Transcript
Chapter 10 Review, pages 678–683 Knowledge
1. (b) 2. (b) 3. (d) 4. (c) 5. (a) 6. (d) 7. False. In a galvanic cell, electrons travel from the anode to the cathode via the external circuit. 8. True 9. True 10. False. The redox reaction in a secondary cell is reversible when a current runs through the cell. 11. False. The rising cost of energy from gasoline has consumers looking at alternative sources of energy. 12. False. Fusion is the nuclear process in which small atoms combine to form larger atoms under conditions of very high temperature and pressure. 13. False. Steel rusts more quickly in salt water than in tap water. 14. False. Metals that readily corrode can often be protected by the application of a thin coating of a metal that is more easily oxidized. 15. True 16. (a) (v) (b) (iii) (c) (vi) (d) (viii) (e) (iv) (f) (i) (g) (vii) (h) (ii) 17. (a) Diagram of a galvanic cell produced using a silver electrode in a silver nitrate solution and a copper electrode in a copper(II) nitrate solution:
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Chapter 10: Electrochemical Cells
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(b) Anode (oxidation) half-reaction equation: Cu(s) → Cu2+(aq) + 2 e− Cathode (reduction) half-reaction equation: Ag+(aq) + e− → Ag(s) (c) Balance electrons by multiplying the half-reaction at the cathode by 2: 2 Ag+(aq) + 2 e− → 2 Ag(s) Add the two half-reactions to give the balanced net ionic equation:
Cu(s) ! Cu 2+ (aq) + 2 e " 2 Ag + (aq) + 2 e " ! 2 Ag(s) Cu(s) + 2 Ag + (aq) ! Cu 2+ (aq) + 2 Ag(s) (d) Cu(s) | Cu2+(aq) | | Ag+(aq) | Ag(s) 18. (a) Oxidation occurs at the anode. (b) The anode is the negative electrode. 19. In the reaction represented by the equation Zn(s) + 2 Cu+(aq) → 2 Cu(s) + Zn2+(aq) zinc metal, Zn(s), is the reducing agent and copper(I) ion, Cu+(aq), is the oxidizing agent. 20. During the operation of a galvanic cell, chemical energy is converted into electrical energy. 21. (a) For the reaction in a galvanic cell represented by the net ionic equation Fe2+(aq) + Al(s) → Al3+(aq) + Fe(s) the standard line notation for the cell is Al(s) | Al3+(aq) | | Fe2+(aq) | Fe(s) (b) For the reaction in a galvanic cell represented by the net ionic equation Cu2+(aq) + Cr2+(aq) → Cu+(aq) + Cr3+(aq) (using inert platinum electrodes) the two half reactions are Anode (oxidation) half-reaction equation: Cr3+(aq) + e → Cr2+(aq) Cathode (reduction) half-reaction equation: Cu2+(aq) + 2 e– → Cu+(aq) so the standard line notation for the cell is Pt(s) | Cr2+(aq) | | Cu2+(aq) | Pt(s) 22. (a) Iron(II) ions are not capable of oxidizing chromium(II) ions to chromium(III) ions because the standard reduction potential for iron(II) ions, –0.44 V, is less than the standard reduction potential for chromium(II) ions, +0.50 V. (b) Iron(II) ions are capable of oxidizing manganese, Mn, because the standard reduction potential for iron(II) ions, –0.44 V, is greater than the standard reduction potential for manganese ions, –1.18 V. (c) Hydrogen gas, H2(g), cannot reduce nickel(II) ions, Ni2+(aq), because the standard reduction potential for hydrogen gas, –0.83 V, is less than the standard reduction potential for nickel(II) ions, –0.23 V. 23. (a) Hydrogen fuel cells are used in the U.S. space program. (b) The product, electrical energy, was used to power NASA spacecraft. 24. Electrolysis occurs when direct current is passed through a cell containing two redox active substances, causing a non-spontaneous redox reaction to occur.
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Chapter 10: Electrochemical Cells
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25. Answers may vary. Sample answer: Three applications of electrolysis are: galvanizing nails to make them weather resistant; plating precious metals like gold and silver onto decorative metal objects; and aluminum recycling. Understanding
26. (a) The salt bridge in a galvanic cell is a porous barrier between two half-cells with different electrolytes. The salt bridge prevents the solutions from mixing but allows ions to flow in both directions, to keep the solution in each half-cell electrically neutral. At the anode, cations are produced and go into solution. This causes a buildup of positive ions in this solution. If this electrical imbalance is not corrected, the reaction cannot continue. The excess positive charge attracts the anions from the salt bridge, thereby keeping the solution electrically neutral. At the cathode, the opposite occurs: as positive ions are removed from solution, the solution becomes overly negative. This attracts the cations from the salt bridge, keeping this side of the cell neutral. (b) The purpose of the wire in a galvanic cell is to create an external circuit through which electrons provided by the oxidation reaction at the anode can flow to the cathode, the site of the reduction reaction. (c) The purpose of the electrolyte in a galvanic cell is to act as an ionic conductor. The two half-cells may use the same electrolyte, or they may use different electrolytes. (d) The purpose of the conductive electrodes in a galvanic cell is to allow oxidation and reduction to occur. The electrodes are the conductors by which the current leaves or returns to the electrolyte. The anion provides electrons and the cathode accepts electrons. 27. A galvanic cell is an electrochemical cell that produces electrical energy from redox reactions that occur spontaneously in the cell. Electrons move from the anode, which undergoes oxidation, through a wire to the cathode, which undergoes reduction. The solutions in which the electrodes are immersed complete the circuit by allowing electrons to flow from the cathode to the anode. The salt bridge, or porous substance in a fruit or vegetable, maintains the neutrality of the solutions by allowing the flow of ions in or between the solutions. The electrical energy produced can be measured by connecting an ammeter or a voltmeter along the wire between the electrodes. 28. Factors That Will Stop a Galvanic Cell from Functioning Factor Explanation The anode is used up. The anode is no longer able to donate electrons to produce current. The cations in the electrolytes are used Current stops because electrons cannot up. move through the electrolyte. The electrolyte in the salt bridge is used There is a buildup of charge in the halfup. cells, so current stops. The anions in the electrolytes are used Current stops because electrons cannot up. move through the electrolyte. 29. (a) If a half-cell containing zinc metal in a zinc nitrate solution and a half-cell containing nickel metal in a nickel nitrate solution were joined to form a galvanic cell, the anode would be the zinc metal. The cathode would be the nickel metal.
