Chapter19

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Electrochemistry Basics …

Electrochemistry Basics
Table of Contents

1. Introduction
2. Voltaic Cells-Galvanic Cells
3. Cell Potential
4. Balancing Redox Reactions
5. Rules for Assigning Oxidation states
6. Additional Materials
6.1. I. Conversion
6.2. II. Free Energy & Cell Potential
6.3. III. Nernst equation
6.4. At Equilibrium
7. Terminology
8. Reference
9. Outside Links
10. Contributors

As the name suggests, electrochemistry is the study of changes that cause electrons to move. This movement of electrons is called electricity. In electrochemistry, electricity can be generated by movements of electrons from one element to another in a reaction known as a redox reaction or oxidation-reduction react

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Transcript

  • 1. Chapter 19 Electrochemistry
  • 2. Contents and Concepts Half−Reactions 1. Balancing Oxidation–Reduction Reactions in Acidic and Basic Solutions Voltaic Cells 2. Construction of Voltaic Cells 3. Notation for Voltaic Cells 4. Cell Potential 5. Standard Cell Potentials and Standard Electrode Potentials Copyright © Cengage Learning. All rights reserved. 19 | 2
  • 3. 6. 7. 8. Equilibrium Constants from Cell Potentials Dependence of Cell Potentials on Concentration Some Commercial Voltaic Cells Electrolytic Cells 9. Electrolysis of Molten Salts 10. Aqueous Electrolysis 11. Stoichiometry of Electrolysis Copyright © Cengage Learning. All rights reserved. 19 | 3
  • 4. Learning Objectives Electrochemistry 1. Balancing Oxidation–Reduction Reactions in Acidic and Basic Solutions a. Learn the steps for balancing oxidation– reduction reactions using the half−reaction method. b. Balance equations by the half−reaction method (acidic solutions). c. Learn the additional steps for balancing oxidation–reduction reactions in basic solution using the half−reaction method. d. Balance equations using the half−reaction method (basic solution). Copyright © Cengage Learning. All rights reserved. 19 | 4
  • 5. 2. Construction of Voltaic Cells a. Define electrochemical cell, voltaic (galvanic cell), electrolytic cell, and half−cell. b. Describe the function of the salt bridge in a voltaic cell. c. State the reaction that occurs at the anode and the cathode in an electrochemical cell. d. Define cell reaction. e. Sketch and label a voltaic cell. Copyright © Cengage Learning. All rights reserved. 19 | 5
  • 6. 3. Notation for Voltaic Cells a. Write the cell reaction from the cell notation. 4. Cell Potential a. Define cell potential and volt. b. Calculate the quantity of work from a given amount of cell reactant. Copyright © Cengage Learning. All rights reserved. 19 | 6
  • 7. 5. Standard Cell Potentials and Standard Electrode Potentials a. Explain how the electrode potential of a cell is an intensive property. b. Define standard cell potential and standard electrode potential. c. Interpret the table of standard reduction potentials. d. Determine the relative strengths of oxidizing and reducing agents. e. Determine the direction of spontaneity from electrode potentials. f. Calculate cell potentials from standard potentials. Copyright © Cengage Learning. All rights reserved. 19 | 7
  • 8. 6. Equilibrium Constants from Cell Potentials a. Calculate the free−energy change from electrode potentials. b. Calculate the cell potential from free−energy change. c. Calculate the equilibrium constant from cell potential. 7. Dependence of Cell Potential on Concentration a. Calculate the cell potential for nonstandard conditions. b. Describe how pH can be determined using a glass electrode. Copyright © Cengage Learning. All rights reserved. 19 | 8
  • 9. 8. Some Commercial Voltaic Cells a. Describe the construction and reactions of a zinc–carbon dry cell, a lithium–iodine battery, a lead storage cell, and a nickel– cadmium cell. b. Explain the operation of a proton−exchange membrane fuel cell. c. Explain the electrochemical process of the rusting of iron. d. Define cathodic protection. 9. Electrolysis of Molten Salts a. Define electrolysis. Copyright © Cengage Learning. All rights reserved. 19 | 9
  • 10. 10. Aqueous Electrolysis a. Learn the half−reactions for water undergoing oxidation and reduction. b. Predict the half−reactions in an aqueous electrolysis. 11. Stoichiometry of Electrolysis a. Calculate the amount of charge from the amount of product in an electrolysis. b. Calculate the amount of product from the amount of charge in an electrolysis. Copyright © Cengage Learning. All rights reserved. 19 | 10
  • 11. A voltaic cell employs a spontaneous oxidation– reduction reaction as a source of energy. It separates the reaction into two half−reactions, physically separating one half−reaction from the other. Copyright © Cengage Learning. All rights reserved. 19 | 11
  • 12. Our first step in studying electrochemical cell is to balance its oxidation–reduction reaction. We will use the half−reaction method from Section 4.6 and extend it to acidic or basic solutions. In this chapter we will focus on electron transfer rather than proton transfer so the hydronium ion, H3O+(aq), will be represented by its simpler notation, H+(aq). Only the notation, not the chemistry, is different. Copyright © Cengage Learning. All rights reserved. 19 | 12
