Electrochemistry

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Electrochemistry

  1. 1. Electrochemistry = the area ofchemistry that deals with theinterconversion of electricalenergy and chemical energy. Electrochemical processes are redox reactions in which the energy released by a spontaneous reaction is converted to electricity or in which electricity is used to drive a non-spontaneous chemical reaction. 1
  2. 2. ELECTROCHEMISTRY Lithium-ion battery. 2
  3. 3. Galvanic or Voltaic cell = anelectrochemical cell that produceselectricity as a result of spontaneouschemical change 3
  4. 4. Why Does a Voltaic Cell Work? The spontaneous reaction occurs as a result of the different abilities of materials (such as metals) to give up their electrons and the ability of the electrons to flow through the circuit. Ecell > 0 for a spontaneous reaction1 Volt (V) = 1 Joule (J)/ Coulomb (C) 4
  5. 5. Example of Voltaic Cell Voltmeter Salt bridge Zn (–) K + (+) Cu Cl– Zn2+ Cu2+ Zn and Cu cell 5
  6. 6. Voltmeter Salt bridgeZn (–) K + (+) Cu Cl– Zn2+ Cu2+ 6
  7. 7. Voltmeter e– Anode Salt bridge Zn (–) K (+) Cu + Cl– Zn2+ Cu2+Oxidation half-reactionZn(s) Zn2+(aq) + 2e– 7
  8. 8. e– 2e– lost per Zn atom oxidized Zn Zn2+ Voltmeter e– Anode Salt bridge Zn (–) K (+) Cu + Cl– Zn2+ Cu2+ Oxidation half-reaction Zn(s) Zn2+(aq) + 2e– 8
  9. 9. e– 2e– lost per Zn atom oxidized Zn Zn2+ Voltmeter e– e– Anode Salt bridge Cathode Zn (–) K (+) Cu + Cl– Zn2+ Cu2+ Oxidation half-reaction Zn(s) Zn2+(aq) + 2e– Reduction half-reaction Cu2+(aq) + 2e– Cu(s) 9
  10. 10. e– 2e– gained per Cu2+ ion 2e– lost reduced per Zn atom oxidized Zn Cu2+ Cu e – Zn2+ Voltmeter e– e– Anode Salt bridge Cathode Zn (–) K (+) Cu + Cl– Zn2+ Cu2+ Oxidation half-reaction Zn(s) Zn2+(aq) + 2e– Reduction half-reaction Cu2+(aq) + 2e– Cu(s) 10
  11. 11. e– 2e– gained per Cu2+ ion 2e– lost reduced per Zn atom oxidized Zn Cu2+ Cu e – Zn2+   1.10 V e– e– Anode Salt bridge Cathode Zn (–) K (+) Cu + Cl– Zn2+ Cu2+ Oxidation half-reaction Zn(s) Zn2+(aq) + 2e– Reduction half-reaction Cu2+(aq) + 2e– Cu(s) Overall (cell) reaction 11 Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)
  12. 12. Figure 21.5 A voltaic cell based on the zinc-copper reaction.Oxidation half-reaction Reduction half-reactionZn(s) Zn2+(aq) + 2e- Cu2+(aq) + 2e- Cu(s) Overall (cell) reaction Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) 12
  13. 13. 13
  14. 14. Symbolic representation of cell diagram 14
  15. 15. Notation for a Voltaic Cell components of components of anode compartment cathode compartment (oxidation half-cell) (reduction half-cell)phase of lower phase of higher phase of higher phase of loweroxidation state oxidation state oxidation state oxidation state phase boundary between half-cellsExamples: Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu (s) Zn(s) Zn2+(aq) + 2e- Cu2+(aq) + 2e- Cu(s)| = represents boundary between an electrode and another phase (solution or gas)|| = signifiess that solutions are joined by a salt bridge 15
  16. 16. * Another way of representing the celldiagram:Zn(s) | Zn2+(aq) | KCl | Cu2+(aq) | Cu (s) 16
  17. 17. Exercise: Write the cell diagram ofthe following electrochemical celland identify the oxidation andreduction couple: Voltmeter Salt bridge Cu (–) K + (+) Ag Cl– Cu(NO3)2 AgNO3 17
  18. 18. Exercise: Write the cell diagram ofthe following electrochemical celland identify the oxidation andreduction couple: Voltmeter Salt bridge Cu (–) K + (+) Ag Cl– Cu(NO3)2 AgNO3 Cell Diagram: Cu (s) | Cu2+(aq) || Ag+(aq) | Ag (s)Oxidation Cu → Cu2+ + 2 e-Reduction (Ag+ + e- → Ag)2 2 Ag+ + Cu → Cu2+ + 2 Ag 18