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Chapter 10: Electrochemical Cells
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(b) From Table 1 on page 646 of the textbook, the standard reduction potential for zinc is −0.76 V, and the standard reduction potential for nickel is −0.23 V. Under standard conditions, the cell potential is !E°r (cell) = E°r (cathode) " E°r (anode) = "0.23 V " ("0.76 V) !E°r (cell) = 0.53 V
30. (a) To determine whether H+(aq) is capable of oxidizing Cu(s) to Cu2+(aq), look up the equations for the two reduction half-cell reactions and their standard reduction potentials in Table 1, Appendix B7: 2 H+(aq) + 2 e– → H2(g) !E°r = 0 V Cu2+(aq) + 2 e– → Cu(s) !E°r = 0.34 V H+(aq) cannot oxidize Cu(s) to Cu2+(aq) because the standard reduction potential for hydrogen ions is less than the standard reduction potential for copper metal. (b) To determine whether Fe3+(aq) is capable of oxidizing I−(aq), look up the equations for the two reduction half-cell reactions and their standard reduction potentials in Table 1, Appendix B7: Fe3+(aq) + e– → Fe2+(aq) !E°r = 0.77 V I2(s) + 2 e– → 2 I–(aq) !E°r = 0.54 V Iron(III) ion can oxidize iodine ion because the standard reduction potential for iron(III) ions is greater than the standard reduction potential for iodine ions. (c) To determine whether H2(g) capable of reducing Ag+(aq), look up the equations for the two reduction half-cell reactions and their standard reduction potentials in Table 1, Appendix B7: H2(g) + 2 e– → 2 H–(aq) !E°r = "2.23 V Ag+(aq) + e– → Ag(s) !E°r = 0.80 V H2(g) is capable of reducing Ag+(aq) because the standard reduction potential for hydrogen gas is less than the standard reduction potential for silver ions (i.e., hydrogen is a much stronger reducing agent than silver). 31. It is impossible to obtain the electric potential for a single half-cell because a half-cell reaction cannot occur on its own. 32. The platinum wire in a standard hydrogen electrode functions as an inert electrode that provides a surface on which redox reactions can occur, but it does not participate in these reactions. 33. The equations for the half-reactions occurring at the electrodes when a half-cell involving zinc metal and aqueous zinc ions is connected to a standard hydrogen electrode are the following: Anode half-reaction: Zn(s) → Zn2+(aq) + 2 e− !E°r = "0.76 V Cathode half-reaction: 2 H+(aq) + 2 e− → H2(g) !E°r = 0 V The hydrogen electrode has the more positive reduction potential, so it acts as the cathode in this cell.
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Chapter 10: Electrochemical Cells
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34. (a) Given: Zn2+(aq) + 2 e− → Zn(s) E°r = !0.76 V Cd2+(aq) + 2 e− → Cd(s) E°r = !0.40 V Required: !E°r (cell) and the net ionic equation for the cell reaction Analysis: Use the equation !E°r (cell) = E°r (cathode) " E°r (anode) to calculate the standard cell potential. Solution: The half-cell reaction with more positive potential is the reduction halfreaction. In this case, the reduction of cadmium ions occurs at the cathode: Cd2+(aq) + 2 e− → Cd(s) Zinc is oxidized at the anode. The equation for this half-cell reaction is written as an oxidization reaction: Zn(s) → Zn2+(aq) + 2 e− The number of electrons is equal in both half-reaction equations, so add the equations for two half-cell reactions to obtain the net ionic equation for this reaction: Zn(s) + Cd2+(aq) → Cd(s) + Zn2+(aq) The standard cell potential is !E°r (cell) = E°r (cathode) " E°r (anode) = "0.40 V " ("0.76 V) !E°r (cell) = 0.36 V
Statement: The net ionic equation and standard cell potential for this cell are Cd2+(aq) + Zn(s) → Cd(s) + Zn2+(aq) and !E°r (cell) = 0.36 V . (b) Given: Ag+(aq) + e− → Ag(s) E°r = 0.80 V Al3+(aq) + 3 e− → Al(s) E°r = !1.66 V Required: !E°r (cell) and the net ionic equation for the cell reaction Analysis: Use the equation !E°r (cell) = E°r (cathode) " E°r (anode) to calculate the standard cell potential. Solution: The half-cell reaction with more positive potential is the reduction halfreaction. In this case, the reduction of silver ions occurs at the cathode: Ag+(aq) + e− → Ag(s) Aluminum is oxidized at the anode. The equation for this half-cell reaction is written as an oxidization reaction: Al(s) → Al3+(aq) + 3 e− To balance electrons in the two half-reactions, multiply the equation for the silver halfreaction by 3. 3 Ag+(aq) + 3 e− → 3 Ag(s)
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Chapter 10: Electrochemical Cells
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Add the equations for the two half-cell reactions to obtain the net ionic equation for this reaction: 3 Ag+(aq) + Al(s) → 3 Ag(s) + Al3+(aq) The standard cell potential is !E°r (cell) = E°r (cathode) " E°r (anode) = 0.80 V " ("1.66 V) !E°r (cell) = 2.46 V
Statement: The net ionic equation and standard cell potential for this cell are Al(s) + 3 Ag+(aq) → 3 Ag(s) + Al3+(aq) and !E°r(cell) = 2.46 V. (c) Given: Fe2+(aq) + 2 e− → Fe(s) E°r = !0.44 V Al3+(aq) + 3 e− → Al(s) E°r = !1.66 V Required: !E°r (cell) and the net ionic equation for the cell reaction Analysis: Use the equation !E°r (cell) = E°r (cathode) " E°r (anode) to calculate the standard cell potential. Solution: The half-cell reaction with more positive potential is the reduction halfreaction. In this case, the reduction of iron ions occurs at the cathode: Fe2+(aq) + 2 e− → Fe(s) Aluminum is oxidized at the anode. The equation for this half-cell reaction is written as an oxidization reaction: Al(s) → Al3+(aq) + 3 e− To balance electrons in the two half-reactions, multiply the iron half-reaction equation by 3 and the aluminum half-reaction equation by 2. 3 Fe2+(aq) + 6 e− → 3 Fe(s) 2 Al(s) → 2 Al3+(aq) + 6 e− Add the equations for the two half-cell reactions to obtain the net ionic equation for this reaction: 3 Fe2+(aq) + 2 Al(s) → 3 Fe(s) + 2 Al3+(aq) The standard cell potential is !E°r (cell) = E°r (cathode) " E°r (anode) = "0.44 V " ("1.66 V) !E°r (cell) = 1.22 V
Statement: The net ionic equation and the standard cell potential for this cell are 3 Fe2+(aq) + 2 Al(s) → 3 Fe(s) + 2 Al3+(aq) and !E°r (cell) = 1.22 V.
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Chapter 10: Electrochemical Cells
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35. (a) Given: 2 H+(aq) + 2 e− → H2(g) Al3+(aq) + 3 e− → Al(s) Required: !E°r (cell) and the net ionic equation for the cell reaction Analysis: Use the equation !E°r (cell) = E°r (cathode) " E°r (anode) to calculate the standard cell potential. Solution: From Table 1, Appendix B7, the reduction potential for hydrogen is 0 V and the reduction potential for aluminum is −1.66 V. The half-cell reaction with more positive potential is the reduction half-reaction. In this case, the reduction of hydrogen ions occurs at the cathode: 2 H+(aq) + 2 e− → H2(g) Aluminum is oxidized at the anode. The equation for this half-cell reaction is written as an oxidization reaction: Al(s) → Al3+(aq) + 3 e− To balance electrons in the two half-reactions, multiply the equation for the hydrogen half-reaction by 3 and the aluminum half-reaction equation by 2. 6 H+(aq) + 6 e− → 3 H2(g) 2 Al(s) → 2 Al3+(aq) + 6 e− Add the two half-cell reactions to obtain the net ionic equation for this reaction: 2 Al(s) + 6 H+(aq) → 3 H2(g) + 2 Al3+(aq) The standard cell potential is !E°r (cell) = E°r (cathode) " E°r (anode) = 0.00 V " ("1.66 V) !E°r (cell) = 1.66 V
Statement: The net ionic equation and standard cell potential for this cell are 2 Al(s) + 6 H+(aq) → 2 Al3+(aq) + 3 H2(g) and ΔE °r (cell) = 1.66 V. (b) Given: Cr2O72−(aq) + 14 H+(aq) + 6 e− → 2 Cr3+(aq) + 7 H2O(l) H2O2(aq) + 2 H+(aq) + 2 e− → 2 H2O(l) Required: !E°r (cell) and the net ionic equation for the cell reaction Analysis: Use the equation !E°r (cell) = E°r (cathode) " E°r (anode) to calculate the standard cell potential.
