1. 21-1 Lecture PowerPoint Chemistry The Molecular Nature of Matter and Change Seventh Edition Martin S. Silberberg and Patricia G. Amateis Copyright  McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

2. 21-2 Chapter 21 Electrochemistry: Chemical Change and Electrical Work

3. 21-3 Electrochemistry: Chemical Change and Electrical Work 21.1Redox Reactions and Electrochemical Cells 21.2Voltaic Cells: Using Spontaneous Reactions to Generate Electrical Energy 21.3Cell Potential: Output of a Voltaic Cell 21.4Free Energy and Electrical Work 21.5Electrochemical Processes in Batteries 21.6Corrosion: An Environmental Voltaic Cell 21.7Electrolytic Cells: Using Electrical Energy to Drive Nonspontaneous Reactions

4. 21-4 Overview of Redox Reactions Oxidation is the loss of electrons and reduction is the gain of electrons. These processes occur simultaneously. The oxidizing agent takes electrons from the substance being oxidized. The oxidizing agent is therefore reduced. The reducing agent gives electrons to the substance being reduced. The reducing agent is therefore oxidized. Oxidation results in an increase in O.N. while reduction results in a decrease in O.N.

5. 21-5 Figure 21.1A summary of redox terminology, as applied to the reaction of zinc with hydrogen ion. Zn(s) + 2H + (aq) → Zn 2+ (aq) + H 2 (g) +100+2

6. 21-6 Half-Reaction Method for Balancing Redox Reactions The half-reaction method divides a redox reaction into its oxidation and reduction half-reactions. - This reflects their physical separation in electrochemical cells. This method does not require assigning O.N.s. The half-reaction method is easier to apply to reactions in acidic or basic solutions.

7. 21-7 Steps in the Half-Reaction Method Divide the skeleton reaction into two half-reactions, each of which contains the oxidized and reduced forms of one of the species. Balance the atoms and charges in each half-reaction. –First balance atoms other than O and H, then O, then H. –Charge is balanced by adding electrons (e - ) to the left side in the reduction half-reaction and to the right side in the oxidation half-reaction. If necessary, multiply one or both half-reactions by an integer so that –number of e - gained in reduction = number of e - lost in oxidation Add the balanced half-reactions, and include states of matter.

8. 21-8 Balancing Redox Reactions in Acidic Solution Cr 2 O 7 2– (aq) + I – (aq) → Cr 3+ (aq) + I 2 (s) Step 1: Divide the reaction into half-reactions. Cr 2 O 7 2 → Cr 3+ I → I 2 Step 2: Balance the atoms and charges in each half-reaction. Cr 2 O 7 2– → 2Cr 3+ Balance atoms other than O and H: Balance O atoms by adding H 2 O molecules: Cr 2 O 7 2– → 2Cr 3+ + 7H 2 O For the Cr 2 O 7 2– /Cr 3+ half-reaction:

9. 21-9 Balance H atoms by adding H + ions: 14H + + Cr 2 O 7 2– → 2Cr 3+ + 7H 2 O Balance charges by adding electrons: 6e – + 14H + + Cr 2 O 7 2- → 2Cr 3+ + 7H 2 O Balance atoms other than O and H: There are no O or H atoms, so we balance charges by adding electrons: For the I – / I 2 half-reaction: 2 I – → I 2 2 I – → I 2 + 2e – This is the reduction half-reaction. Cr 2 O 7 2- is reduced, and is the oxidizing agent. The O.N. of Cr decreases from +6 to +3. This is the oxidation half-reaction. I – is oxidized, and is the reducing agent. The O.N. of I increases from –1 to 0.

10. 21-10 Step 3: Multiply each half-reaction, if necessary, by an integer so that the number of e - lost in the oxidation equals the number of e - gained in the reduction. 3(2 I – → I 2 + 2e – ) The reduction half-reaction shows that 6e - are gained; the oxidation half-reaction shows only 2e - being lost and must be multiplied by 3: 6 I – → 3 I 2 + 6e – Step 4: Add the half-reactions, canceling substances that appear on both sides, and include states of matter. Electrons must always cancel. 6e – + 14H + + Cr 2 O 7 2– → 2Cr 3+ + 7H 2 O 6 I - → 3 I 2 + 6e – 6 I - (aq) + 14H + (aq) + Cr 2 O 7 2- (aq) → 3 I 2 (s) + 7H 2 O(l) + 2Cr 3+ (aq)

11. 21-11 Balancing Redox Reactions in Basic Solution An acidic solution contains H + ions and H 2 O. We use H + ions to balance H atoms. A basic solution contains OH - ions and H 2 O. To balance H atoms, we proceed as if in acidic solution, and then add one OH – ion to both sides of the equation. For every OH – ion and H + ion that appear on the same side of the equation we form an H 2 O molecule. Excess H 2 O molecules are canceled in the final step, when we cancel electrons and other common species.

12. 21-12 Sample Problem 21.1 Balancing a Redox Reaction in Basic Solution PROBLEM:Permanganate ion reacts in basic solution with oxalate ion to form carbonate ion and solid manganese dioxide. Balance the skeleton ionic equation for the reaction between NaMnO 4 and Na 2 C 2 O 4 in basic solution: MnO 4 – (aq) + C 2 O 4 2– (aq) → MnO 2 (s) + CO 3 2– (aq)[basic solution] PLAN:We follow the numbered steps as described in the text, and proceed through step 4 as if this reaction occurs in acidic solution. Then we add the appropriate number of OH - ions and cancel excess H 2 O molecules. SOLUTION: Step 1: Divide the reaction into half-reactions. MnO 4 – → MnO 2 C 2 O 4 2– → CO 3 2–

13. 21-13 Sample Problem 21.1 Step 2: Balance the atoms and charges in each half-reaction. Balance atoms other than O and H: Balance O atoms by adding H 2 O molecules: MnO 4 – → MnO 2 C 2 O 4 2– → 2CO 3 2– MnO 4 – → MnO 2 + 2H 2 O2H 2 O + C 2 O 4 2– → 2CO 3 2– Balance H atoms by adding H + ions: 4H + + MnO 4 – → MnO 2 + 2H 2 O2H 2 O + C 2 O 4 2– → 2CO 3 2– + 4H + Balance charges by adding electrons: 3e – + 4H + + MnO 4 – → MnO 2 + 2H 2 O [reduction] 2H 2 O + C 2 O 4 2– → 2CO 3 2– + 4H + + 2e – [oxidation]