  • 13. We will be studying more complex situations, so our initial analysis is key. First, we need to identify what is being oxidized and what is being reduced. Then, we determine if the reaction is in acidic or basic conditions. Copyright © Cengage Learning. All rights reserved. 19 | 13
  • 14. For example, consider the skeleton reaction: +2 +5 +3 +2 Fe2+(aq) + NO3−(aq) → Fe3+(aq) + NO(g) acidic solution We determine the oxidation number of Fe and N in each substance. Fe2+(aq) is oxidized from +2 to +3 in Fe3+(aq). N in NO3−(aq) is reduced from +5 to +2 in NO(g). The reaction is in acidic conditions (HNO3). Copyright © Cengage Learning. All rights reserved. 19 | 14
  • 15. Now we split the skeleton reaction into half−reactions and balance each half. 1. Balance all atoms except H and O. 2. Balance O atoms by adding H2O to the side of the equation that needs O. 3. Balance H by adding H+ to the side of the equation that needs H. 4. Balance the electric charge by adding electrons (e−) to the more positive side of the equation. Copyright © Cengage Learning. All rights reserved. 19 | 15
  • 16. Oxidation half−reaction Fe2+(aq) → Fe3+(aq) + e− Reduction half−reaction First we balance N, then O: NO3−(aq) → NO(aq) + 2H2O(l) Next we balance H: 4H+(aq) + NO3−(aq) → NO(g) + 2H2O(l) Then we balance e−: 3e− + 4H+(aq) + NO3−(aq) → NO(aq) + 2H2O(l) Copyright © Cengage Learning. All rights reserved. 19 | 16
  • 17. Finally, we combine the two half−reactions. 1. Multiply each half−reaction by a factor so that each has the same number of electrons. 2. Add the two half−reactions (the electrons should cancel). 3. Simplify the overall reaction by canceling species that occur on both sides and reducing coefficients to the smallest whole numbers. 4. Check that the reaction is balanced. Copyright © Cengage Learning. All rights reserved. 19 | 17
  • 18. 3Fe2+(aq) → 3Fe3+(aq) + 3e− 3e− + 4H+(aq) + NO3−(aq) → NO(g) + 2H2O(l) 4H+(aq) + 3Fe2+(aq) + NO3−(aq) → 3Fe3+(aq) + NO(g) + 2H2O(l) Check that the reaction is balanced, in terms of both atoms and charge. 4H; 3Fe; 1N; 3O; charge is +9 Copyright © Cengage Learning. All rights reserved. 19 | 18
  • 19. ? Dichromate ion in acidic solution is an oxidizing agent. When it reacts with zinc, the metal is oxidized to Zn2+, and nitrate is reduced. Assume that dichromate ion is reduced to Cr3+. Write the balanced ionic equation for this reaction using the half−reaction method. Copyright © Cengage Learning. All rights reserved. 19 | 19
  • 20. First we determine the oxidation numbers of N and Zn: +6 +3 0 +2 Cr2O72−(aq) → Cr3+(aq) and Zn(s) → Zn2+(aq) Cr was reduced from +6 to +3. Zn was oxidized from 0 to +2. Now we balance the half−reactions. Copyright © Cengage Learning. All rights reserved. 19 | 20
  • 21. Oxidation half−reaction Zn(s) → Zn2+(aq) + 2e− Reduction half−reaction First we balance Cr and O: Cr2O72−(aq) → 2Cr3+(aq) + 7H2O(l) Next we balance H: 14H+(aq) + Cr2O72−(aq) → 2Cr3+(aq) + 7H2O(l) Then we balance e−: 6e− + 14H+(aq) + Cr2O72−(aq) → 2Cr3+(aq) + 7H2O(l) Copyright © Cengage Learning. All rights reserved. 19 | 21
  • 22. Now we combine the two half−reactions by multiplying the oxidation half reaction by 3. 3Zn(s) → 3Zn2+(aq) + 6e− 6e− + 14H+(aq) + Cr2O72−(aq) → 2Cr3+(aq) + 7H2O(l) 3Zn(s) + 14H+(aq) + Cr2O72−(aq) → 3Zn2+(aq) + 2Cr3+(aq) + 7H2O(l) Check that atoms and charge are balanced. 3Zn; 14H; 2Cr; 7O; charge +12 Copyright © Cengage Learning. All rights reserved. 19 | 22
  • 23. To balance a reaction in basic conditions, first follow the same procedure as for acidic solution. Then 1. Add one OH− to both sides for each H+. 2. When H+ and OH− occur on the same side, combine them to form H2O. 3. Cancel water molecules that occur on both sides. Copyright © Cengage Learning. All rights reserved. 19 | 23
  • 24. ? Lead(II) ion, Pb2+, yields the plumbite ion, Pb(OH)3−, in basic solution. In turn, this ion is oxidized in basic hypochlorite solution, ClO−, to lead(IV) oxide, PbO2. Balance the equation for this reaction using the half−reaction method. The skeleton equation is Pb(OH)3− + ClO− → PbO2 + Cl− Copyright © Cengage Learning. All rights reserved. 19 | 24
  • 25. First we determine the oxidation number of Pb and Cl in each species. +2 +1 +4 −1 Pb(OH)3− + ClO− → PbO2 + Cl− Pb is oxidized from +2 to +4. Cl is reduced from +1 to −1. Now we balance the half−reactions. Copyright © Cengage Learning. All rights reserved. 19 | 25
  • 26. Oxidation half−reaction First we balance Pb and O: Pb(OH)3−(aq) → PbO2(s) + H2O(l) Next we balance H: Pb(OH)3−(aq) → PbO2(s) + H2O(l) + H+(aq) Then we balance e−: Pb(OH)3−(aq) → PbO2(s) + H2O(l) + H+(aq) + 2e− Copyright © Cengage Learning. All rights reserved. 19 | 26
  • 27. Reduction half−reaction First we balance Cl and O: ClO−(aq) → Cl−(aq) + H2O(l) Next we balance H: 2H+(aq) + ClO−(aq) → Cl−(aq) + H2O(l) Finally we balance e−: 2e− + 2H+(aq) + ClO−(aq) → Cl−(aq) + H2O(l) Copyright © Cengage Learning. All rights reserved. 19 | 27
  • 28. Now we combine the half−reactions. Pb(OH)3−(aq) → PbO2(s) + H2O(l) + H+(aq) + 2e− 2e− + 2H+(aq) + ClO−(aq) → Cl−(aq) + H2O(l) H+(aq) + Pb(OH)3−(aq) + ClO−(aq) → PbO2(s) + Cl−(aq) + 2H2O(l) Copyright © Cengage Learning. All rights reserved. 19 | 28
  • 29. H+(aq) + Pb(OH)3−(aq) + ClO−(aq) → PbO2(s) + Cl−(aq) + 2H2O(l) To convert to basic solution, we add OH− to each side, converting H+ to H2O. H2O(l) + Pb(OH)3−(aq) + ClO−(aq) → PbO2(s) + Cl−(aq) + 2H2O(l) + OH−(aq) Finally, we cancel the H2O that is on both sides. Pb(OH)3−(aq) + ClO−(aq) → PbO2(s) + Cl−(aq) + H2O(l) + OH−(aq) Copyright © Cengage Learning. All rights reserved. 19 | 29