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Chapter 10: Electrochemical Cells
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Solution: From Table 1, Appendix B7, the reduction potential for the dichromate ion is +1.33 V and the reduction potential for hydrogen peroxide is +1.78 V. The half-cell reaction with more positive potential is the reduction half-reaction. In this case, the reduction of hydrogen peroxide occurs at the cathode: H2O2(aq) + 2 H+(aq) + 2 e− → 2 H2O(l) Dichromate ion is oxidized at the anode. The equation for this half-cell reaction is written as an oxidization reaction: 2 Cr3+(aq) + 7 H2O(l) → Cr2O72−(aq) + 14 H+(aq) + 6 e− To balance electrons in the two half-reactions, multiply the equation for the hydrogen peroxide half-reaction by 3. 3 H2O2(aq) + 6 H+(aq) + 6 e− → 6 H2O(l) Add the equations for the two half-cell reactions to obtain the net ionic equation for this reaction: 2 Cr3+(aq) + 7 H2O(l) + 3 H2O2(aq) + 6 H+(aq) → Cr2O72−(aq) + 14 H+(aq) + 6 H2O(l) Eliminate redundant hydrogen ions and water molecules. 2 Cr3+(aq) + H2O(l) + 3 H2O2(aq) → Cr2O72−(aq) + 8 H+(aq) The standard cell potential is !E°r (cell) = E°r (cathode) " E°r (anode) = 1.78 V " 1.33 V !E°r (cell) = 0.45 V
Statement: The net ionic equation and the standard cell potential for this cell are 2 Cr3+(aq) + H2O(l) + 3 H2O2(aq) → Cr2O72−(aq) + 8 H+(aq) and !E°r (cell) = 0.45 V. 36. (a) Given: Pb2+(aq) + Cu(s) → Pb(s) + Cu2+(aq) Required: !E°r (cell) and whether the reaction is spontaneous as written Analysis: Use Table 1 in Appendix B7 to predict reaction spontaneity. Use the equation !E°r (cell) = E°r (cathode) " E°r (anode) to calculate the standard cell potential. Solution: Cathode half-reaction equation: Pb2+(aq) + 2 e− → Pb(s) Anode half-reaction equation: Cu(s) → Cu2+(aq) + 2 e− From Table 1, Appendix B7, the reduction potential for the lead(II) ion is −0.13 V and the reduction potential for copper is +0.34 V. The standard cell potential is: !E°r (cell) = E°r (cathode) " E°r (anode) = "0.13 V " 0.34 V !E°r (cell) = "0.47 V
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Chapter 10: Electrochemical Cells
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Statement: The standard cell potential for this cell is !E°r (cell) = "0.47 V . Because the standard cell potential is negative, the reaction is not spontaneous. (b) Given: Au3+(aq) + 3 Ag(s) → 3 Ag+(aq) + Au(s) Required: !E°r (cell) and whether the reaction is spontaneous as written Analysis: Use Table 1 in Appendix B7 to predict reaction spontaneity. Use the equation !E°r (cell) = E°r (cathode) " E°r (anode) to calculate the standard cell potential. Solution: Cathode half-reaction equation: Au3+(aq) + 3 e− → Au(s) Anode half-reaction equation: Ag(s) → Ag+(aq) + e− From Table 1, Appendix B7, the reduction potential for the gold is 1.50 V and the reduction potential for silver is 0.80 V. The standard cell potential is !E°r (cell) = E°r (cathode) " E°r (anode) = 1.50 V " 0.80 V !E°r (cell) = 0.70 V
Statement: The standard cell potential for this cell is !E°r (cell) = 0.70 V . Because the standard cell potential is positive, the reaction is spontaneous. (c) Given: 2 Cu+(aq) → Cu(s) + Cu2+(aq) Required: !E°r (cell) and whether the reaction is spontaneous as written Analysis: Use Table 1 in Appendix B7 to predict reaction spontaneity. Use the equation !E°r (cell) = E°r (cathode) " E°r (anode) to calculate the standard cell potential. Solution: Cathode half-reaction equation: Cu+(aq) + e− → Cu(s) Anode half-reaction equation: Cu+(aq) → Cu2+(aq) + e− From Table 1, Appendix B7, the reduction potential for the copper(I) ion is 0.52 V and the reduction potential for copper(II) ion is 0.16 V. The standard cell potential is !E°r (cell) = E°r (cathode) " E°r (anode) = 0.52 V " 0.16 V !E°r (cell) = 0.36 V
Statement: The standard cell potential for this cell is !E°r (cell) = 0.36 V . Because the standard cell potential is positive, the reaction is spontaneous.
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Chapter 10: Electrochemical Cells
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37. (a) Given: MnO4−(aq) + I−(aq) → I2(s) + Mn2+(aq) Required: !E°r (cell) , whether the reaction is spontaneous as written, and the balanced equation in acid Analysis: Use Table 1 in Appendix B7 to predict reaction spontaneity. Use the equation !E°r (cell) = E°r (cathode) " E°r (anode) to calculate the standard cell potential. Solution: Manganese gains 5 electrons and iodine loses 1 electron. Therefore, manganese is reduced and iodine is oxidized. Cathode half-reaction equation: MnO4−(aq) + 5 e− → Mn2+(aq) Anode half-reaction equation: 2 I−(aq) → I2(aq) + 2 e− From Table 1, Appendix B7, the reduction potential for the permanganate ion is 1.51 V and the reduction potential for iodine is 0.54 V. The standard cell potential is !E°r (cell) = E°r (cathode) " E°r (anode) = 1.51 V " 0.54 V !E°r (cell) = 0.97 V
To balance the equation in acid, first balance electrons by multiplying the equation for the permanganate ion half-reaction by 2 and the iodine half-reaction equation by 5: 2 MnO4−(aq) + 10 e− → 2 Mn2+(aq) 10 I−(aq) → 5 I2(aq) + 10 e− Add the half-reaction equations: 2 MnO4−(aq) + 10 I−(aq) → 2 Mn2+(aq) + 5 I2(aq) Balance oxygen atoms by adding water. 2 MnO4−(aq) + 10 I−(aq) → 2 Mn2+(aq) + 5 I2(aq) + 8 H2O(l) Balance hydrogen atoms by adding hydrogen ions. 2 MnO4−(aq) + 10 I−(aq) + 16 H+(aq) → 2 Mn2+(aq) + 5 I2(aq) + 8 H2O(l) Statement: The standard cell potential for this cell is !E°r (cell) = 0.97 V . Because the standard cell potential is positive, the reaction is spontaneous. The balanced equation in acid is 2 MnO4−(aq) + 10 I−(aq) + 16 H+(aq) → 2 Mn2+(aq) + 5 I2(aq) + 8 H2O(l). (b) Given: MnO4−(aq) + F−(aq) → F2(g) + Mn2+(aq) Required: !E°r (cell) , whether the reaction is spontaneous as written, and the balanced equation in acid Analysis: Use Table 1 in Appendix B7 to predict reaction spontaneity. Use the equation !E°r (cell) = E°r (cathode) " E°r (anode) to calculate the standard cell potential.
