3+ 2  0 0 3+ 2  4Al(s) + 3O 2 (g) → 2Al 2 O 3 (s) 4Al(s) + 3O 2 (g) → 2Al 2 O 3 (s) Reduction and Oxidizing agent Oxidation and Reducing agent

27. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 14-27 Activity Solutions: Identifying Oxidation-Reduction Reactions 4. Mg(s) + Zn(NO 3 ) 2 (aq) → Mg(NO 3 ) 2 (aq) + Zn(s) This is an oxidation-reduction reaction because the oxidation numbers change. The oxidation numbers for the reaction are: 0 2+ 5+ 2– 2+ 5+ 2– 0 0 2+ 5+ 2– 2+ 5+ 2– 0 Mg(s) + Zn(NO 3 ) 2 (aq) → Mg(NO 3 ) 2 (aq) + Zn(s) Reduction and Oxidizing agent Oxidation and Reducing agent

28. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 14-28 Oxidation and Reduction Half reactions represent either the oxidation or the reduction that occurs in separate compartments of a voltaic cell. Half reactions represent either the oxidation or the reduction that occurs in separate compartments of a voltaic cell. Each compartment is called a half-cell. Each compartment is called a half-cell. Oxidation half-reaction Oxidation half-reaction Zn(s) → Zn 2+ (aq) + 2e – Reduction half-reaction Reduction half-reaction Cu 2+ (aq) + 2e – → Cu(s) Notice that because the electrons exist on either side of the arrow in the two reactions, they cancel out. Notice that because the electrons exist on either side of the arrow in the two reactions, they cancel out. The separation of the oxidation and reduction processes into half-cells harnesses the energy released from a chemical reaction to generate electricity. The separation of the oxidation and reduction processes into half-cells harnesses the energy released from a chemical reaction to generate electricity.

29. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 14-29 Activity: Oxidation and Reduction in Batteries The reaction that occurs in most camera batteries is: The reaction that occurs in most camera batteries is: Zn(s) + Ag 2 O(s) → ZnO(s) + 2Ag(s) What is the oxidizing agent? What is the oxidizing agent? What is the reducing agent? What is the reducing agent?

30. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 14-30 Activity Solutions: Oxidation and Reduction in Batteries The reaction that occurs in most camera batteries is: The reaction that occurs in most camera batteries is: 0 1+ 2–2+ 2– 0 0 1+ 2–2+ 2– 0 Zn(s) + Ag 2 O(s) → ZnO(s) + 2Ag(s) Zinc is oxidized and thus, the reducing agent. Zinc is oxidized and thus, the reducing agent. Silver is reduced and thus, the oxidizing agent. Silver is reduced and thus, the oxidizing agent.

31. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 14-31 14.4 Balancing Simple Oxidation- Reduction Equations The reaction between silver nitrate and copper can be written simply as: The reaction between silver nitrate and copper can be written simply as: Cu(s) + Ag + (aq) → Cu 2+ (aq) + Ag(s) Does this equation appear balanced to you? Do the atoms balance? Does the charge balance? Does this equation appear balanced to you? Do the atoms balance? Does the charge balance?

32. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 14-32 14.4 Balancing Simple Oxidation- Reduction Equations The reaction between silver nitrate and copper can be written simply as: The reaction between silver nitrate and copper can be written simply as: Cu(s) + Ag + (aq) → Cu 2+ (aq) + Ag(s) Does this equation seem balanced to you? Do the atoms balance? Does the charge balance? Does this equation seem balanced to you? Do the atoms balance? Does the charge balance? You probably noticed that the total of the charges of the reactants is 1+, and the total of the charges of the products is 2+. Electrons lost do not equal electrons gained. You probably noticed that the total of the charges of the reactants is 1+, and the total of the charges of the products is 2+. Electrons lost do not equal electrons gained. The equation is NOT balanced. The equation is NOT balanced.

33. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 14-33 Balancing Simple Equations Cu(s) + Ag + (aq) → Cu 2+ (aq) + Ag(s) Separating the reaction into half-reactions: Separating the reaction into half-reactions: Oxidation: Cu(s) → Cu 2+ (aq) + 2e – Reduction: Ag + (aq) + e – → Ag(s) The electrons lost must equal the electrons gained, so we need to multiply the reduction half-reaction by 2: The electrons lost must equal the electrons gained, so we need to multiply the reduction half-reaction by 2: 2[Ag + (aq) + e – → Ag(s)] 2Ag + (aq) + 2e – → 2Ag(s)

34. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 14-34 Balancing Simple Equations Now we can add the two half-reactions: Now we can add the two half-reactions: Cu(s) → Cu 2+ (aq) + 2e – Cu(s) → Cu 2+ (aq) + 2e – 2 Ag + (aq) + 2e – → 2Ag(s) 2 Ag + (aq) + 2e – → 2Ag(s) Cu(s) + 2Ag + (aq) + 2e – → Cu 2+ (aq) + 2e – + 2Ag(s) Cu(s) + 2Ag + (aq) + 2e – → Cu 2+ (aq) + 2e – + 2Ag(s) Notice that 2 electrons appear as a reactant and a product. They cancel, and the overall reaction is: Notice that 2 electrons appear as a reactant and a product. They cancel, and the overall reaction is: Cu(s) + 2Ag + (aq) → Cu 2+ (aq) + 2Ag(s)

35. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 14-35 Balancing Simple Equations Cu(s) + 2Ag + (aq) → Cu 2+ (aq) + 2Ag(s) Double-checking the numbers of atoms and the overall charge: ReactantsProducts Cu11 Ag22 Charge2+2+

36. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 14-36 Balancing Simple Equations Figure 14.17

37. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 14-37 Balancing Simple Equations Some questions to guide you through the balancing process: Some questions to guide you through the balancing process: 1. Which substance is oxidized and which is reduced? To answer this question, assign oxidation numbers to all the elements in the reaction. 2. What are the half-reactions for the oxidation and reduction processes? For each half-reaction, a) Which element changes in oxidation number? b) Can any spectator ions be ignored until the final balancing step? c) How many electrons must be added to the appropriate side of the equation to account for the change in oxidation number?

38. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 14-38 Balancing Simple Equations Some questions to guide you through the balancing process: Some questions to guide you through the balancing process: 3. What factor can be used to multiply each coefficient in the balanced half-reactions to equalize the number of electrons gained or lost? 4. When adding the two half-reactions, can any substances that are present in equal amounts on both sides of the equation be canceled out? After answering these questions, you should have a balanced overall equation. After answering these questions, you should have a balanced overall equation. Make sure the sum of the atoms and the sum of the charges equal on either side of the equation. Make sure the sum of the atoms and the sum of the charges equal on either side of the equation.

39. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 14-39 Activity: Balancing Simple Equations Magnesium metal reacts with aqueous chromium(III) nitrate to produce magnesium nitrate, Mg(NO 3 ) 2, and solid chromium. Write a balanced equation for this reaction.

40. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 14-40 Activity Solutions: Balancing Simple Equations Magnesium metal reacts with aqueous chromium(III) nitrate to produce magnesium nitrate, Mg(NO 3 ) 2, and solid chromium. Write a balanced equation for this reaction. We begin by writing formulas for reactant and products and writing a skeletal equation: Mg(s) + Cr(NO 3 ) 3 (aq) → Mg(NO 3 ) 2 (aq) + Cr(s) To determine which substances are oxidized and reduced, we write oxidation numbers for the reactants and products: 0 3+ 5+ 2– 2+ 5+ 2–0 0 3+ 5+ 2– 2+ 5+ 2–0 Mg(s) + Cr(NO 3 ) 3 (aq) → Mg(NO 3 ) 2 (aq) + Cr(s)

41. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 14-41 Activity Solutions: Balancing Simple Equations Since the nitrate, NO 3 –, is not involved in the oxidation-reduction reaction, we can write a skeletal ionic equation: Mg(s) + Cr 3+ (aq) → Mg 2+ (aq) + Cr(s) Next, we can write half-reactions for the oxidation and reduction reactions: Oxidation: Mg(s) → Mg 2+ (aq) + 2e – Oxidation: Mg(s) → Mg 2+ (aq) + 2e – Reduction: Cr 3+ (aq) + 3e – → Cr(s) Notice: The electrons lost do not equal the electrons gained.

42. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 14-42 Activity Solutions: Balancing Simple Equations The magnesium atom loses 2 electrons, but the chromium atom gains 3 electrons. We must multiply the oxidation reaction by 3 and the reduction reaction by 2: 3[Mg(s) → Mg 2+ (aq) + 2e – ] 2[Cr 3+ (aq) + 3e – → Cr(s)] 3Mg(s) → 3Mg 2+ (aq) + 6e – 2Cr 3+ (aq) + 6e – → 2Cr(s) Now electrons lost equal electrons gained.

43. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 14-43 Activity Solutions: Balancing Simple Equations Add the two reactions together: 3Mg(s) → 3Mg 2+ (aq) + 6e – 3Mg(s) → 3Mg 2+ (aq) + 6e – 2Cr 3+ (aq) + 6e – → 2Cr(s) 2Cr 3+ (aq) + 6e – → 2Cr(s) 3 Mg(s) + 2Cr 3+ (aq) + 6e – → 2Cr(s) + 3Mg 2+ (aq) + 6e – The balanced ionic equation is: 3Mg(s) + 2Cr 3+ (aq) → 2Cr(s) + 3Mg 2+ (aq) Returning the nitrates to the equation: 3Mg(s) + 2Cr(NO 3 ) 3 (aq) → 2Cr(s) + 3Mg(NO 3 ) 2 (aq)

44. Copyright © McGraw-Hill Education. Permission required for reproduction or display. 14-44 Activity Solutions: Balancing Simple Equations 3Mg(s) + 2Cr(NO 3 ) 3 (aq) → 2Cr(s) + 3Mg(NO 3 ) 2 (aq) Double-check the number of atoms and charges: ReactantsProducts Mg33 Cr22 N66 O1818 Charge00