14. 21-14 Step 3: Multiply each half-reaction, if necessary, by an integer so that the number of e - lost in the oxidation equals the number of e - gained in the reduction. 6e – + 8H + + 2MnO 4 – → 2MnO 2 + 4H 2 Ox 2 6H 2 O + 3C 2 O 4 2– → 6CO 3 2– + 12H + + 6e – x 3 Step 4: Add the half-reactions, canceling substances that appear on both sides. 6e – + 8H + + 2MnO 4 – → 2MnO 2 + 4H 2 O 6H 2 O + 3C 2 O 4 2– → 6CO 3 2– + 12H + + 6e – 2MnO 4 – + 2H 2 O + 3C 2 O 4 2– → 2MnO 2 + 6CO 3 2– + 4H + 24 Sample Problem 21.1

15. 21-15 2MnO 4 – + 2H 2 O + 3C 2 O 4 2– + 4OH – → 2MnO 2 + 6CO 3 2– + 4H 2 O Sample Problem 21.1 Basic. Add OH – to both sides of the equation to neutralize H +, and cancel H 2 O. 2MnO 4 – + 2H 2 O + 3C 2 O 4 2– + 4OH – → 2MnO 2 + 6CO 3 2– + [4H + + 4OH – ] 2 Including states of matter gives the final balanced equation: 2MnO 4 – (aq) + 3C 2 O 4 2– (aq) + 4OH – (aq) → 2MnO 2 (s) + 6CO 3 2– (aq) + 2H 2 O(l)

16. 21-16 Electrochemical Cells A voltaic cell uses a spontaneous redox reaction (  G < 0) to generate electrical energy. - The system does work on the surroundings. A electrolytic cell uses electrical energy to drive a nonspontaneous reaction (  G > 0). - The surroundings do work on the system. Both types of cell are constructed using two electrodes placed in an electrolyte solution. The anode is the electrode at which oxidation occurs. The cathode is the electrode at which reduction occurs.

17. 21-17 Figure 21.2General characteristics of (A) voltaic and (B) electrolytic cells.

18. 21-18 Spontaneous Redox Reactions A strip of zinc metal in a solution of Cu 2+ ions will react spontaneously: Cu 2+ (aq) + 2e – → Cu(s)[reduction] Zn(s) → Zn 2+ (aq) + 2e – [oxidation] Cu 2+ (aq) + Zn(s) → Zn 2+ (aq) + Cu(s) Zn is oxidized, and loses electrons to Cu 2+. Although e – are being transferred, electrical energy is not generated because the reacting substances are in the same container.

19. 21-19 Figure 21.3The spontaneous reaction between zinc and copper(II) ion.

20. 21-20 Construction of a Voltaic Cell Each half-reaction takes place in its own half-cell, so that the reactions are physically separate. Each half-cell consists of an electrode in an electrolyte solution. The half-cells are connected by the external circuit. A salt bridge completes the electrical circuit.

21. 21-21 Figure 21.4AA voltaic cell based on the zinc-copper reaction. Oxidation half-reaction Zn(s) → Zn 2+ (aq) + 2e – Reduction half-reaction Cu 2+ (aq) + 2e – → Cu(s) Overall (cell) reaction Zn(s) + Cu 2+ (aq) → Zn 2+ (aq) + Cu(s)

22. 21-22 Operation of the Voltaic Cell Oxidation (loss of e – ) occurs at the anode, which is therefore the source of e –. Zn(s) → Zn 2+ (aq) + 2e – Over time, the Zn(s) anode decreases in mass and the [Zn 2+ ] in the electrolyte solution increases. Reduction (gain of e – ) occurs at the cathode, where the e – are used up. Cu 2+ (aq) + 2e – → Cu(s) Over time, the [Cu 2+ ] in this half-cell decreases and the mass of the Cu(s) cathode increases.

23. 21-23 Figure 21.4BA voltaic cell based on the zinc-copper reaction. Zn(s) + Cu 2+ (aq) → Zn 2+ (aq) + Cu(s) Oxidation half-reaction Zn(s) → Zn 2+ (aq) + 2e – After several hours, the Zn anode weighs less as Zn is oxidized to Zn 2+. Reduction half-reaction Cu 2+ (aq) + 2e – → Cu(s) The Cu cathode gains mass over time as Cu 2+ ions are reduced to Cu.

24. 21-24 Charges of the Electrodes The anode produces e - by the oxidation of Zn(s). The anode is the negative electrode in a voltaic cell. Zn(s) + Cu 2+ (aq) → Zn 2+ (aq) + Cu(s) Electrons flow through the external wire from the anode to the cathode, where they are used to reduce Cu 2+ ions. The cathode is the positive electrode in a voltaic cell.

25. 21-25 The Salt Bridge The salt bridge completes the electrical circuit and allows ions to flow through both half-cells. As Zn is oxidized at the anode, Zn 2+ ions are formed and enter the solution. Cu 2+ ions leave solution to be reduced at the cathode. The salt bridge maintains electrical neutrality by allowing excess Zn 2+ ions to enter from the anode, and excess negative ions to enter from the cathode. A salt bridge contains nonreacting cations and anions, often K + and NO 3 –, dissolved in a gel.

26. 21-26 Flow of Charge in a Voltaic Cell Zn(s) → Zn 2+ (aq) + 2e – Cu 2+ (aq) + 2e – → Cu(s) Electrons flow through the wire from anode to cathode. Cations move through the salt bridge from the anode solution to the cathode solution. Zn 2+ Anions move through the salt bridge from the cathode solution to the anode solution. SO 4 2– By convention, a voltaic cell is shown with the anode on the left and the cathode on the right.

27. 21-27 Active and Inactive Electrodes An inactive electrode provides a surface for the reaction and completes the circuit. It does not participate actively in the overall reaction. - Inactive electrodes are necessary when none of the reaction components can be used as an electrode. An active electrode is an active component in its half- cell and is a reactant or product in the overall reaction. Inactive electrodes are usually unreactive substances such as graphite or platinum.

28. 21-28 Figure 21.5A voltaic cell using inactive electrodes. Reduction half-reaction MnO 4 – (aq) + 8H + (aq) + 5e – → Mn 2+ (aq) + 4H 2 O(l) Oxidation half-reaction 2 I – (aq) → I 2 (s) + 2e – Overall (cell) reaction 2MnO 4 – (aq) + 16H + (aq) + 10 I – (aq) → 2Mn 2+ (aq) + 5 I 2 (s) + 8H 2 O(l)

29. 21-29 Notation for a Voltaic Cell Zn(s)│Zn 2+ (aq)║Cu 2+ (aq) │Cu(s) The anode components are written on the left. The cathode components are written on the right. The single line shows a phase boundary between the components of a half-cell. The double line shows that the half- cells are physically separated. The components of each half-cell are written in the same order as in their half-reactions. If needed, concentrations of dissolved components are given in parentheses. (If not stated, it is assumed that they are 1 M.)

30. 21-30 graphite│ I – (aq)│ I 2 (s)║MnO 4 – (aq), H + (aq), Mn 2+ (aq)│graphite Notation for a Voltaic Cell The inert electrode is specified. A comma is used to show components that are in the same phase.