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Chapter 10: Electrochemical Cells
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Solution: Manganese gains 5 electrons and fluorine loses 1 electron. Therefore, manganese is reduced and fluorine is oxidized. Cathode half-reaction equation: MnO4−(aq) + 5 e− → Mn2+(aq) Anode half-reaction equation: 2 F−(aq) → F2(g) + 2 e− From Table 1, Appendix B7, the reduction potential for the permanganate ion is 1.51 V and the reduction potential for fluorine is 2.87 V. The standard cell potential is !E°r (cell) = E°r (cathode) " E°r (anode) = 1.51 V " 2.87 V !E°r (cell) = "1.36 V
To balance the equation in acid, first balance electrons by multiplying the equation for the permanganate ion half-reaction by 2 and the fluorine half-reaction equation by 5: 2 MnO4−(aq) + 10 e− → 2 Mn2+(aq) 10 F−(aq) → 5 F2(g) + 10 e− Add the half-reaction equations: 2 MnO4−(aq) + 10 F−(aq) → 2 Mn2+(aq) + 5 F2(g) Balance oxygen atoms by adding water. 2 MnO4−(aq) + 10 F−(aq) → 2 Mn2+(aq) + 5 F2(g) + 8 H2O(l) Balance hydrogen atoms by adding hydrogen ions. 2 MnO4−(aq) + 10 F−(aq) + 16 H+(aq) → 2 Mn2+(aq) + 5 F2(g) + 8 H2O(l) Statement: The standard cell potential for this cell is !E°r (cell) = "1.36 V . Because the standard cell potential is negative, the reaction is not spontaneous. The balanced equation in acid is 2 MnO4−(aq) + 10 F−(aq) + 16 H+(aq) → 2 Mn2+(aq) + 5 F2(g) + 8 H2O(l). 38. (a) The electrolyte in a typical alkaline dry cell is potassium hydroxide. (b) The electrolytes of alkaline dry cells are corrosive. When handling a corroded alkaline dry cell I would recommend wearing gloves to prevent your skin coming into contact with the electrolytes. 39. (a) The two substances that must always be present for rusting to occur are oxygen and water. (b) Rust forms more readily in a damp environment than in dry conditions because water is needed to complete the cathode half-cell reaction with oxygen as follows: O2(g) + 2 H2O(l) + 4 e– → 4 OH–(aq)
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Chapter 10: Electrochemical Cells
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40. Examples may vary. Sample answer:
41. (a) Since aluminum is a light metal, its use instead of steel for automobile parts will decrease the mass of the automobile, resulting in increased fuel efficiency. (b) One positive impact on the environment of using aluminum instead of steel for automobile parts is a decrease in the greenhouse gas emissions from automobiles, since less fuel will be used to drive the same distance. Two negative impact on the environment are the increase in greenhouse gas production and the release of toxic ions and metals in the production and purification of aluminum. 42. (a) When molten potassium fluoride, KF(s), is electrolyzed using inert electrodes: Anode half-reaction equation: 2 F–(l) → F2(g) + 2 e– Cathode half-reaction equation: 2 K+(l) + 2 e– → 2 K(l) (b) When molten copper(II) chloride, CuCl2(s), is electrolyzed using inert electrodes: Anode half-reaction equation: 2 Cl–(l) → Cl2(g) + 2 e– Cathode half-reaction equation: Cu2+ + 2 e– → Cu(l) (c) When molten magnesium iodide, MgI2(s), is electrolyzed using inert electrodes: Anode half-reaction equation: 2 I–(l) → I2(l) + 2 e– Cathode half-reaction equation: Mg2+(l) + 2 e– → Mg(l) Analysis and Application
43. Answers may vary. Sample answers: (a) A potential difference of 1.2 V could be supplied by a galvanic cell consisting of chlorine gas and copper ions. (b) Cathode half-reaction equation: Cl2(g) + 2 e− → 2 Cl−(aq) Anode half-reaction equation: Cu+(aq) → Cu2+(aq) + e−
E°r = 1.36 V
E°r = 0.16 V
!E°r (cell) = E°r (cathode) " E°r (anode) = 1.36 V " 0.16 V !E°r (cell) = 1.20 V
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Chapter 10: Electrochemical Cells
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(c) Line notation for the cell: Pt(s) | Cu2+(aq) | | Cl−(aq) | Pt(s) 44. (a) The student could build galvanic cells using 6 combinations of 2 electrodes— Ag−Zn, Ag−Cu, Ag−Sn, Zn−Cu, Zn−Sn, Cu−Sn—and their corresponding electrolytes and compare their potential differences. (b) To carry out this task, the student would require 4 beakers for the electrolytes, a U-tube and cotton plugs for a salt bridge, a connecting wire with alligator clips, and a voltmeter to measure the potential difference. (c) Sketch of the setup necessary to determine experimentally which combination of metal strips and solutions will produce the maximum potential difference:
Silver nitrate will be a source of Ag+ ions, so from Table 1 in Appendix B7, the standard reduction potential for the copper metal strip will be E°r = +0.80 V. Zinc nitrate will be a source of Zn2+ ions, so from Table 1 the standard reduction potential for the copper metal strip will be E°r = !0.76 V. Copper(II) nitrate will be a source of Cu2+ ions, so from Table 1 the standard reduction potential for the copper metal strip will be E°r = +0.34 V. Tin(II) nitrate will be a source of Sn2+ ions, so from Table 1 the standard reduction potential for the copper metal strip will be E°r = !0.14 V. Based on these values, the greatest cell potential would be achieved by using the metals that have the greatest potential difference, resulting in the highest positive number for !E°r (cell) , that is, by using silver as the cathode and zinc as the anode. The resulting standard cell potential would be !E°r (cell) = E°r (cathode) " E°r (anode) = 0.80 V " ("0.76 V) !E°r (cell) = 1.56 V
45. (a) Given: Ag(s), Ag+(aq), Fe2+(aq), Pt(s) Required: balanced ionic equation, !E°r (cell) Analysis: Use the equation !E°r (cell) = E°r (cathode) " E°r (anode) to calculate the standard cell potential.