  • 30. The next several topics describe battery cells or voltaic cells (galvanic cells). An electrochemical cell is a system consisting of electrodes that dip into an electrolyte and in which a chemical reaction either uses or generates an electric current. Copyright © Cengage Learning. All rights reserved. 19 | 30
  • 31. A voltaic or galvanic cell is an electrochemical cell in which a spontaneous reaction generates an electric current. An electrolytic cell is an electrochemical cell in which an electric current drives an otherwise nonspontaneous reaction. Copyright © Cengage Learning. All rights reserved. 19 | 31
  • 32. Physically, a voltaic cell consists of two half−cells that are electrically connected. Each half−cell is the portion of the electrochemical cell in which a half−reaction takes place. The electrical connections allow the flow of electrons from one electrode to the other. The cells must also have an internal cell connection, such as a salt bridge, to allow the flow of ions. Copyright © Cengage Learning. All rights reserved. 19 | 32
  • 33. The Cu2+ ions gain The zinc metal atom two electrons, loses two electrons, forming solid copper. forming Zn2+ ions. The electrons flow through the external circuit from the zinc electrode to the copper electrode. Ions flow through the salt bridge to maintain charge balance. Copyright © Cengage Learning. All rights reserved. 19 | 33
  • 34. A salt bridge is a tube of electrolyte in a gel that is connected to the two half−cells of the voltaic cell. It allows the flow of ions but prevents the mixing of the different solutions that would allow direct reactions of the cell reactants. In the salt bridge, cations move toward the cathode and anions move toward the anode. Copyright © Cengage Learning. All rights reserved. 19 | 34
  • 35. The electrode at which oxidation takes place is called the anode. The electrode at which reduction takes place is called the cathode. Electrons flow through the external circuit from the anode to the cathode. Copyright © Cengage Learning. All rights reserved. 19 | 35
  • 36. Copyright © Cengage Learning. All rights reserved. 19 | 36
  • 37. This figure below illustrates the same reaction, replacing the light bulb with a voltmeter. The solution on the right is now blue from the Cu2+ ion formed. Copyright © Cengage Learning. All rights reserved. 19 | 37
  • 38. You construct one half−cell of a voltaic cell by inserting a copper metal strip into a solution of copper(II) sulfate. You construct another half−cell by inserting an aluminum metal strip in a solution of aluminum nitrate. You now connect the half−cells by a salt bridge. When connected to an external circuit, the aluminum is oxidized. Sketch the resulting voltaic cell. Label the anode and the cathode, showing the corresponding half−reactions. Indicate the direction of electron flow in the external circuit. ? Copyright © Cengage Learning. All rights reserved. 19 | 38
  • 39. e− cathode Cu e− Ca2+ NO3− anode Al Cu2+ Al3+ SO42− NO3− Cu2+ + 2e− → Cu Copyright © Cengage Learning. All rights reserved. Al → Al3+ + 3e− 19 | 39
  • 40. Voltaic cell notation is a shorthand method of describing a voltaic cell. The oxidation half−cell, the anode, is written on the left. The reduction half−cell, the cathode, is written on the right. Copyright © Cengage Learning. All rights reserved. 19 | 40
  • 41. The cell terminal is written on the outside: left for the anode and right for the cathode. Phase boundaries are shown with a single vertical bar, |. For example, at the anode, the oxidation of Cu(s) to Cu2+(aq) is shown as Cu(s) | Cu2+(aq). At the cathode, the reduction of Zn2+(aq) to Zn(s) is shown as Zn2+(aq) | Zn(s). Copyright © Cengage Learning. All rights reserved. 19 | 41
  • 42. Between the anode and cathode the salt bridge is represented by two vertical bars, ||. The complete notation for the reaction is Cu(s) | Cu2+(aq) || Zn2+(aq) | Zn(s) Copyright © Cengage Learning. All rights reserved. 19 | 42
  • 43. When the half−reaction involves a gas, the electrode is an inert material such as platinum, Pt. It is included as a third substance in the half−cell. For example, the half−reaction of Cl2 being reduced to Cl− is written as follows: Cl2(g) | Cl−(aq) | Pt Because this is a reduction, the electrode appears on the far right. For the oxidation of H2(g) to H+(aq), the notation is Pt | H2(g) | H+(aq) In this case, the electrode appears on the far left. Copyright © Cengage Learning. All rights reserved. 19 | 43
  • 44. To fully specify a voltaic cell, it is necessary to give the concentrations of solutions or ions and the pressures of gases. In the cell notation, these values are written within parentheses for each species. For example, the oxidation of Cu(s) to Cu2+(aq) at the anode and the reduction of F2(g) to F−(aq) at the cathode is written as follows: Cu(s) | Cu2+(1.0 M) || F2(1.0 atm) | F−(1.0 M) | Pt Copyright © Cengage Learning. All rights reserved. 19 | 44
  • 45. ? The cell notation for a voltaic cell is Al(s) | Al3+(aq) || Cu2+(aq) | Cu(s) Write the cell reaction. Our strategy is to write and balance each half−reaction, and then to combine the half−reactions to give the overall cell reaction. The skeleton oxidation reaction is Al(s) → Al3+(aq). The skeleton reduction reaction is Cu2+(aq) → Cu(s). Copyright © Cengage Learning. All rights reserved. 19 | 45