31. 21-31 Sample Problem 21.2 Describing a Voltaic Cell with Diagram and Notation PROBLEM: Draw a diagram, show balanced equations, and write the notation for a voltaic cell that consists of one half-cell with a Cr bar in a Cr(NO 3 ) 3 solution, another half-cell with an Ag bar in an AgNO 3 solution, and a KNO 3 salt bridge. Measurement indicates that the Cr electrode is negative relative to the Ag electrode. PLAN:From the given contents of the half-cells, we write the half- reactions. To determine which is the anode compartment (oxidation) and which is the cathode (reduction), we note the relative electrode charges. Electrons are released into the anode during oxidation, so it has a negative charge. Since Cr is negative, it must be the anode, and Ag is the cathode.

32. 21-32 SOLUTION: Sample Problem 21.2 Ag + (aq) + e – → Ag(s)[reduction; cathode] Cr(s) → Cr 3+ (aq) + 3e – [oxidation; anode] 3Ag + + Cr(s) → 3Ag(s) + Cr 3+ (aq) The half-reactions are: The balanced overall equation is: The cell notation is given by: Cr(s)│Cr 3+ (aq)║Ag + (aq)│Ag(s) The cell diagram shows the anode on the left and the cathode on the right.

33. 21-33 Electrical Potential and the Voltaic Cell When the switch is closed and no reaction is occurring, each half-cell is in an equilibrium state: Zn(s) Zn 2+ (aq) + 2e – (in Zn metal) Cu(s) Cu 2+ (aq) + 2e – (in Cu metal) Zn is a stronger reducing agent than Cu, so the position of the Zn equilibrium lies farther to the right. Zn has a higher electrical potential than Cu. When the switch is closed, e – flow from Zn to Cu to equalize the difference in electrical potential. The spontaneous reaction occurs as a result of the different abilities of these metals to give up their electrons.

34. 21-34 Cell Potential A voltaic cell converts the  G of a spontaneous redox reaction into the kinetic energy of electrons. The cell potential (E cell ) of a voltaic cell depends on the difference in electrical potential between the two electrodes. Cell potential is also called the voltage of the cell or the electromotive force (emf). E cell > 0 for a spontaneous process.

35. 21-35 Table 21.1 Voltages of Some Voltaic Cells Voltaic CellVoltage (V) Common alkaline flashlight battery 1.5 Lead-acid car battery (6 cells ≈ 12 V) 2.1 Calculator battery (mercury) 1.3 Lithium-ion laptop battery 3.7 Electric eel (~5000 cells in 6-ft eel = 750 V) 0.15 Nerve of giant squid (across cell membrane) 0.070

36. 21-36 Figure 21.6Measuring the standard cell potential of a zinc- copper cell. The standard cell potential is designated E° cell and is measured at a specified temperature with no current flowing and all components in their standard states.

37. 21-37 Standard Electrode Potentials The standard electrode potential (E° half-cell ) is the potential of a given half-reaction when all components are in their standard states. By convention, all standard electrode potentials refer to the half-reaction written as a reduction. The standard cell potential depends on the difference between the abilities of the two electrodes to be reduced (i.e., to act as oxidizing agents). E° cell = E° cathode (reduction) – E° anode (oxidation)

38. 21-38 Sample Problem 21.3 Using E° half-cell to Find E° cell PROBLEM: From the following skeleton ionic equation, write a balanced equation and calculate E° cell to decide whether the reaction: Mn 2+ (aq) + Br 2 (l) → MnO 4 – (aq) + Br – (aq)[acidic solution] is spontaneous. PLAN: We balance the skeleton reaction. Then, we look up the E° half-cell values in Appendix D and use Equation 21.3 to find E° cell. If E° cell is positive, the reaction is spontaneous. SOLUTION: Balancing the skeleton reaction: 2[Mn 2+ (aq) + 4H 2 O(l) → MnO 4 – (aq) + 8H + (aq) + 5e – ][oxidation] 5[Br 2 (l) + 2e – → 2Br – (aq)][reduction] 2Mn 2+ (aq) + 5Br 2 (l) + 8H 2 O(l) → 2MnO 4 – (aq) + 10Br – (aq) + 16H + (aq) [overall] Using E° half-cell values to find E° cell : E° cell = E° cathode (reduction) − E° anode (oxidation) = 1.07 V − 1.51 V = −0.44 V The negative sign implies the cell reaction is not spontaneous.

39. 21-39 The Standard Hydrogen Electrode Half-cell potentials are measured relative to a standard reference half-cell. The standard hydrogen electrode has a standard electrode potential defined as zero (E° reference = 0.00 V). This standard electrode consists of a Pt electrode with H 2 gas at 1 atm bubbling through it. The Pt electrode is immersed in 1 M strong acid. 2H + (aq; 1 M) + 2e – H 2 (g; 1 atm)E° ref = 0.00V

40. 21-40 Figure 21.7Determining an unknown E° half-cell with the standard reference (hydrogen) electrode. Oxidation half-reaction Zn(s) → Zn 2+ (aq) + 2e − Reduction half-reaction 2H 3 O + (aq) + 2e – → H 2 (g) + 2H 2 O(l) Overall (cell) reaction Zn(s) + 2H 3 O + (aq) → Zn 2+ (aq) + H 2 (g) + 2H 2 O(l)

41. 21-41 Sample Problem 21.4 Calculating an Unknown E° half-cell from E° cell PROBLEM:A voltaic cell houses the reaction between aqueous bromine and zinc metal: Br 2 (aq) + Zn(s) → Zn 2+ (aq) + 2Br – (aq) E° cell = 1.83 V. Calculate E° bromine, given that E° zInc = –0.76 V PLAN:E° cell is positive, so the reaction is spontaneous as written. By dividing the reaction into half-reactions, we see that Br 2 is reduced and Zn is oxidized; thus, the zinc half-cell contains the anode. We can use the equation for E° cell to calculate E° bromine. SOLUTION: Br 2 (aq) + 2e – → 2Br - (aq)[reduction; cathode] Zn(s) → Zn 2+ (aq) + 2e – [oxidation; anode] E° zinc = –0.76 V

42. 21-42 Sample Problem 21.4 E° cell = E° cathode − E° anode 1.83 = E° bromine – (–0.76) 1.83 + (–0.76) = E° bromine E° bromine = 1.07 V

43. 21-43 Comparing E° half-cell values Standard electrode potentials refer to the half-reaction as a reduction. E° values therefore reflect the ability of the reactant to act as an oxidizing agent. The more positive the E° value, the more readily the reactant will act as an oxidizing agent. The more negative the E° value, the more readily the product will act as a reducing agent.