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Chapter 10: Electrochemical Cells
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Solution: From Table 1 in Appendix B7, the half-reaction equations and standard reduction potentials are Ag+(aq) + e− → Ag(s) E°r = 0.80 V Fe3+(aq) + e− → Fe2+(aq) E°r = 0.77 V The half-cell reaction with more positive potential is the reduction half-reaction. In this case, the reduction of silver ion occurs at the cathode: Ag+(aq) + e− → Ag(s) Iron(II) ion is oxidized at the anode. The equation for this half-cell reaction is written as an oxidization reaction: Fe2+(aq) → Fe3+(aq) + e− The net ionic equation is Ag+(aq) + Fe2+(aq) → Fe3+(aq) + Ag(s) The standard cell potential is !E°r (cell) = E°r (cathode) " E°r (anode) = 0.80 V " 0.77 V !E°r (cell) = 0.03 V
Statement: The net ionic equation for this reaction is Ag+(aq) + Fe2+(aq) → Fe3+(aq) + Ag(s) The standard cell potential is !E°r (cell) = 0.03 V. (b) Given: Au(s), Au3+(aq), Cu+(aq), Cu2+(aq), Pt(s) Required: balanced ionic equation, !E°r (cell) Analysis: Use the equation !E°r (cell) = E°r (cathode) " E°r (anode) to calculate the standard cell potential. Solution: From Table 1 in Appendix B7, the half-reaction equations and standard reduction potentials are Au3+(aq) + 3 e− → Au(s) E°r = 1.50 V Cu2+(aq) + e− → Cu+(aq)
E°r = 0.16 V
The half-cell reaction with more positive potential is the reduction half-reaction. In this case, the reduction of gold ion occurs at the cathode: Au3+(aq) + 3 e− → Au(s) Copper(I) ion is oxidized at the anode. The equation for this half-cell reaction is written as an oxidization reaction: Cu+(aq) → Cu2+(aq) + e− To balance electrons, multiply the equation for the oxidation reaction by 3: 3 Cu+(aq) → 3 Cu2+(aq) + 3 e−
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Chapter 10: Electrochemical Cells
10-15
The net ionic equation is Au3+(aq) + 3 Cu+(aq) → 3 Cu2+(aq) + Au(s) The standard cell potential is !E°r (cell) = E°r (cathode) " E°r (anode) = 1.50 V " 0.16 V !E°r (cell) = 1.34 V
Statement: The net ionic equation for this reaction is Au3+(aq) + 3 Cu+(aq) → 3 Cu2+(aq) + Au(s) The standard cell potential is !E°r (cell) = 1.34 V. (c) Given: Cd(s), Cd2+(aq), VO2+(aq), H+(aq), Pt(s) Required: balanced ionic equation, !E°r (cell) Analysis: Use the equation !E°r (cell) = E°r (cathode) " E°r (anode) to calculate the standard cell potential. Solution: From Table 1 in Appendix B7, the half-reaction equations and standard reduction potentials are Cd2+(aq) + 2 e− → Cd(s) E°r = !0.40 V VO2+(aq) + 2 H+(aq) + e− → VO2+(aq) + H2O(l) E°r = 1.00 V The half-cell reaction with more positive potential is the reduction half-reaction. In this case, the reduction of VO2+(aq) occurs at the cathode: VO2+(aq) + 2 H+(aq) + e− → VO2+(aq) + H2O(l) Cadmium is oxidized at the anode. The equation for this half-cell reaction is written as an oxidization reaction: Cd(s) → Cd2+(aq) + 2 e− To balance electrons, multiply the equation for the reduction reaction by 2: 2 VO2+(aq) + 4 H+(aq) + 2 e− → 2 VO2+(aq) + 2 H2O(l) The net ionic equation is Cd(s) + 2 VO2+(aq) + 4 H+(aq) → Cd2+(aq) + 2 VO2+(aq) + 2 H2O(l) The standard cell potential is !E°r (cell) = E°r (cathode) " E°r (anode) = 1.00 V " ("0.40 V) !E°r (cell) = 1.40 V
Statement: The net ionic equation for this reaction is Cd(s) + 2 VO2+(aq) + 4 H+(aq) → Cd2+(aq) + 2 VO2+(aq) + 2 H2O(l) The standard cell potential is !E°r (cell) = 1.40 V.
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Chapter 10: Electrochemical Cells
10-16
46. (a) Given: IO3−(aq) + Fe2+(aq) → Fe3+(aq) + I2(aq) Required: sketch of cell, balanced equation, !E°r (cell) Analysis: Use the equation !E°r (cell) = E°r (cathode) " E°r (anode) to calculate the standard cell potential. Solution: From Table 1 in Appendix B7 and the given unbalanced equation, the halfreaction equations and standard reduction potentials are IO3−(aq) + 6 H+(aq) + 5 e−→
1 I2(s) E°r = 1.20 V 2
Fe2+(aq) → Fe3+(aq) + e− E°r = 0.77 V The iodate ion half-reaction has the more positive reduction potential, so iodate ion is reduced at the cathode and iron(II) ion is oxidized at the anode. To balance electrons, multiply the equation for the iron(II) ion half-reaction by 5. 5 Fe2+(aq) → 5 Fe3+(aq) + 5 e− Add the two half-reaction equations. IO3−(aq) + 6 H+(aq) + 5 Fe2+(aq) →
1 I2(s) + 5 Fe3+(aq) 2
Add water to balance oxygen atoms.
1 I2(s) + 5 Fe3+(aq) + 3 H2O(l) 2 Multiply all entities by 2 to obtain the net ionic equation. 2 IO3−(aq) + 12 H+(aq) + 10 Fe2+(aq) → I2(s) + 10 Fe3+(aq) + 6 H2O(l) IO3−(aq) + 6 H+(aq) + 5 Fe2+(aq) →
!E°r (cell) = E°r (cathode) " E°r (anode) = 1.20 V " 0.77 V !E°r (cell) = 0.43 V
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Chapter 10: Electrochemical Cells
10-17
Statement: The net ionic equation for this cell is 2 IO3−(aq) + 12 H+(aq) + 10 Fe2+(aq) → I2(s) + 10 Fe3+(aq) + 6 H2O(l). The standard cell potential is !E°r (cell) = 0.43 V. (b) Given: Zn(s) + Ag+(aq) → Zn2+(aq) + Ag(s) Required: sketch of cell, balanced equation, !E°r (cell) Analysis: Use the equation !E°r (cell) = E°r (cathode) " E°r (anode) to calculate the standard cell potential. Solution: From Table 1 in Appendix B7 and the given unbalanced equation, the halfreaction equations and standard reduction potentials are Zn(s) → Zn2+(aq) + 2 e− E°r = !0.76 V Ag+(aq) + e− → Ag(s) E°r = +0.80 V The silver ion half-reaction has the more positive reduction potential, so silver ion is reduced at the cathode and zinc is oxidized at the anode. To balance electrons, multiply the equation for the silver half-reaction by 2. 2 Ag+(aq) + 2 e− → 2 Ag(s) Add the two half-reaction equations to obtain the net ionic equation. Zn(s) + 2 Ag+(aq) → Zn2+(aq) + 2 Ag(s) !E°r (cell) = E°r (cathode) " E°r (anode) = 0.80 V " ("0.76 V) !E°r (cell) = 1.56 V
Statement: The net ionic equation for this cell is Zn(s) + 2 Ag+(aq) → Zn2+(aq) + 2 Ag(s). The standard cell potential is !E°r (cell) = 1.56 V.
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Chapter 10: Electrochemical Cells
10-18
47. (a) Given: !E°r (cell) = 2.00 V , Cu(s), Cu(NO3)2(aq) Required: identity of metal Analysis: Use the equation !E°r (cell) = E°r (cathode) " E°r (anode) to calculate the reduction potential of the unknown metal. Solution: From Table 1 in Appendix B7, the half-reaction equation and reduction potential for copper metal are Cu2+(aq) + 2 e− → Cu(s) E°r = 0.34 V The standard cell potential and the reduction potential of copper are both positive, so copper must be the cathode.