  • 46. Balanced oxidation half−reaction Al(s) → Al3+(aq) + 3e− Balanced reduction half−reaction Cu2+(aq) + 2e− → Cu(s) The common multiple of the electrons is 6. Combine the half−reactions by multiplying the oxidation half−reaction by 2 and the reduction half−reaction by 3. Copyright © Cengage Learning. All rights reserved. 19 | 46
  • 47. 2Al(s) → 2Al3+(aq) + 6e− 3Cu2+(aq) + 6e− → 3Cu(s) 2Al(s) + 3Cu2+(aq) → 2Al3+(aq) + 3Cu(s) Copyright © Cengage Learning. All rights reserved. 19 | 47
  • 48. Now we shift our focus to the movement of the electrons in the oxidation–reduction reaction. To better understand this movement, we can compare the flow of electrons to the flow of water. Copyright © Cengage Learning. All rights reserved. 19 | 48
  • 49. Just as work is required to pump water from one point to another, so work is required to move electrons. Water flows from areas of high pressure to areas of low pressure. Similarly, electrons flow from high electric potential to low electric potential. Electric potential can be thought of as electric pressure. Copyright © Cengage Learning. All rights reserved. 19 | 49
  • 50. Potential difference is the difference in electric potential (electrical pressure) between two points. Potential difference is measured in the SI unit volt (V). Electrical work = charge x potential difference The SI units for this are J=C×V J V= C Copyright © Cengage Learning. All rights reserved. 19 | 50
  • 51. The magnitude of charge on one mole of electrons is given by the Faraday constant, F. It equals 9.6485 × 103 C per mole of electrons. 1 F = 9.6485 × 103 C Substituting this into the equation for work: w = –F × potential difference The term is negative because the cell is doing work in creating the current (that is, electron flow). Copyright © Cengage Learning. All rights reserved. 19 | 51
  • 52. Normally the potential across the voltaic cell electrodes is less than the maximum possible. One reason for this discrepancy is that it takes energy (work) to drive the current through the cell: The greater the current, the less the voltage. The cell has its maximum voltage only when no current flows. Copyright © Cengage Learning. All rights reserved. 19 | 52
  • 53. The situation for electrons is analogous to that seen with water. The difference between water pressure in the tap and the pressure of the outside atmosphere is at its maximum when the tap is off; once the tap is on, the pressure difference decreases. Copyright © Cengage Learning. All rights reserved. 19 | 53
  • 54. The maximum potential difference between the electrodes of a voltaic cell is called the cell potential or electromotive force (emf) of the cell, or Ecell. Ecell can be measured with a digital voltmeter. The anode of a voltaic cell has negative polarity; the cathode has positive polarity. Copyright © Cengage Learning. All rights reserved. 19 | 54
  • 55. The maximum work is given by the following equation: Wmax = –nFEcell Copyright © Cengage Learning. All rights reserved. 19 | 55
  • 56. ? The emf of a particular cell is 0.500 V. The cell reaction is 2Al(s) + 3Cu2+(aq) → 2Al3+(aq) + 3Cu(s) Calculate the maximum electrical work of this cell obtained from 1.00 g of aluminum. Copyright © Cengage Learning. All rights reserved. 19 | 56
  • 57. To determine the value of n (that is, the number of moles of electrons involved in either half−cell reaction), we need to examine the half−reactions. 2Al(s) → 2Al3+(aq) + 6e− 3Cu2+(aq) + 6e− → 3Cu(s) n=6 F = 96,485 C/mol Ecell = 0.500 V = 0.500 J/C wmax = –nFEcell wmax = –(6 mol)(96,485 C/mol)(0.500 J/C) wmax = 2.89 × 105 J Copyright © Cengage Learning. All rights reserved. 19 | 57
  • 58. This is the energy corresponding to the balanced reaction, so we need to convert to one gram of aluminum. wmax 2.89 × 105 J 1 mol Al = × = 5.36 × 103 J 2 mol Al 26.98 g Al wmax = 5.36 kJ Copyright © Cengage Learning. All rights reserved. 19 | 58
  • 59. A cell potential is a measure of the driving force of the cell reaction. It is composed of the oxidation potential for the oxidation half−reaction at the anode and the reduction potential, for the reduction half−reaction at the cathode. Ecell = oxidation potential + reduction potential The oxidation potential for a half−reaction = –(reduction potential for the reverse half−reaction) Copyright © Cengage Learning. All rights reserved. 19 | 59
  • 60. Because the oxidation and reduction potentials are opposites, we need to calculate only one or the other. By convention, reduction potentials are calculated and are called electrode potentials, E (with no subscript). Electrode potentials are a measure of the oxidizing ability of the reactant. Table 19.1 shows the increasing strength of the oxidizing agents. Copyright © Cengage Learning. All rights reserved. 19 | 60
  • 61. Copyright © Cengage Learning. All rights reserved. 19 | 61
  • 62. Copyright © Cengage Learning. All rights reserved. 19 | 62
  • 63. Ecell = Eoxidation + Ereduction Because the electrode potentials are for reduction: Eoxidation = –Eanode We can rewrite the equation for the cell: Ecell = − Eanode + Ecathode Ecell = Ecathode – Eanode Copyright © Cengage Learning. All rights reserved. 19 | 63