44. 21-44 Table 21.2 Selected Standard Electrode Potentials (298 K) Half-ReactionEº(V) +2.87 −3.05 +1.36 +1.23 +0.96 +0.80 +0.77 +0.40 +0.34 0.00 −0.23 −0.44 −0.83 −2.71 strength of reducing agent strength of oxidizing agent F 2 (g) + 2e − 2F − (aq) Cl 2 (g) + 2e − 2Cl − (aq) MnO 2 (s) + 4H + (aq) + 2e − Mn 2+ (aq) + 2H 2 O(l) NO 3 - (aq) + 4H + (aq) + 3e − NO(g) + 2H 2 O(l) Ag + (aq) + e − Ag(s) Fe 3+ (aq) + e − Fe 2+ (aq) O 2 (g) + 2H 2 O(l) + 4e − 4OH − (aq) Cu 2+ (aq) + 2e − Cu(s) N 2 (g) + 5H + (aq) + 4e − N 2 H 5 + (aq) Fe 2+ (aq) + 2e − Fe(s) 2H 2 O(l) + 2e − H 2 (g) + 2OH − (aq) Na + (aq) + e − Na(s) Li + (aq) + e − Li(s) 2H + (aq) + 2e − H 2 (g) −076 Zn 2+ (aq) + 2e − Zn(s)

45. 21-45 Writing Spontaneous Redox Reactions Each half-reaction contains both a reducing agent and an oxidizing agent. The stronger oxidizing and reducing agents react spontaneously to form the weaker ones. A spontaneous redox reaction (E° cell > 0) will occur between an oxidizing agent and any reducing agent that lies below it in the emf series (i.e., one that has a less positive value for E°). The oxidizing agent is the reactant from the half-reaction with the more positive E° half-cell.

46. 21-46 Using half-reactions to write a spontaneous redox reaction: Sn 2+ (aq) + 2e – → Sn(s)E° tin = –0.14 V Ag + (aq) + e – → Ag(s) E° silver = 0.80 V Step 1: Reverse one of the half-reactions into an oxidation step so that the difference between the E° values will be positive. Here the Ag + /Ag half-reaction has the more positive E° value, so it must be the reduction. This half-reaction remains as written. We reverse the Sn 2+ /Sn half-reaction, but we do not reverse the sign: Sn(s) → Sn 2+ (aq) + 2e – E° tin = -0.14 V

47. 21-47 Step 2: Multiply the half-reactions if necessary so that the number of e – lost is equal to the number of e – gained. 2Ag + (aq) + 2e – → 2Ag(s)E° silver = 0.80 V Note that we multiply the equation but not the value for E°. Sn(s) → Sn 2+ (aq) + 2e – E° tin = –0.14 V 2Ag + (aq) + 2e – → 2Ag(s) E° silver = 0.80 V Step 3: Add the reactions together, cancelling common species. Calculate E° cell = E° cathode – E° anode. Sn(s) + 2Ag + (aq) → 2Ag(s) + Sn 2+ (aq)E° cell = 0.94 V E° cell = E° silver – E° tin = 0.80 – (–0.14) = 0.94 V

48. 21-48 Sample Problem 21.5 Writing Spontaneous Redox Reactions and Ranking Oxidizing and Reducing Agents by Strength PROBLEM:(a) Combine the following three half-reactions into three balanced equations for spontaneous reactions (A, B, and C), and calculate E° cell for each. (b) Rank the relative strengths of the oxidizing and reducing agents. (1) NO 3 – (aq) + 4H + (aq) + 3e – → NO(g) + 2H 2 O(l) E° = 0.96 V (2) N 2 (g) + 5H + (aq) + 4e – → N 2 H 5 + (aq) E° = –0.23 V (3) MnO 2 (s) + 4H + (aq) + 2e – → Mn 2+ (aq) + 2H 2 O(l) E° = 1.23 V PLAN:To write the redox equations, we combine the possible pairs of half-reactions. In each case the half-reaction with the less positive value for E° will be reversed. We make e - lost equal to e - gained, add the half-reactions and calculate E° cell. We can then rank the relative strengths of the oxidizing and reducing agents by comparing E° values.

49. 21-49 Sample Problem 21.5 SOLUTION: (a) For (1) and (2), equation (2) has the smaller, less positive E° value: (1) NO 3 – (aq) + 4H + (aq) + 3e – → NO(g) + 2H 2 O(l) E° = 0.96 V (2) N 2 H 5 + (aq) → N 2 (g) + 5H + (aq) + 4e – E° = –0.23 V We multiply equation (1) by 4 and equation (2) by 3: (1) 4NO 3 - (aq) + 16H + (aq) + 12e – → 4NO(g) + 8H 2 O(l) E° = 0.96 V (2) 3N 2 H 5 + (aq) → 3N 2 (g) + 15H + (aq) + 12e – E° = -0.23 V (A) 4NO 3 – (aq) + 3N 2 H 5 + (aq) + H + (aq) → 3N 2 (g) + 4NO(g) + 8H 2 O(l) E° cell = 0.96 V – (– 0.23 V) = 1.19 V

50. 21-50 Sample Problem 21.5 For (1) and (3), equation (1) has the smaller, less positive E° value: (1) NO(g) + 2H 2 O(l) → NO 3 – (aq) + 4H + (aq) + 3e – E° = 0.96 V (3) MnO 2 (s) + 4H + (aq) + 2e – → Mn 2+ (aq) + 2H 2 O(l) E° = 1.23 V We multiply equation (1) by 2 and equation (3) by 3: (1) 2NO(g) + 4H 2 O(l) → 2NO 3 – (aq) + 8H + (aq) + 6e – E° = 0.96 V (3) 3MnO 2 (s) + 12H + (aq) + 6e – → 3Mn 2+ (aq) + 6H 2 O(l) E° = 1.23 V (B) 3MnO 2 (s) + 4H + (aq) + 2NO(g) → 3Mn 2+ (aq) + 2NO 3 – (aq) + 2H 2 O(l) E° cell = 1.23 V – (0.96 V) = 0.27 V

51. 21-51 Sample Problem 21.5 For (2) and (3), equation (2) has the smaller, less positive E° value: (2) N 2 H 5 + (aq) → N 2 (g) + 5H + (aq) + 4e – E° = –0.23 V (3) MnO 2 (s) + 4H + (aq) + 2e – → Mn 2+ (aq) + 2H 2 O(l) E° = 1.23 V We multiply equation (3) by 2: (2) N 2 H 5 + (aq) → N 2 (g) + 5H + (aq) + 4e – E° = –0.23 V (3) 2MnO 2 (s) + 8H + (aq) + 4e – → 2Mn 2+ (aq) + 4H 2 O(l) E° = 1.23 V (C) N 2 H 5 + (aq) + 2MnO 2 (s) + 3H + (aq) → N 2 (g) + 2Mn 2+ (aq) + 4H 2 O(l) E° cell = 1.23 V – (–0.23 V) = 1.46 V