!E°r (cell) = E°r (cathode) " E°r (anode) E°r (anode) = E°r (cathode) " !E°r (cell) = 0.34 V " 2.00 V E°r (anode) = "1.66 V From Table 1, the unknown metal is aluminum. Statement: The unknown metal is aluminum. (b) The half-reaction equation for aluminum is Al3+(aq) + 3 e− → Al(s) This reaction occurs at the anode, so it is written as an oxidation reaction. Al(s) → Al3+(aq) + 3 e− To balance electrons, multiply equation for the copper half-reaction by 3 and the aluminum half-reaction equation by 2. 3 Cu2+(aq) + 6 e− → 3 Cu(s) 2 Al(s) → 2 Al3+(aq) + 6 e− Add the two half-reaction equations. The net ionic equation is 3 Cu2+(aq) + 2 Al(s) → 2 Al3+(aq) + 3 Cu(s)
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Chapter 10: Electrochemical Cells
10-19
48. The half-reactions and reduction potentials of the given entities (in boldface) are !! Na+(aq) + e– # E°r = –2.71 V !" ! Na(s) – !! Cl2(g) = 2 e– # !" ! 2 Cl (aq)
E°r = 1.36 V
!! Ag+(aq) + e– # !" ! Ag(s)
E°r = 0.80 V
!! Zn2+(aq) + 2 e– # !" ! Zn(s)
E°r = !0.76 V
!! Pb2+(aq) + 2 e– # E°r = !0.13 V !" ! Pb(s) Arrange the half-reactions in order of decreasing strength of oxidizing agents. – !! Cl2(g) = 2 e– # !" ! 2 Cl (aq) E°r = 1.36 V !! Ag+(aq) + e– # !" ! Ag(s)
E°r = 0.80 V
!! Pb2+(aq) + 2 e– # !" ! Pb(s)
E°r = !0.13 V
!! Zn2+(aq) + 2 e– # !" ! Zn(s)
E°r = !0.76 V
!! Na+(aq) + e– # E°r = –2.71 V !" ! Na(s) (a) The strongest oxidizing agent is silver ion, Ag+(aq), because it appears highest in the table of reduction potentials on the reactant side of the equation. (b) The strongest reducing agent is zinc metal, Zn(s), because it appears lowest in the table of reduction potentials on the product side of the equation. (c) The reduction potential of SO42–(aq) is 0.20 V, so the entities that can be oxidized by sulfate ions in acid are lead metal, Pb(s), and zinc metal, Zn(s), because they appear lower than sulfate ions on the product side of the equation. (d) The reduction potential of aluminum metal, Al(s), is –1.66 V, so Ag+(aq), and zinc ion, Zn2+(aq) can be reduced by aluminum metal, because they appear above aluminum metal in the table of reduction potentials on the reactant side of the equation. 49. The half-reactions and reduction potentials of the given entities (in boldface) are Br–(aq), Br2(g), H+(aq), H2(g), La3+(aq), Ca(s), Cd(s) − !! Br2(l) + 2 e− # E°r = 1.09 V !" ! 2 Br (aq) !! 2 H+(aq) + 2 e– # !" ! H2(g)
E°r = 0 V
!! La3+(aq) + 3 e– # !" ! La(s)
E°r = !2.37 V
!! Ca2+(aq) + 2 e– # !" ! Ca(s)
E°r = !2.76 V
!! Cd2+(aq) + 2 e– # E°r = !0.40 V !" ! Cd(s) Arrange the half-reactions in order of decreasing strength of oxidizing agents. − !! Br2(l) + 2 e− # E°r = 1.09 V !" ! 2 Br (aq) !! 2 H+(aq) + 2 e– # !" ! H2(g)
E°r = 0 V
!! Cd2+(aq) + 2 e– # !" ! Cd(s)
E°r = !0.40 V
!! La3+(aq) + 3 e– # !" ! La(s)
E°r = !2.37 V
!! Ca2+(aq) + 2 e– # !" ! Ca(s)
E°r = !2.76 V
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Chapter 10: Electrochemical Cells
10-20
(a) The strongest oxidizing agent is Br2(l) because it appears highest in the table of reduction potentials on the reactant side of the equation. (b) The strongest reducing agent is Ca(s) because it appears lowest in the table of reduction potentials on the product side of the equation. (c) The reduction potential of MnO4–(aq) is 1.68 V, so the entities that can be oxidized by permanganate ions in acid are Br−(aq), H2(g), Cd(s), and Ca(s) because they appear lower than permanganate ions on the product side of the equation. (d) The reduction potential of zinc metal, Zn(s), is –0.76 V, so the entities that can be reduced by zinc metal are H+(aq) and Br2(l), because they appear above zinc metal in the table of reduction potentials on the reactant side of the equation. 50. Answers may vary. Sample answers: (a) A reagent that, at SATP in acidic solution, will oxidize bromide ions to bromine gas, Br2(g), but not oxidize chloride ions to chloride gas is Cr2O72– (aq). (b) A reagent that, at SATP in acidic solution, will oxidize magnesium to magnesium(II) ions, but not oxidize iron to iron(II) ions is Zn2+(aq). (c) A reagent that, at SATP in acidic solution, will reduce copper(II) ions to copper metal, but not to copper(I) ions is Ag(s), Cl–(aq). 51. (a) Given: Cu2+(aq) + 2 e− → Cu(s) E°r (SCE) = 0.242 V Required: !E°r (cell) , and whether SCE is the cathode or the anode Analysis: Use the equation !E°r (cell) = E°r (cathode) " E°r (anode) to calculate the standard cell potential. Solution: From Table 1 in Appendix B7, the reduction potential for copper(II) ions is 0.34 V. This reduction potential is more positive than the reduction potential for SCE; therefore, copper ion is reduced at the cathode and SCE is oxidized at the anode. !E°r (cell) = E°r (cathode) " E°r (anode) = 0.34 V " 0.242 V !E°r (cell) = 0.098 V = 0.10 V
Statement: SCE is the anode in this cell. The standard reduction potential of this cell is !E°r (cell) = 0.10 V. (b) Given: Fe3+(aq) + e− → Fe2+(aq) E°r (SCE) = 0.242 V Required: !E°r (cell) , and whether SCE is the cathode or the anode Analysis: Use the equation !E°r (cell) = E°r (cathode) " E°r (anode) to calculate the standard cell potential.
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Chapter 10: Electrochemical Cells
10-21
Solution: From Table 1 in Appendix B7, the reduction potential for iron(III) ions is 0.77 V. This reduction potential is more positive than the reduction potential for SCE; therefore, iron(III) ion is reduced at the cathode and SCE is oxidized at the anode. !E°r (cell) = E°r (cathode) " E°r (anode) = 0.77 V " 0.242 V !E°r (cell) = 0.53 V
Statement: SCE is the anode in this cell. The standard reduction potential of this cell is !E°r (cell) = 0.53 V. (c) Given: AgCl(aq) + e− → Ag(s) + Cl−(aq) E°r (SCE) = 0.242 V Required: !E°r (cell) , and whether SCE is the cathode or the anode Analysis: Use the equation !E°r (cell) = E°r (cathode) " E°r (anode) to calculate the standard cell potential. Solution: From Table 1 in Appendix B7, the reduction potential for chloride ions is 0.22 V. This reduction potential is less positive than the reduction potential for SCE; therefore, chloride ion is oxidized at the anode and SCE is reduced at the cathode. !E°r (cell) = E°r (cathode) " E°r (anode) = 0.242 V " 0.22 V !E°r (cell) = 0.02 V
Statement: SCE is the cathode in this cell. The standard reduction potential of this cell is !E°r (cell) = 0.02 V. (d) Given: Al3+(aq) + 3 e− → Al(s) E°r (SCE) = 0.242 V Required: !E°r (cell) , and whether SCE is the cathode or the anode Analysis: Use the equation !E°r (cell) = E°r (cathode) " E°r (anode) to calculate the standard cell potential. Solution: From Table 1 in Appendix B7, the reduction potential for aluminum ions is −1.66 V. This reduction potential is less positive than the reduction potential for SCE; therefore, aluminum ion is oxidized at the anode and SCE is reduced at the cathode. !E°r (cell) = E°r (cathode) " E°r (anode) = 0.242 V " ("1.66 V) !E°r (cell) = 1.90 V
Statement: SCE is the cathode in this cell. The standard reduction potential of this cell is !E°r (cell) = 1.90 V.