  • 64. Let’s find Ecell for the following cell: Al(s) | Al3+(aq) || Cu2+(aq) | Cu(s) At the cathode, Cu2+(aq) is reduced to Cu(aq). E = 0.34 V At the anode, Al is oxidized to Al3+. E = –[electrode potential for Al3+(aq)] E = –(–1.66 V) = 1.66 V Ecell = 0.34 V + 1.66 V Ecell = 2.00 V Copyright © Cengage Learning. All rights reserved. 19 | 64
  • 65. The standard electrode potential, E°, is the electrode potential when the concentrations of solutes are 1 M, the gas pressures are 1 atm, and the temperature has a specified value (usually 25°C). The superscript degree sign (°) signifies standard−state conditions. Copyright © Cengage Learning. All rights reserved. 19 | 65
  • 66. A fuel cell is simply a voltaic cell that uses a continuous supply of electrode materials to provide a continuous supply of electrical energy. A fuel cell employed by NASA on spacecraft uses hydrogen and oxygen under basic conditions to produce electricity. The water produced in this way can be used for drinking. The net reaction is ? 2H2(g) + O2(g) → 2H2O(g) Calculate the standard emf of the oxygen–hydrogen fuel cell. 2H2O(l) + 2e− ⇄ H2(g) + 2OH−(aq) E° = –0.83 V O2(g) + 2H2O(l) + 4e− ⇄ 4OH−(aq) E° = 0.40 V Copyright © Cengage Learning. All rights reserved. 19 | 66
  • 67. Anode reaction: 2H2O(l) + 2e− ⇄ H2(g) + 2OH−(aq) Cathode reaction: O2(g) + 2H2O(l) + 4e− ⇄ 4OH−(aq) E°red = –0.83 V E°red = 0.40 V Overall reaction: Ecell = Ecathode – Eanode Ecell = 0.40 V – (–0.83 V) Ecell = 1.23 V Copyright © Cengage Learning. All rights reserved. 19 | 67
  • 68. Comparing Oxidizing Strengths The oxidizing agent is itself reduced and is the species on the left of the reduction half−reaction. Consequently, the strongest oxidizing agent is the product of the half−reaction with the largest (most positive) E° value. Copyright © Cengage Learning. All rights reserved. 19 | 68
  • 69. Comparing Reducing Strengths The reducing agent is itself oxidized and is the species on the right of the reduction half−reaction. Consequently, the strongest reducing agent is the reactant in the half−reaction with the smallest (most negative) E° value. Copyright © Cengage Learning. All rights reserved. 19 | 69
  • 70. ? Which is the stronger reducing agent under standard conditions: Sn2+ (to Sn4+) or Fe (to Fe2+)? Which is the stronger oxidizing agent under standard conditions: Cl2 or MnO4−? The stronger reducing agent will be oxidized and has the more negative electrode potential. The standard (reduction) potentials are Sn2+ to Sn4+ E = 0.15 V Fe to Fe2+ E = –0.41 V The stronger reducing agent is Fe (to Fe 2+). Copyright © Cengage Learning. All rights reserved. 19 | 70
  • 71. The stronger oxidizing agent will be reduced. The standard (reduction) potentials are Cl2 to Cl− E = 1.36 V MnO4− to Mn2+ E = 1.49 V The stronger oxidizing agent is MnO4−. Copyright © Cengage Learning. All rights reserved. 19 | 71
  • 72. Predicting the Direction of Reaction You can predict the direction of reaction by comparing the relative oxidizing (or reducing) strengths. The stronger oxidizing agent will be reduced. (The stronger reducing agent will be oxidized.) Copyright © Cengage Learning. All rights reserved. 19 | 72
  • 73. ? Will dichromate ion oxidize manganese(II) ion to permanganate ion in acid solution under standard conditions? Standard potentials Cr2O72− to Cr3+ MnO4− to Mn2+ E° = 1.33 V E° = 1.49 V MnO4− has a larger reduction potential; it will oxidize Cr3+ to Cr2O72−. Cr2O72− will not oxidize Mn2+ to MnO4−. Copyright © Cengage Learning. All rights reserved. 19 | 73
  • 74. ? Let us define the reduction of I 2 to I− ions, I2(s) + 2e− → 2I−(aq), as the standard reduction reaction with E° = 0.00 V. We then construct a new standard reduction table based on this definition. a. What would be the new standard reduction potential of H+? Copyright © Cengage Learning. All rights reserved. 19 | 74
  • 75. b. Would using a new standard reduction table change the measured value of freshly prepared voltaic cell made from Cu and Zn? (Assume you have the appropriate solutions and equipment to construct the cell.) c. Would the calculated voltage for the cell in part b be different if your were using the values given in Table 19.1? Do the calculations to justify your answer. Copyright © Cengage Learning. All rights reserved. 19 | 75
  • 76. a. When the H+(aq) E° = 0.000 V, the I2 E° = 0.54 V. So, if the new I2 potential is 0.000 V, then the H+ potential would be –0.54 V. b. The measured voltaic cell voltage will be unchanged. c. No. The calculated voltage is the same using either standard because it is a difference is potential. Copyright © Cengage Learning. All rights reserved. 19 | 76
  • 77. The free−energy change, ∆G, for a reaction equals the maximum useful work of the reaction. ∆G° = wmax = –nFE° Copyright © Cengage Learning. All rights reserved. 19 | 77
  • 78. ? Calculate the standard free−energy change for the net reaction used in the hydrogen–oxygen fuel cell: 2H2(g) + O2(g) → 2H2O(l) The cell potential is 1.23 V. How does this compare with ∆Gf° of H2O(l)? Copyright © Cengage Learning. All rights reserved. 19 | 78
  • 79. First we determine the number of electrons involved in the reaction. Four H are oxidized to H+, and two O are reduced to O2−. Thus four electrons are involved. n=4 E° = 1.23 V ∆G° = –nFE° ∆G = –(4 mol)(96,485 C/mol)(1.23 J/C) ∆G° = –4.75 × 105 J ∆G° = –475 kJ ∆Gf° = –285.8 kJ/mol; –550 kJ for 2 mol Copyright © Cengage Learning. All rights reserved. 19 | 79