52. 21-52 Sample Problem 21.5 (b)We first rank the oxidizing and reducing agents within each equation, then we can compare E° cell values. Equation (A) Oxidizing agents: NO 3 – > N 2 Reducing Agents: N 2 H 5 + > NO Equation (B) Oxidizing agents: MnO 2 > NO 3 – Reducing Agents: N 2 H 5 + > NO Equation (C) Oxidizing agents: MnO 2 > N 2 Reducing Agents: N 2 H 5 + > Mn 2+ Comparing the relative strengths from the E° cell values: Oxidizing agents:MnO 2 > NO 3 – > N 2 Reducing agents: N 2 H 5 + > NO > Mn 2+

53. 21-53 The Activity Series of the Metals Metals that can displace H 2 from acid are metals that are stronger reducing agents than H 2. 2H + (aq) + 2e – → H 2 (g) E° = 0.00V Fe(s) → Fe 2+ + 2e – E° = –0.44 V Fe(s) + 2H + (aq) → Fe 2+ (aq) + H 2 (g) E° cell = 0.44 V The lower the metal is in the list of half-cell potentials, the more negative its E° half-cell, and the stronger it is as a reducing agent. The more negative the E° half-cell of a metal, the more positive its E° cell for the reduction of H +, and the more active a metal it is.

54. 21-54 The Activity Series of the Metals Metals that cannot displace H 2 from acid are metals that are weaker reducing agents than H 2. 2H + (aq) + 2e – → H 2 (g) E° = 0.00V 2Ag(s) → 2Ag + + 2e – E° = 0.80 V 2Ag(s) + 2H + (aq) → 2Ag + (aq) + H 2 (g) E° cell = –0.80 V The higher the metal is in the list of half-cell potentials, the more positive its E° half-cell, and the weaker it is as a reducing agent. The more positive the E° half-cell of a metal, the less positive its E° cell for the reduction of H +, and the less active a metal it is.

55. 21-55 The Activity Series of the Metals Metals that can displace H 2 from water are metals whose half-reactions lie below that of H 2 O: 2H 2 O(l) + 2e – → H 2 (g) + 2OH – (aq) E° = –0.42 V 2Na(s) → 2Na + (aq) + 2e – E° = –2.17 V 2Na(s) + 2H 2 O(l) → 2Na + (aq) + H 2 (g) + 2OH – (aq) E° cell = 2.29 V In order to reduce H 2 O to H 2, the metal should have a more negative E° half-cell than H 2 O.

56. 21-56 The Activity Series of the Metals We can also predict whether one metal can displace another from solution. Any metal that is lower in the list of electrode potentials (i.e., has a more negative E° value) will reduce the ion of a metal higher up the list. Zn(s) → Zn 2+ (aq) + 2e – E° = –0.76V Fe 2+ (aq) + 2e – → Fe(s) E° = –0.44V Zn (s) + Fe 2+ (aq) → Zn 2+ (aq) + Fe(s) E° cell = 0.32 V

57. 21-57 Figure 21.8The reaction of calcium in water. Overall (cell) reaction Ca(s) + 2H 2 O(l) → Ca(OH) 2 (aq) + H 2 (g) Oxidation half-reaction Ca(s) → Ca 2+ (aq) + 2e – Reduction half-reaction 2H 2 O(l) + 2e – → H 2 (g) + 2OH – (aq)

58. 21-58 Figure 21.9A dental “voltaic cell.” Biting down with a filled tooth on a scrap of aluminum foil will cause pain. The foil acts as an active anode (E° aluminum = –1.66 V), saliva as the electrolyte, and the filling as an inactive cathode as O 2 is reduced to H 2 O.

59. 21-59 Free Energy and Electrical Work For a spontaneous redox reaction,  G 0.  G = –nFE cell n = mol of e – transferred F is the Faraday constant = 9.65x10 4 J/V·mol e – Under standard conditions,  G° = –nFE° cell and E° cell = ln K RT nF or E° cell = log K 0.0592 V n for T = 298.15 K

60. 21-60 Figure 21.10 The interrelationship of  G°, E° cell, and K. Reaction Parameters at the Standard State G°G° KE° cell Reaction at standard- state conditions < 0> 1> 0Spontaneous 0 1 0At equilibrium > 0< 1< 0Nonspontaneous  G° E° cell K  G° = –nFE° cell  G = –RT ln K E° cell = ln K RT nF Copyright  The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

61. 21-61 Sample Problem 21.6 Calculating K and  G° from E° cell PROBLEM:Lead can displace silver from solution, and silver occurs in trace amounts in some ores of lead. Pb(s) + 2Ag + (aq) → Pb 2+ (aq) + 2Ag(s) As a consequence, silver is a valuable byproduct in the industrial extraction of lead from its ore. Calculate K and  G° at 298.15 K for this reaction. PLAN:We divide the spontaneous redox reaction into the half-reactions and use values from Appendix D to calculate E° cell. From this we can find K and  G°. SOLUTION: Writing the half-reactions with their E° values: (1) Ag + (aq) + e − → Ag(s) E° = 0.80 V (2) Pb 2+ (aq) + 2e − → Pb(s) E° = −0.13 V

62. 21-62 Sample Problem 21.6 We need to reverse equation (2) and multiply equation (1) by 2: (1) 2Ag + (aq) + 2e − → 2Ag(s) E° = 0.80 V (2) Pb(s) → Pb 2+ (aq) + 2e − E° = -0.13 V 2Ag + (aq) + Pb(s) → 2Ag(s) + Pb 2+ (aq)E cell = 0.80 – (–0.13) = 0.93 V E° cell = ln K RT nF = 0.0592 V 2 log K = 0.93 V 0.93 V x 2 0.0592 V log K = = 31.42 K = 2.6x10 31  G° = −nFE° cell x 96.5 kJ V·mol e − = − 2 mol e − mol rxn x 0.93 V = −1.8x10 2 kJ/mol rxn

63. 21-63 Cell Potential and Concentration When Q [product], ln Q E° cell When Q = 1, [reactant] = [product], ln Q = 0, so E cell = E° cell When Q > 1, [reactant] 0, so E cell < E° cell E cell = E° cell −log Q 0.0592 V n We can simplify the equation as before for T = 298.15 K: Nernst Equation E cell = E° cell −ln Q RT nF

64. 21-64 Sample Problem 21.7 Using the Nernst Equation to Calculate E cell PROBLEM:In a test of a new reference electrode, a chemist constructs a voltaic cell consisting of a Zn/Zn 2+ half-cell and an H 2 /H + half-cell under the following conditions: [Zn 2+ ] = 0.010 M [H + ] = 2.5 M P = 0.30 atm H2H2 Calculate E cell at 298.15 K. PLAN:To apply the Nernst equation and determine E cell, we must know E° cell and Q. We write the equation for the spontaneous reaction and calculate E° cell from standard electrode potentials. We must convert the given pressure to molarity in order to have consistent units. SOLUTION: (1) 2H + (aq) + 2e − → H 2 (g) E° = 0.00 V (2) Zn(s) → Zn 2+ (aq) + 2e − E° = −0.76 V 2H + (aq) + Zn(s) → H 2 (g) + Zn 2+ (aq)E° cell = 0.00 − (−0.76) = 0.76 V