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Chapter 10: Electrochemical Cells
10-22
(e) Given: Ni2+(aq) + 2 e− → Ni(s) E°r (SCE) = 0.242 V Required: !E°r (cell) , and whether SCE is the cathode or the anode Analysis: Use the equation !E°r (cell) = E°r (cathode) " E°r (anode) to calculate the standard cell potential. Solution: From Table 1 in Appendix B7, the reduction potential for nickel(II) ions is −0.23 V. This reduction potential is less positive than the reduction potential for SCE; therefore, nickel(II) ion is oxidized at the anode and SCE is reduced at the cathode. !E°r (cell) = E°r (cathode) " E°r (anode) = 0.242 V " ("0.23 V) !E°r (cell) = 0.47 V
Statement: SCE is the cathode in this cell. The standard reduction potential of this cell is !E°r (cell) = 0.47 V. 52. (a) Given: 2 NaClO2(aq) + Cl2(g) → 2 ClO2(g) + 2 NaCl(aq) Required: !E°r (cell) for the production of ClO2(g) Analysis: Use the equation !E°r (cell) = E°r (cathode) " E°r (anode) to calculate the standard cell potential. Solution: From Table 1 in Appendix B7, the half-reaction equations and reduction potentials for chlorine gas and chlorine dioxide gas are Cl2(g) + 2 e− → 2 Cl−(aq) E°r = 1.36 V ClO2(g) + e− → ClO2−(aq) E°r = 0.954 V The chlorine gas half-reaction has the more positive reduction potential, so chlorine gas is reduced at the cathode and chlorine dioxide is oxidized at the anode. !E°r (cell) = E°r (cathode) " E°r (anode) = 1.36 V " 0.954 V !E°r (cell) = 0.41 V
Statement: The standard cell potential for the production of chlorine dioxide gas is 0.41 V. (b) Electrolysis is not required for the production of chlorine dioxide gas because the standard cell potential is a positive value, so the reaction is spontaneous. 53. When jump starting a dead battery, the ground jumper connection should be attached to a remote part of the engine block so that any spark that is produced when the cable is disconnected will not cause the battery to explode. The reaction in the battery when starting produces gaseous hydrogen and oxygen, which can ignite in the presence of a spark.
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Chapter 10: Electrochemical Cells
10-23
54. Both fusion and fission are nuclear reactions that release huge amounts of energy; however, fusion has two significant advantages. While fission produces radioactive products, the products of fusion are not radioactive or toxic. While fission relies on fuels such as uranium, fusion uses readily available hydrogen as a fuel. Fusion also has one very significant disadvantage—it only occurs at very high temperatures, which researchers have not yet succeeded in reproducing in a format that could be used for commercial energy generation. 55. The dissolved carbon dioxide in natural water affects the rusting of metallic iron as follows: the hydrogen ions in dissolved carbon dioxide acidify the water, which results in rusting of metallic iron occurring more quickly. 56. When iron rusts it loses electrons. Galvanized nails have a zinc coating that forms a protective barrier between the air and the iron nail. Galvanized nails will not rust because zinc holds its electrons more loosely than iron does, so it will donate electrons instead of iron. In this way, zinc prevents the rusting of the nails. Ungalvanized nails exposed to the elements will begin to rust. The iron nails hammered through the aluminum siding did not rust because aluminum donated electrons instead of iron so, like the zinc on the galvanized nails, the aluminum prevented the rusting of the nails. 57. (a) Answers may vary. Sample answer: The metals that rarely corrode in air include gold, platinum, and silver. These are called the noble metals because people recognized that they were unreactive and suitable for use in highly esteemed products such as coins and jewellery. (b) Answers may vary. Sample answer: The noble metals do not generally corrode because they have stable electron orbitals with full d orbitals. 58. Sterling silver corrodes more readily than pure silver or gold because it contains copper. The reduction potential of copper ( E°r = +0.34 V) is lower than that of silver ( E°r = +0.80 V) and of gold ( E°r = +0.1.50 V). 59. A balanced chemical equation for the reaction of gold in aqua regia is the following: Au(s) + HNO3(aq) + 4 HCl(aq) → AuCl4–(aq) + NO(g) + 2 H2O(l) + H+(aq) 60. (a) When 1.0 mol/L nickel bromide solution, NiBr2(aq) is used in an electrolytic cell: Anode half-reaction equation: 2 Br–(aq) → Br2(g) + 2 e– Cathode half-reaction equation: Ni2+(aq) + 2 e– → Ni(s) The reaction would result in the metal being plated with nickel. (b) When 1.0 mol/L aluminum fluoride solution, AlF3(aq) is used in an electrolytic cell: Anode half-reaction equation: 2 F–(aq) → F2(g) + 2 e– Cathode half-reaction equation: Al3+(aq) + 3 e– → Al(s) The reaction would result in the metal being plated with aluminum. (c) When 1.0 mol/L manganese iodide solution, MnI2(aq) is used in an electrolytic cell: Anode half-reaction equation: 2 I–(aq) → I2(l) + 2 e– Cathode half-reaction equation: Mn2+(aq) + 2 e– → Mn(s) The reaction would result in the metal being plated with manganese.
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Chapter 10: Electrochemical Cells
10-24
Evaluation
61. Given: ClO−(aq) + H2O(l) + 2 e− → 2 OH−(aq) + Cl−(aq) E°r = 0.90 V N2H4(aq) + 2 H2O(l) + 2 e− → 2 NH3(g) + 2 OH−(aq) E°r = !0.10 V Required: !E°r (cell) Analysis: Use the equation !E°r (cell) = E°r (cathode) " E°r (anode) to calculate the standard cell potential. Solution: Determine whether mixing the given solutions results in a spontaneous reaction. Since the reduction potential for hypochlorite ion is more positive than the reduction potential for hydrazine, the reduction of hypochlorite ions occurs at the cathode and hydrazine is oxidized at the anode. !E°r (cell) = E°r (cathode) " E°r (anode) = 0.90 V " ("0.10 V) !E°r (cell) = 1.00 V
The standard cell potential is a positive value, so the reaction is spontaneous. Statement: Since combining hydrazine with hypochlorite ion results in a spontaneous reaction, mixing household bleach (a solution of sodium hypochlorite, NaClO(aq)) with household ammonia or glass cleaners that contain ammonia is not safe. 62. According to the redox table, the half-reaction equation associated with the reaction of mercury metal with stomach acid is: Hg2Cl2(aq) + 2 e− → 2 Hg(l) + 2 Cl−(aq) E°r = 0.34 V The positive reduction potential indicates that this half-reaction is spontaneous in the forward direction, which means that it is non-spontaneous in the reverse direction. From this result, we can conclude that the reaction of mercury metal with stomach acid is not a spontaneous reaction, so swallowing dental fillings containing elemental mercury poses minimal health risks, as asserted by the Canadian Dental Association. 63. Answers will vary. Sample answer: Although the scientific meaning of the word “battery” is quite different from its everyday use, I don’t think that the scientific definition of a battery should be changed, because the term is the accepted term in science and has been in use for hundreds of years. There are many terms in everyday language that are used inaccurately. Studying science teaches us the correct meaning of these terms. 64. (a) Corrosion is an example of an galvanic process. (It is spontaneous.) (b) The corrosion of steel involves the oxidation of iron coupled with the reduction of oxygen. (c) Steel rusts more easily in humid climates than in dry ones. This is correct. (d) Salting roads in the winter has the added cost of increasing the corrosion of steel. (The salts in solution act as electrolytes.) (e) The key to cathodic protection is to connect a metal more easily oxidized than iron to the surface to be protected. This is correct.