  • 80. ? A voltaic cell consists of one half−cell with Fe dipping into an aqueous solution of 1.0 M FeCl2 and the other half−cell with Ag dipping into an aqueous solution of 1.0 M AgNO3. Obtain the standard free−energy change for the cell reaction using the standard free−energy change for the reaction (found using standard free energies of formation). The standard free energies of formation of the ions are Ag+(aq), 77 kJ/mol, and Fe2+(aq), –85 kJ/mol. What is the cell potential? Copyright © Cengage Learning. All rights reserved. 19 | 80
  • 81. We first determine the reaction by using reduction potentials. Fe2+(aq), Fe(s) E° = –0.41 V Ag+(aq), Ag(s) E° = 0.80 V Fe is the anode (oxidation). Ag is the cathode (reduction). E° = 0.80 – (–0.41) = 1.21 V The half−reactions are Fe(s)  Fe2+(aq) + 2e− 2Ag+(aq) + 2e− → 2Ag(s) The overall reaction is Fe(s) + 2Ag+(aq) → Fe2+(aq) + 2Ag(s) Copyright © Cengage Learning. All rights reserved. 19 | 81
  • 82. Fe(s) + 2Ag+(aq) → Fe2+(aq) + 2Ag(s) ∆G° = 1 mol(–85 kJ/mol) – 2 mol(77 kJ/mol) ∆G° = –85 kJ – 154 kJ ∆G° = –239 kJ Copyright © Cengage Learning. All rights reserved. 19 | 82
  • 83. Now we can use this value to find E°. ΔG ° E° = − nF J − 239,000 mol E° = − J  (2 mol) 96,485  V  E ° = 1.24 V This answer compares favorably to the 1.21 V determined using standard potentials. Copyright © Cengage Learning. All rights reserved. 19 | 83
  • 84. Equilibrium Constants from Cell Potentials nFE° = RT ln K Rearranging: RT E = ln K nF We could now convert from ln to log: o cell 2.303 RT E = log K nF 0.0592 o Ecell = log K n o cell At 25°C: Copyright © Cengage Learning. All rights reserved. 19 | 84
  • 85. ? Calculate the equilibrium constant K at 25°C for the following reaction for the standard cell potential: Pb2+(aq) + Fe(s) ⇄ Pb(s) + Fe2+(aq) First, we find the cell potential. E°cathode = –0.13 V (Pb2+) E°anode = –0.41 V (Fe2+) E°cell = E°cathode – E°anode E°cell = –0.13 V – (–0.41 V) E°cell = +0.28 V Now we can find K. Copyright © Cengage Learning. All rights reserved. 19 | 85
  • 86. o cell nFE ln K = RT C  J  ( 2 mol )  96,485  0.28  mol  C  ln K = J    8.31 ( 298 K ) mol K   ln K = 21.82 K = 3.0 × 109 Copyright © Cengage Learning. All rights reserved. 19 | 86
  • 87. Summary of Relationship sAmong Variables Copyright © Cengage Learning. All rights reserved. 19 | 87
  • 88. Dependence of Cell Potential on Concentration: Nernst Equation ∆G = ∆G° + RT ln Q Q is the thermodynamic equilibrium constant. This equation allows us to relate E to E°. Ecell = E Copyright © Cengage Learning. All rights reserved. o cell RT – ln Q nF 19 | 88
  • 89. At 25°C, this reduces to Ecell = E Copyright © Cengage Learning. All rights reserved. o cell 0.02568 – ln Q n 19 | 89
  • 90. Determination of pH Cell measurements can allow us to determine the pH of a solution by determining [H+]. This is the basis of the pH meter. Copyright © Cengage Learning. All rights reserved. 19 | 90
  • 91. On the left a small, commercial electrode is pictured. On the right is a sketch showing the construction of a glass electrode for measuring hydrogen concentrations. Copyright © Cengage Learning. All rights reserved. 19 | 91
  • 92. Consider a voltaic cell, Fe(s) | Fe2+(aq) || Cu2+(aq) | Cu(s), being run under standard conditions. a. Is ∆G° positive or negative for this process? ? b. Change the concentrations from their standard values in such a way that Ecell is reduced. Write your answer using the shorthand notation. Copyright © Cengage Learning. All rights reserved. 19 | 92
  • 93. a. ∆G° = –nFE° E° = 0.36 – (–0.41) = 0.75 V Since E° is positive, ∆G° is negative. b. Overall reaction: Fe(s) + Cu2+(aq) → Fe2+(aq) + Cu(s) To reduce Ecell, Q must be > 1. Fe2+  Q =  2+  So, [Fe2+] must be > [Cu2+]. Cu    Larger Smaller Fe(s) | Fe2+(aq) (1.1 M) || Cu2+(aq) (0.50 M) | Cu(s) Copyright © Cengage Learning. All rights reserved. 19 | 93
  • 94. A pH meter is constructed using hydrogen gas bubbling over an inert platinum electrode (the hydrogen electrode) at a pressure of 1.2 atm. other electrode is aluminum metal immersed in The a 0.20 M Al3+ solution. What is the cell potential when the hydrogen electrode is immersed in a sample of acid rain with a pH of 4.0 at 25°C? If the electrode is placed in a sample of shampoo solution and the cell potential is 1.17 V, what is the pH of the shampoo solution? The reaction is 2Al(s) + 6H+(aq) ⇄ 2Al3+(aq) + 3H2(g) ? Copyright © Cengage Learning. All rights reserved. 19 | 94
  • 95. E =E o Q= 3 H2 [ P Al ] 3+ 2 [H ] + 6 o cell RT − ln Q nF (1.2 atm) ( 0.20 M ) = 3 (10 -4 M ) 2 6 = 6.91 × 1022 ln Q = 52.6 0.02568 ( 52.6 ) E = 1.66 – 6 E = 1.66 – 0.23 E = 1.43 V Copyright © Cengage Learning. All rights reserved. 19 | 95
  • 96. E =E o o cell RT − ln Q nF RT o ln Q = Ecell − E o nF RT ln Q = 1.66 V − 1.17 V = 0.49 V nF nF ln Q = 0.49 V × RT C   ( 6)  96,485  J mol   ln Q = 0.49 × J  C   8.3145 ( 298 K ) mol K   Copyright © Cengage Learning. All rights reserved. 19 | 96
  • 97. ln Q = 114.5 Q = 5.26 × 1049 Q= [H ] + 6 = 3 H2 [ P Al Q ] 3+ 2 3 H2 [ P Al ] 3+ 2 [H ] + 6 (1.2 atm) ( 0.20 M ) = 3 2 (5.26 × 10 ) 49 [H+] 6 = 1.31 × 10−51 [H+] = 3.31 × 10−9 M pH = 8.48 Copyright © Cengage Learning. All rights reserved. 19 | 97
  • 98. Commercial Voltaic Cells We next look at several commercial voltaic cells. Zinc–Carbon Dry Cell: Leclanché Anode: Cathode: Zn(s) → Zn2+(aq) + 2e− 2NH4+(aq) + 2MnO2(s) + 2e− → Mn2O3(s) + H2O(l) + 2NH3(aq) The initial voltage is about 1.5 V, but decreases and deteriorates rapidly in cold weather. Copyright © Cengage Learning. All rights reserved. 19 | 98