65. 21-65 Sample Problem 21.7 Converting pressure to molarity: nVnV = P RT = 0.30 atm atm·L mol·K 0.0821 x 298.15 K = 1.2x10 −2 M [H 2 ][Zn 2+ ] [H + ] 2 Q = 0.012 x 0.010 (2.5) 2 = = 1.9x10 −5 Solving for E cell at 25°C (298.15 K), with n = 2: E cell = E° cell −log Q 0.0592 V n = 0.76 V − 0.0592 V 2 log(1.9x10 −5 ) = 0.76 − (−0.14 V) = 0.90 V

66. 21-66 Figure 21.11AThe relation between E cell and log Q for the zinc- copper cell. If the reaction starts with [Zn 2+ ] < [Cu 2+ ] (Q < 1), E cell is higher than the standard cell potential. As the reaction proceeds, [Zn 2+ ] decreases and [Cu 2+ ] increases, so E cell drops. Eventually the system reaches equilibrium and the cell can no longer do work.

67. 21-67 Figure 21.11BThe relation between E cell and log Q for the zinc- copper cell. A summary of the changes in E cell as any voltaic cell operates.

68. 21-68 Concentration Cells A concentration cell exploits the effect of concentration changes on cell potential. The cell has the same half-reaction in both cell compartments, but with different concentrations of electrolyte: Cu(s) → Cu 2+ (aq; 0.10 M) + 2e − [anode; oxidation] Cu 2+ (aq; 1.0 M) + 2e − → Cu(s) [cathode; reduction] Cu 2+ (aq; 1.0 M) → Cu 2+ (aq; 0.10 M) As long as the concentrations of the solutions are different, the cell potential is > 0 and the cell can do work.

69. 21-69 E cell > 0 as long as the half-cell concentrations are different. The cell is no longer able to do work once the concentrations are equal. Figure 21.12A concentration cell based on the Cu/Cu 2+ half-reaction. Overall (cell) reaction Cu 2+ (aq,1.0 M) → Cu 2+ (aq, 0.10 M) Oxidation half-reaction Cu(s) → Cu 2+ (aq, 0.10 M) + 2e − Reduction half-reaction Cu 2+ (aq, 1.0 M) + 2e − → Cu(s) Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

70. 21-70 Sample Problem 21.8 Calculating the Potential of a Concentration Cell PROBLEM:A concentration cell consists of two Ag/Ag + half-cells. In half-cell A, the electrolyte is 0.010 M AgNO 3 ; in half-cell B, it is 4.0x10 −4 M AgNO 3. What is the cell potential at 298.15 K? PLAN:The standard half-cell potentials are identical, so E° cell is zero, and we find E cell from the Nernst equation. Half-cell A has a higher [Ag + ], so Ag + ions are reduced and plate out on electrode A, which is therefore the cathode. In half-cell B, Ag atoms of the electrode are oxidized and Ag + ions enter the solution. Electrode B is thus the anode. As for all voltaic cells, the cathode is positive and the anode is negative.

71. 21-71 Sample Problem 21.8 = 0.0828 V E cell = E° cell −log 0.0592 V 1 [Ag + ] dil [Ag + ] conc = 0.0 V − 0.0592 log 4.0x10 −4 0.010 SOLUTION: The [Ag + ] decreases in half-cell A and increases in half- cell B, so the spontaneous reaction is: Ag + (aq; 0.010 M) [half-cell A] → Ag + (aq; 4.0x10 −4 M) [half-cell B]

72. 21-72 Figure 21.13Laboratory measurement of pH. The operation of a pH meter illustrates an important application of concentration cells. The glass electrode monitors the [H + ] of the solution relative to its own fixed internal [H + ]. An older style of pH meter includes two electrodes. Modern pH meters use a combination electrode.

73. 21-73 Table 21.3 Some Ions Measured with Ion-Specific Electrodes Species DetectedTypical Sample NH 3 /NH 4 + Industrial wastewater, seawater CO 2 /HCO 3 − Blood, groundwater F−F− Drinking water, urine, soil, industrial stack gases Br − Grain, plant tissue I−I− Milk, pharmaceuticals NO 3 − Soil, fertilizer, drinking water K+K+ Blood serum, soil, wine H+H+ Laboratory solutions, soil, natural waters

74. 21-74 Figure 21.14Minimicroanalysis. A microelectrode records electrical impulses of a single neuron in a monkey’s visual cortex. The electrical potential of a nerve cell is due to the difference in concentration of [Na + ] and [K + ] ions inside and outside the cell.

75. 21-75 Electrochemical Processes in Batteries A primary battery cannot be recharged. The battery is “dead” when the cell reaction has reached equilibrium. A secondary battery is rechargeable. Once it has run down, electrical energy is supplied to reverse the cell reaction and form more reactant. A battery consists of self-contained voltaic cells arranged in series, so their individual voltages are added.

76. 21-76 Alkaline battery.Figure 21.15 Anode (oxidation): Zn(s) + 2OH − (aq) → ZnO(s) + H 2 O(l) + 2e − Cathode (reduction): MnO 2 (s) + 2H 2 O(l) + 2e − → Mn(OH) 2 (s) + 2OH − (aq) Overall (cell) reaction: Zn(s) + MnO 2 (s) + H 2 O(l) → ZnO(s) + Mn(OH) 2 (s) E cell = 1.5 V

77. 21-77 Silver button battery.Figure 21.16 Anode (oxidation): Zn(s) + 2OH − (aq) → ZnO(s) + H 2 O(l) + 2e − Cathode (reduction): Ag 2 O(s) + H 2 O(l) + 2e − → 2Ag(s) + 2OH − (aq) Overall (cell) reaction: Zn(s) + Ag 2 O(s) → ZnO(s) + 2Ag(s) E cell = 1.6 V The mercury battery uses HgO as the oxidizing agent instead of Ag 2 O and has cell potential of 1.3 V.

78. 21-78 Figure 21.17Lithium battery. Anode (oxidation): 3.5Li(s) → 3.5Li + + 3.5e − Cathode (reduction): AgV 2 O 5.5 + 3.5Li + + 3.5e − → Li 3.5 V 2 O 5.5 Overall (cell) reaction: AgV 2 O 5.5 + 3.5Li(s) → Li 3.5 V 2 O 5.5 The primary lithium battery is widely used in watches, implanted medical devices, and remote-control devices.

79. 21-79 Lead-acid battery.Figure 21.18 The lead-acid car battery is a secondary battery and is rechargeable.