Copyright © 2012 Nelson Education Ltd.
Chapter 10: Electrochemical Cells
10-25
65. Answers may vary. Sample answer: Place four steel strips in shallow pools of salt water. Before doing this, galvanize one with zinc, attach one to a block of zinc by an electrode, and apply a DC current to one. The fourth will act as a control. Allow the steel strips to sit for two days, then observe and record the amount of corrosion on each strip. Clean the steel strips, remove any corrosion, and determine the mass of each strip. If the methods are as effective as I predict they will be, the control will have the most corrosion and will be the lightest. The galvanized strip will have no visible rust but may have additional mass from zinc oxides that formed, likely the most effective protection. The steel attached to a block of zinc will not change mass and will show only minor signs of corrosion. The impressed current strip will have reduced corrosion compared to the control. Materials required: four steel strips, four shallow non-reactive containers, salt water, zinc to galvanize, zinc block, electrolytic cell, source of DC current, balance. 66. (a) Pure aluminum is difficult to produce because it is difficult to extract from the aluminum oxides found in bauxite ore. The Hall-Héroult process uses molten cryolite, Na3AlF6(l), as a solvent for the aluminum oxide. (b) The Hall-Héroult process made it less expensive to produce aluminum, which enabled aluminum to be used in many applications, including in the steel industry. Reflect on Your Learning
67. Answers may vary. Sample answer: I do not find the use of an arbitrary reference electrode challenging. Instead, I find it helpful. It helps scientists to compare different cells using one standard. Many measured quantities have an arbitrary “zero”, for example, temperature measured in degrees Celsius or degrees Fahrenheit; time measured in years CE or BCE; and longitude measured east of west of the Prime Meridian. 68. Answers may vary. Sample answer: Before studying this chapter, I knew commercial batteries contained some substance that produced electricity by undergoing a chemical reaction, but I thought when the battery went dead the chemicals inside had all been released, and were no longer inside the battery. I now understand that the chemicals do not leave the battery, but they transform from reactants to products. Recharging a battery reverses the chemical reactions so that the battery once again contains reactants and can produce energy by undergoing the reaction again. 69. Answers may vary. Students should include examples of items they use in their lives that involve spontaneous electrochemical reactions, such as battery-powered phones and corrosion of vehicles; as well as examples of items they use that were created by nonspontaneous electrochemical reactions, such as metal plated items and recharged batteries. 70. Answers may vary. Students should discuss the importance of making sure that results are accurate so they do not mislead other scientists, as well as the competitive nature of science that motivates scientists to want to be the first to announce new discoveries.
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Chapter 10: Electrochemical Cells
10-26
Research
71. Answers may vary, and should be presented according to question guidelines. Plating with gold, nickel, zinc, or other metals may be accomplished at home by purchasing kits designed for this purpose. These kits include most of the materials required for the procedure such as electrodes, chemicals, wires, and detailed instructions. Students may research one of these kits and make a plan for electroplating the ornament based on the included materials and instructions. Safety precautions, disposal methods, and appropriate attire such as gloves and goggles should also be taken into consideration, because some materials used in metal-plating, such as cyanide and chromic acid, are toxic. Some toxic or hazardous chemicals may not be appropriate to use at home under any circumstances. 72. Answers may vary, and should be presented according to question guidelines. One type of battery that students may research is the common alkaline battery, used in devices such as flashlights and remote controls. This type of battery contains a zinc metal anode, manganese dioxide cathode, and potassium hydroxide as the electrolyte. A cell consists of a cylinder with the manganese dioxide along the inner surface, separated from the zinc at the centre. The ions in the electrolyte solution can move freely through the porous separator. Alkaline batteries are cost-effective, easily available, and provide good performance. However, the alkaline electrolyte solution is corrosive and prone to leakage. Alkaline batteries are generally not rechargeable and should be recycled. Facilities for recycling will vary among municipalities. 73. Answers may vary. Sample answer: Advantages and Disadvantages of Lithium-ion Batteries Advantages Disadvantages rechargeable expensive large cell potential risk of explosion if overheated small mass performance decreases with use and age variety of shapes and sizes
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Chapter 10: Electrochemical Cells
10-27
74. Answers may vary. Sample answer: Lead storage batteries contain lead and sulfuric acid, and are classified as hazardous waste; they should be disposed of properly in order to minimize their environmental impact.
75. Lead storage batteries have been used in vehicles for the past century. These rechargeable batteries contain lead-based electrodes and sulfuric acid as an electrolyte. Lithium-ion batteries are frequently used in electric cars. Their electrodes may be composed of various materials between which lithium ions move. Lead storage batteries are durable and relatively inexpensive but there are environmental concerns associated with them because of their lead content. Lithium ion batteries have a larger cell potential, are relatively small and lightweight, and include environmentally safe components; a major disadvantage, however, is their high cost. 76. Answers may vary, and should be presented according to question guidelines. A “cradle to grave” analysis of new batteries involves assessing the entire life cycle of the batteries and their components to determine environmental consequences. This includes the environmental impact associated with acquisition of raw materials, manufacturing, distribution, usage, and disposal of the batteries. Students should evaluate the benefits of such an analysis and state an opinion about whether it should be necessary to conduct this type of analysis on new batteries prior to their introduction.
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Chapter 10: Electrochemical Cells
10-28
77. Answers may vary. Sample answer: Benefits and Consequences of Recycling Aluminum Benefits Consequences costs less than manufacturing from raw none materials lower energy consumption lower greenhouse gas emissions reduced landfill The percentage of aluminum that is recycled can be increased by raising awareness about common recyclable aluminum products such as beverage cans and by increasing the availability of recycling programs. Global aluminum recycling rates are between 70 % (for beverage cans) and 90 % (for construction and transportation uses of aluminum). 78. Answers may vary, and should include at least two “greener” electrolytic reactions. One example of an industrial process that produces a harmful byproduct is the cyanidation process used in the recovery of gold from ore: 4 Au(s) + 8 NaCN(aq) + O2(g) + 2 H2O(l) → 4 NaAu(CN) 2(aq) + 4 NaOH(aq) This redox reaction produces a soluble gold-cyanide complex, which is recovered in a subsequent precipitation reaction. However, the process solution contains free cyanide ions and byproducts such as cyanide metal complexes. Accidental spills can reach bodies of water and pose a hazard to the environment. A “greener” electrolytic reaction is the thiosulphate leaching of gold. The process also involves ammonia and copper catalysts, but a simplified overall reaction is given below: 4 Au(s) + 8 S2O32–(aq) + O2(g) + 2 H2O(l) → 4 [Au(S2O3)2]3–(aq) + 4 OH–(aq) This process does not involve cyanide and is a less-toxic alternative.
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Chapter 10: Electrochemical Cells
10-29