  • 99. Copyright © Cengage Learning. All rights reserved. 19 | 99
  • 100. Zinc–Carbon Dry Cell: Alkaline Anode: Zn(s) + 2OH−(aq) → Zn(OH)2(s) + 2e−Cathode: 2MnO2(s) + H2O(l) + 2e− → Mn2O3(s) + 2OH− (aq) This cell performs better under current drain and in cold weather. It isn’t truly “dry” but rather uses an aqueous paste. Copyright © Cengage Learning. All rights reserved. 19 | 100
  • 101. Copyright © Cengage Learning. All rights reserved. 19 | 101
  • 102. Lithium–Iodine Battery In this solid−state battery, the electrodes are separated by a thin crystalline layer of lithium iodide. Diffusion of the Li+ ion carries the current. The cell has high resistance and, therefore, low voltage. It is very reliable and is used for pacemakers. Copyright © Cengage Learning. All rights reserved. 19 | 102
  • 103. Lead Storage Cell The electrodes are lead alloy grids: one is packed with a spongy lead to form the anode, and the other is packed with lead dioxide to form the cathode. Both electrodes are in an aqueous solution of H 2SO4. Anode: Cathode: Pb(s) + HSO4−(aq) → PbSO4(s) + H+(aq) + 2e− PbO2(s) + 3H+(aq) + HSO4−(aq) + 2e− → PbSO4(s) + 2H2O(l) Unlike dry cells, after discharge, lead storage cells can be recharged. Copyright © Cengage Learning. All rights reserved. 19 | 103
  • 104. Each of the six cells in the lead storage cell generates 2 V, yielding 12 V. During discharge, white PbSO4(s) coats each electrode. To recharge the cell, an external current is used, reversing the previous reactions. Some water decomposes into hydrogen and oxygen gas, so more water may need to be added. Newer batteries use electrodes with calcium in the lead, which resists decomposition by water. These versions are maintenance free. Copyright © Cengage Learning. All rights reserved. 19 | 104
  • 105. Copyright © Cengage Learning. All rights reserved. 19 | 105
  • 106. Nickel–Cadmium Cell Anode: Cd(s) + 2OH−(aq) → Cd(OH)2(s) + 2e− Cathode: NiOOH(s) + H2O(l) + e− → Ni(OH)2(s) + OH−(aq) These cells are used in calculators, portable power tools, shavers, and toothbrushes. During recharge, the reactions are reversed, which can be done many times. When cadmium is replaced with a metal hydride (MH), nickel metal hydride and lithium hydride cells result. They are less toxic. Copyright © Cengage Learning. All rights reserved. 19 | 106
  • 107. Fuel Cell Fuel cells require a continuous supply of reactants (fuel). Anode: H2(g) → 2H+(aq) + 2e− Cathode: O2(g) + 4H+(aq) + 4e− → 2H2O(l) Fuel cells were originally used in space applications, but are now being explored for more uses. Copyright © Cengage Learning. All rights reserved. 19 | 107
  • 108. Copyright © Cengage Learning. All rights reserved. 19 | 108
  • 109. Corrosion Control: Cathodic Protection Voltaic cells can be used to control corrosion of underground pipelines and tanks. Rusting occurs when water comes in contact with iron. The edge of the water drop, when exposed to air, becomes one pole of a voltaic cell where oxygen is reduced to hydroxide. Anode: Cathode: Fe(s) → Fe2+(aq) + 2e− O2(g) + 2H2O(l) + 4e− → 4OH−(aq) Copyright © Cengage Learning. All rights reserved. 19 | 109
  • 110. Copyright © Cengage Learning. All rights reserved. 19 | 110
  • 111. When the buried metal is connected to a more active metal such as magnesium, the magnesium becomes the anode and the iron becomes the cathode. The iron is, therefore, protected from oxidation. This phenomenon is called cathodic protection. Copyright © Cengage Learning. All rights reserved. 19 | 111
  • 112. Copyright © Cengage Learning. All rights reserved. 19 | 112
  • 113. Cathodic protection is illustrated below. The nails are in a gel containing phenolphthalein indicator and potassium ferricyanide. Iron corrosion yields Fe2+, which reacts with ferricyanide ion to give a dark blue precipitate. Where OH− forms, phenolphthalein appears pink. Unprotected Copyright © Cengage Learning. All rights reserved. Protected 19 | 113
  • 114. ? Keeping in mind that seawater contains a number of ions, explain why seawater corrodes iron much faster than freshwater. Many of the ions in seawater have very high reduction potentials—higher than the reduction potential of Fe(s). This means spontaneous electrochemical reactions will occur with Fe(s), causing the iron to form ions and go into solution. At the same time, the ions in the sea will be reduced and plate out on the surface of the iron. Copyright © Cengage Learning. All rights reserved. 19 | 114
  • 115. Electrolytic Cell An electrolytic cell is an electrochemical cell in which an electric current drives an otherwise nonspontaneous reaction. The process of producing a chemical change in an electrolytic cell is called electrolysis. Many important substances are produced commercially by electrolysis—for example, aluminum and chlorine. Copyright © Cengage Learning. All rights reserved. 19 | 115
  • 116. Downs Cell A Downs cell is an electrolytic cell used to obtain sodium metal by electrolysis of sodium chloride. The products must be kept separated or they would react. Anode: Cl−(l) → ½Cl2(g) + e− Cathode: Na+(l) + e− → Na(l) Copyright © Cengage Learning. All rights reserved. 19 | 116
  • 117. Copyright © Cengage Learning. All rights reserved. 19 | 117
  • 118. Copyright © Cengage Learning. All rights reserved. 19 | 118