80. 21-80 Anode (oxidation): Pb(s) + HSO 4 − (aq) → PbSO 4 (s) + H + (aq) + 2e − Cathode (reduction): PbO 2 (s) + 3H + (aq) + HSO 4 − (aq) + 2e − → PbSO 4 (s) + 2H 2 O(l) Overall (cell) reaction (discharge): PbO 2 (s) + Pb(s) + H 2 SO 4 (aq) → 2PbSO 4 (s) + 2H 2 O(l) E cell = 2.1 V The reactions in a lead-acid battery: Overall (cell) reaction (recharge): 2PbSO 4 (s) + 2H 2 O(l) → PbO 2 (s) + Pb(s) + H 2 SO 4 (aq) The cell generates electrical energy when it discharges as a voltaic cell.

81. 21-81 Nickel-metal hydride batteryFigure 21.19 Anode (oxidation): MH(s) + OH − (aq) → M(s) + H 2 O(l) + e − Cathode (reduction): NiO(OH)(s) + H 2 O(l) + e − → Ni(OH) 2 (s) + OH − (aq) Overall (cell) reaction: MH(s) + NiO(OH)(s) → M(s) + Ni(OH) 2 (s) E cell = 1.4 V

82. 21-82 Lithium-ion battery.Figure 21.20 Anode (oxidation): Li x C 6 (s) → xLi + + xe − + C 6 (s) Cathode (reduction): Li 1-x Mn 2 O 4 (s) + xLi + + xe − → LiMn 2 O 4 (s) Overall (cell) reaction: Li x C 6 (s) + Li 1-x Mn 2 O 4 (s) → LiMn 2 O 4 (s) E cell = 3.7 V The secondary (rechargeable) lithium-ion battery is used to power laptop computers, cell phones, and camcorders.

83. 21-83 In a fuel cell, also called a flow cell, reactants enter the cell and products leave, generating electricity through controlled combustion. Fuel Cells Reaction rates are lower in fuel cells than in other batteries, so an electrocatalyst is used to decrease the activation energy.

84. 21-84 Figure 21.21Hydrogen fuel cell. Anode (oxidation): 2H 2 (g) → 4H + (aq) + 4e − Cathode (reduction): O 2 (g) + 4H + (aq) + 4e − → 2H 2 O(g) Overall (cell) reaction: 2H 2 (g) + O 2 (g) → 2H 2 O(g) E cell = 1.2 V

85. 21-85 Corrosion is the process whereby metals are oxidized to their oxides and sulfides. Corrosion: an Environmental Voltaic Cell The rusting of iron is a common form of corrosion. - Rusting requires moisture, and occurs more quickly at low pH, in ionic solutions, and when the iron is in contact with a less active metal. - Rust is not a direct product of the reaction between Fe and O 2, but arises through a complex electrochemical process.

86. 21-86 The Rusting of Iron Fe(s) → Fe 2+ (aq) + 2e − [anodic region; oxidation] O 2 (g) + 4H + (aq) + 4e − → 2H 2 O(l) [cathodic region; reduction] The loss of iron: 2Fe(s) + O 2 (g) + 4H + (aq) → 2Fe 2+ (aq) + 2H 2 O(l)[overall] The rusting process: 2Fe 2+ (aq) + ½O 2 (g) + (2 + n)H 2 O(l) → Fe 2 O 3 ·nH 2 O(s) + 4H + (aq) Overall reaction: H + ions are consumed in the first step, so lowering the pH increases the overall rate of the process. H + ions act as a catalyst, since they are regenerated in the second part of the process.

87. 21-87 Figure 21.22The corrosion of iron.

88. 21-88 Figure 21.23Enhanced corrosion at sea. The high ion concentration of seawater enhances the corrosion of iron in hulls and anchors.

89. 21-89 Figure 21.24The effect of metal-metal contact on the corrosion of iron. Fe in contact with Cu corrodes faster. Fe in contact with Zn does not corrode. The process is known as cathodic protection.

90. 21-90 Figure 21.25The use of sacrificial anodes to prevent iron corrosion. In cathodic protection, an active metal, such as zinc, magnesium, or aluminum, acts as the anode and is sacrificed instead of the iron.

91. 21-91 Electrolytic Cells An electrolytic cell uses electrical energy from an external source to drive a nonspontaneous redox reaction. Cu(s) → Cu 2+ (aq) + 2e − [anode; oxidation] Sn 2+ (aq) + 2e − → Sn(s) [cathode; reduction] Cu(s) + Sn 2+ (aq) → Cu 2+ (aq) + Sn(s) E° cell = −0.48 V and ΔG° = 93 kJ As with a voltaic cell, oxidation occurs at the anode and reduction takes place at the cathode. An external source supplies the cathode with electrons, which is negative, and removes them from the anode, which is positive.

92. 21-92 Figure 21.26The tin-copper reaction as the basis of a voltaic and an electrolytic cell. voltaic cell Sn(s) → Sn 2+ (aq) + 2e − Cu 2+ (aq) + 2e − → Cu(s) Cu 2+ (aq) + Sn(s) → Cu(s) + Sn 2+ (aq) electrolytic cell Cu(s) → Cu 2+ (aq) + 2e − Sn 2+ (aq) + 2e − → Sn(s) Sn 2+ (aq) + Cu(s) → Sn(s) + Cu 2+ (aq) Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

93. 21-93 Figure 21.27The processes occurring during the discharge and recharge of a lead-acid battery. VOLTAIC (discharge) ELECTROLYTIC (recharge) Switch

94. 21-94 Table 21.4 Comparison of Voltaic and Electrolytic Cells Cell Type GG E cell Electrode NameProcessSign Voltaic < 0 > 0 Anode Cathode Oxidation Reduction − + Electrolytic > 0 < 0 Anode Cathode Oxidation Reduction− +

95. 21-95 Products of Electrolysis Electrolysis is the splitting (lysing) of a substance by the input of electrical energy. During electrolysis of a pure, molten salt, the cation will be reduced and the anion will be oxidized. During electrolysis of a mixture of molten salts - the more easily oxidized species (stronger reducing agent) reacts at the anode, and - the more easily reduced species (stronger oxidizing agent) reacts at the cathode.

96. 21-96 Sample Problem 21.9 Predicting the Electrolysis Products of a Molten Salt Mixture PROBLEM:A chemical engineer melts a naturally occurring mixture of NaBr and MgCl 2 and decomposes it in an electrolytic cell. Predict the substance formed at each electrode, and write balanced half-reactions and the overall cell reaction. PLAN:We need to determine which metal and nonmetal will form more easily at the electrodes. We list the ions as oxidizing or reducing agents. If a metal holds its electrons more tightly than another, it has a higher ionization energy (IE). Its cation will gain electrons more easily, and it will be the stronger oxidizing agent. If a nonmetal holds its electrons less tightly than another, it has a lower electronegativity (EN). Its anion will lose electrons more easily, and it will be the reducing agent.