  • 119. Aqueous Electrolysis For a molten salt, the possible reactions are limited to those involving the ions from the salt. In aqueous situations, however, the possible reactions of water must also be included. 2H2O(l) + 2e− → H2(g) + 2OH−(aq) 2H2O(l) → O2(g) + 4H+(aq) + 4e− Copyright © Cengage Learning. All rights reserved. 19 | 119
  • 120. Unlike voltaic cell reactions, which maximize the cell potential, the reaction requiring the smallest voltage will be the one that occurs. To determine the reaction, it is necessary to consider all possible half−reactions that might occur. First, examine the possible oxidation reactions. The one with the least negative E° value will occur. Next examine the possible reduction reactions. The one with the more positive E° value will occur. Copyright © Cengage Learning. All rights reserved. 19 | 120
  • 121. Electrolysis of Sulfuric Acid Solutions Possible oxidation reactions: 2SO42−(aq) → S2O82−(aq) + 2e− E°red = 2.01 V 2H2O(l) → O2(g) + 4H+(aq) + 4e− E°red = 1.23 V Smaller reduction potential Possible reduction reactions: H+(aq) + e− → ½ H2(g) E°red = 0.00 V 2H2O(l) + 2e− → H2(g) + 2OH−(aq) E°red = –0.83 V Larger (more positive) reduction potential Copyright © Cengage Learning. All rights reserved. 19 | 121
  • 122. 2H2O(l) → O2(g) + 4H+(aq) + 4e− 4H+(aq) + 4e− → 2H2(g) 2H2O(l) → 2H2(g) + O2(g) Ecell°= E°cathode – E°anode Ecell°= 0.00 – 1.23 Ecell°= –1.23 V Copyright © Cengage Learning. All rights reserved. 19 | 122
  • 123. Electrolysis of Sodium Chloride Solution Smaller Possible oxidation reactions: reduction potential 2Cl−(aq) → Cl2(g) + 2e− E°red = 1.36 V 2H2O(l) → O2(g) + 4H+(aq) + 4e− E°red = 1.23 V Possible reduction reactions: Na+(aq) + e− → Na(s) E°red = –2.71 V H+(aq) + e− → ½H2(g) E°red = 0.00 V Larger (more positive) value Copyright © Cengage Learning. All rights reserved. 19 | 123
  • 124. Cl2(g) + 2e− → 2Cl−(aq) 2H+(aq) + 2e− → H2(g) 2H+(aq) + Cl2 (g) → H2(g) + 2Cl−(aq) Ecell°= E°cathode – E°anode Ecell° = 0.00 – 1.36 Ecell° = –1.36 V Copyright © Cengage Learning. All rights reserved. 19 | 124
  • 125. ? Describe what you expect to happen at the electrodes when an aqueous solution of sodium iodide is electrolyzed. Copyright © Cengage Learning. All rights reserved. 19 | 125
  • 126. Electrolysis of Sodium Iodide Solution Smaller reduction potential Possible oxidation reactions: 2I−(aq) → I2(g) + 2e− E°red = 0.54 V 2H2O(l) → O2(g) + 4H+(aq) + 4e− E°red = 1.23 V Possible reduction reactions: Na+(aq) + e− → Na(s) E°red = –2.71 V H+(aq) + e− → ½H2(g) E°red = 0.00 V Larger (more positive) value Copyright © Cengage Learning. All rights reserved. 19 | 126
  • 127. I2(g) + 2e− → 2I−(aq) 2H+(aq) + 2e− → H2(g) 2H+(aq) + I2 (g) → H2(g) + 2I−(aq) Ecell°= E°cathode – E°anode Ecell°= 0.00 – 0.54 Ecell°= –0.54 V Copyright © Cengage Learning. All rights reserved. 19 | 127
  • 128. The electrolysis of sodium chloride is the basis of the chlor−alkali industry, which produces chlorine and sodium hydroxide. Copyright © Cengage Learning. All rights reserved. 19 | 128
  • 129. The chlor−alkali mercury cell uses mercury metal as the cathode. Sodium is reduced, rather than water, forming a sodium mercury amalgam. Copyright © Cengage Learning. All rights reserved. 19 | 129
  • 130. Metals can also be protected from corrosion by plating them with other metals. For example, zinc can be plated on steel by electrogalvanizing to prevent rusting. Copyright © Cengage Learning. All rights reserved. 19 | 130
  • 131. Metals can also be purified by electrolysis. Here we see copper being purified. Copyright © Cengage Learning. All rights reserved. 19 | 131
  • 132. Stoichiometry of Electrolysis In the 1830s, Michael Faraday showed that the total charge that flows in a circuit is related to the amount of substance released at the electrodes. One faraday of charge is the charge on one mole of electrons and is equal to 96,485 coulombs. Copyright © Cengage Learning. All rights reserved. 19 | 132
  • 133. For quantitative considerations, we also need to know the magnitude of the current and the time it has flowed. Electric charge = current × time Coulombs = amperes × seconds The ampere, A, is the base unit of current. The coulomb, C, is equal to an ampere−second. Copyright © Cengage Learning. All rights reserved. 19 | 133
  • 134. ? What electric charge is required to plate a piece of automobile molding with 1.00 g of chromium metal using a chromium(III) ion solution? If the electrolysis current is 2.00 A, how long does the plating take? Copyright © Cengage Learning. All rights reserved. 19 | 134
  • 135. First, we convert mass to moles Cr; then to moles e−; then, using current, to seconds. 1 mol Cr 3 mol e - 96,485 C 1s 1.00 g Cr × × × × 52.00 g Cr 1 mol Cr 1 mol e 2.00 C = 2.78 × 103 s = 46.4 min Copyright © Cengage Learning. All rights reserved. 19 | 135
  • 136. ? A solution of nickel salt is electrolyzed to nickel metal by a current of 2.43 A. If this current flows for 10.0 min, how many coulombs is this? How much nickel metal is deposited in the electrolysis? Copyright © Cengage Learning. All rights reserved. 19 | 136
  • 137. We first find the charge: C 60 s 2.43 × 10.0 min × s 1 min = 1458 C Given the half−reaction: Ni2+(aq) + 2e− → Ni(s) 1 mol e− 1 mol Ni 58.69 g 1458 C × × × − 96,485 C 2 mol e 1 mol Ni = 0.443 g Ni Copyright © Cengage Learning. All rights reserved. 19 | 137