97. 21-97 Sample Problem 21.9 SOLUTION: Possible oxidizing agents: Na +, Mg 2+ Possible reducing agents: Br −, Cl − Mg is to the right of Na in Period 3. IE increases from left to right across the period, so Mg has the higher IE and gives up its electrons less easily. The Mg 2+ ion has a greater attraction for e - than the Na + ion. Br is below Cl in Group 7A. EN decreases down the group, so Br accepts e - less readily than Cl. The Br - ion will lose its e - more easily, so it is more easily oxidized. Mg 2+ (l) + 2e − → Mg(l)[cathode; reduction] 2Br - (l) → Br 2 (g) + 2e − [anode; oxidation] The overall cell reaction is:Mg 2+ (l) + 2Br − (l) → Mg(l) + Br 2 (g)

98. 21-98 Figure 21.28The electrolysis of water. Oxidation half-reaction 2H 2 O(l) → O 2 (g) + 4H + (aq) + 4e − Reduction half-reaction 2H 2 O(l) + 2e − → H 2 (g) + 2OH − (aq) Overall (cell) reaction 2H 2 O(l) → 2H 2 (g) + O 2 (g)

99. 21-99 Electrolysis of Aqueous Salt Solutions When an aqueous salt solution is electrolyzed - The strongest oxidizing agent (most positive electrode potential) is reduced, and - The strongest reducing agent (most negative electrode potential) is oxidized. Overvoltage is the additional voltage needed (above that predicted by E° values) to produce gases at metal electrodes. Overvoltage needs to be taken into account when predicting the products of electrolysis for aqueous solutions. Overvoltage is 0.4 – 0.6 V for H 2 (g) or O 2 (g).

100. 21-100 Cations of less active metals (Au, Ag, Cu, Cr, Pt, Cd) are reduced to the metal. Cations of more active metals are not reduced. H 2 O is reduced instead. Anions that are oxidized, because of overvoltage from O 2 formation, include the halides, except for F −. Anions that are not oxidized include F − and common oxoanions. H 2 O is oxidized instead. Summary of the Electrolysis of Aqueous Salt Solutions

101. 21-101 Sample Problem 21.10 Predicting the Electrolysis Products of Aqueous Salt Solutions PROBLEM:What products form at which electrode during electrolysis of aqueous solution of the following salts? (a) KBr (b) AgNO 3 (c) MgSO 4 PLAN:We identify the reacting ions and compare their electrode potentials with those of water, taking the 0.4 – 0.6 V overvoltage into account. The reduction half-reaction with the less negative E° occurs at the cathode, while the oxidation half-reaction with the less positive E° occurs at the anode. Despite the overvoltage, which gives E for the reduction of water a value between −0.8 and −1.0 V, H 2 O is still easier to reduce than K +, so H 2 (g) forms at the cathode. SOLUTION: (a) KBr K + (aq) + e − → K(s) E° = −2.93 2H 2 O(l) + 2e − → H 2 (g) + 2OH − (aq) E° = −0.42V

102. 21-102 (b) AgNO 3 Ag + (aq) + e − → Ag(s) E° = 0.80 V 2H 2 O(l) + 2e − → H 2 (g) + 2OH - (aq) E° = −0.42V Sample Problem 21.10 2Br - (aq) → Br 2 (l) + 2e − E° = 1.07 V 2H 2 O(l) → O 2 (g) + 4H + (aq) + 4e − E° = 0.82 V The overvoltage places E for the oxidation of water at between 1.2 and 1.4 V. Br − is therefore easier to oxidize than water, so Br 2 (l) forms at the anode. As the cation of an inactive metal, Ag + is a better oxidizing agent than H 2 O, so Ag(s) forms at the cathode. NO 3 − cannot be oxidized, because N is already in its highest (+5) oxidation state. Thus O 2 (g) forms at the anode: 2H 2 O(l) → O 2 (g) + 4H + (aq) + 4e −

103. 21-103 (c) MgSO 4 Mg 2+ (aq) + 2e − → Mg(s) E° = −2.37 V 2H 2 O(l) + 2e − → H 2 (g) + 2OH − (aq) E° = −0.42V Sample Problem 21.10 Mg 2+ is a much weaker oxidizing agent than H 2 O, so H 2 (g) forms at the cathode. SO 4 2− cannot be oxidized, because S is already in its highest (+6) oxidation state. Thus O 2 (g) forms at the anode: 2H 2 O(l) → O 2 (g) + 4H + (aq) + 4e −

104. 21-104 Stoichiometry of Electrolysis Faraday’s law of electrolysis states that the amount of substance produced at each electrode is directly proportional to the quantity of charge flowing through the cell. The current flowing through the cell is the amount of charge per unit time. Current is measured in amperes. Current x time = charge

105. 21-105 Figure 21.29A summary diagram for the stoichiometry of electrolysis. MASS (g) of substance oxidized or reduced AMOUNT (mol) of substance oxidized or reduced CHARGE (C) CURRENT (A) AMOUNT (mol) of electrons transferred M (g/mol) balanced half-reaction Faraday constant (C/mol e − ) time (s)

106. 21-106 Sample Problem 21.11 Applying the Relationship Among Current, Time, and Amount of Substance PROBLEM:A technician plates a faucet with 0.86 g of Cr metal by electrolysis of aqueous Cr 2 (SO 4 ) 3. If 12.5 min is allowed for the plating, what current is needed? PLAN:To find the current, we divide charge by time, so we need to find the charge. We write the half-reaction for Cr 3+ reduction to get the amount (mol) of e - transferred per mole of Cr. We convert mass of Cr needed to amount (mol) of Cr. We can then use the Faraday constant to find charge and current. divide by M mass (g) of Cr needed mol of Cr 3 mol e − = 1 mol Cr divide by time in s mol e − transferred Charge (C) 1 mol e − = 9.65x10 4 C current (A)

107. 21-107 Sample Problem 21.11 SOLUTION: Cr 3+ (aq) + 3e − → Cr(s) 0.86 g Cr x 1 mol Cr 52.00 g Cr x 3 mol e - 1 mol Cr = 0.050 mol e − Charge (C) = 0.050 mol e − x 9.65x10 4 C 1 mol e − = 4.8x10 3 C Current (A) = charge (C) time (s) = 4.8x10 3 C 12.5 min x 1 min 60 s = 6.4 C/s = 6.4 A

108. 21-108 Figure B21.1The mitochondrion Chemical Connections Mitochondria are subcellular particles outside the cell nucleus that control the electron-transport chain, an essential part of energy production in living organisms.

109. 21-109 Figure B21.2The main energy-yielding steps in the electron- transport chain (ETC). Chemical Connections

110. 21-110 Figure B21.3Coupling electron transport to proton transport to ATP synthesis. Chemical Connections