  19. 19. Eo = Standard Electrode Potential = is based on the tendency for a reduction process to occur at the electrode 19
  20. 20. Rules:•By convention, electrode potentials are written as reductions. The Eovalues apply to the half-cell reactions as read in the forward direction (leftto right).•The more positive Eo, the greater the tendency to be reduced.•The half-cell reactions are reversible, (sign changes). The reduction half-cell potential and the oxidation half-cell potential are added to obtain theE0cell.•Eo values are unaffected by multiplying half-equations by constantcoefficients. 20
  21. 21. Table 21.2 Selected Standard Electrode Potentials (298K) Half-Reaction E0(V) F2(g) + 2e- 2F-(aq) +2.87 Cl2(g) + 2e- 2Cl-(aq) +1.36 MnO2(g) + 4H+(aq) + 2e- Mn2+(aq) + 2H2O(l) +1.23 NO3-(aq) + 4H+(aq) + 3e- NO(g) + 2H2O(l) +0.96 strength of reducing agent Ag+(aq) + e- Ag(s) +0.80strength of oxidizing agent Fe3+(g) + e- Fe2+(aq) +0.77 O2(g) + 2H2O(l) + 4e- 4OH-(aq) +0.40 Cu2+(aq) + 2e- Cu(s) +0.34 2H+(aq) + 2e- H2(g) 0.00 N2(g) + 5H+(aq) + 4e- N2H5+(aq) -0.23 Fe2+(aq) + 2e- Fe(s) -0.44 2H2O(l) + 2e- H2(g) + 2OH-(aq) -0.83 Na+(aq) + e- Na(s) -2.71 Li+(aq) + e- Li(s) -3.05 21
  22. 22. Example: Cu (s) | Cu2+(aq) || Ag+(aq) | Ag (s)Oxidation Cu → Cu2+ + 2 e- - 0.34Reduction (Ag+ + e- → Ag)2 0.80 2 Ag+ + Cu → Cu2+ + 2 Ag 0.46 V*reaction is spontaneous 22
  23. 23. Problem:1) Write the cell reactions for the electrochemical cellsdiagrammed below and calculate Eo cell for eachreaction. Will these reactions occur spontaneously ornon-spontaneously?a) Zn(s) | Zn2+(aq) | Na2SO4 | Sn2+(aq) | Sn (s)b) Pt(s) | Fe2+(aq), Fe3+(aq) | | Sn+4(aq), Sn2+(aq) | Pt(s)c) Al(s) | Al3+(aq) | NH4NO3 | Cu2+(aq) | Cu (s) 23
  24. 24. Answers:a) Ox: Zn → Zn2+ + 2e- 0.76 V Red: Sn2+ + 2e- → Sn -0.14 VZn + Sn2+ → Zn2+ + Sn Eo = 0.62 V, spontaneousb) Ox: 2 (Fe2+ → Fe3+ + 1e-) -0.77 V Red: Sn4+ + 2e- → Sn2+ 0.15 V2 Fe2+ + Sn4+ → 2 Fe3+ + Sn2+ Eo = -0.62 V, non-spoc) Ox: 2 (Al → Al3+ + 3e-) 1.66 V Red: 3 (Cu2+ + 2e- → Cu) 0.34 V2 Al + 3 Cu2+ → 3 Cu + 2 Al3+ Eo = 2.00 V,spontaneous 24
  25. 25. Problem:2) Give the cell diagram and determine the Eo cell forthe reaction: Zn(s) + Cl2(g) ZnCl2(aq).Answers:Cell diagram: Zn(s) | Zn2+(aq) | | Cl2(g), Cl-(aq) | Pt (s)Ox: Zn → Zn2+ + 2e- 0.76 VRed: Cl2 + 2e- → 2 Cl- 1.36 VZn + Cl2 → Zn2+ + 2 Cl- 2.12 V = Eo 25
  26. 26. Free Energy = ▲G▲G = ▲H - T▲S ▲G = 0 , the system is at equilibrium ▲G < 0 , spontaneous reaction ▲G > 0 , non-spontaneous reaction 26
  27. 27. Free Energy = ▲G ▲G = ▲H - T▲S ▲G = ( - ), always a spontaneous processin ELECTROCHEMISTRY ▲G = Wmax (max. amt. of work that can be done) 27
  28. 28. in a voltaic cell, chemical energy is convertedinto electrical energy Voltages of Some Voltaic or Galvanic Cells Voltaic Cell Voltage (V) Common alkaline battery 1.5 Lead-acid car battery (6 cells = 12V) 2.0 Calculator battery 1.3 Electric eel 0.15 28
  29. 29. Electrical energy = volts X coulombs = JoulesThe total charge is determined by the n of electrons that pass through the circuit. By definition:total charge = n F (1 F = 96,500 C/mol)since 1 J = 1 C X 1 VWe can also express the units of Faraday as 1 F = 96,500 J/V.mol 29
  30. 30. The total work done is theproduct of 3 terms:1. The emf (voltage) of the cell2. The n of e-s transferred between the electrodes3. The electric charge per mole of e-s. Welec = (n) (F) (Eocell) 30
  31. 31. Welec = (n) (F) (Eocell)Since for a spontaneous reaction ▲G < 0 ▲G = - (n) (F) (Eocell) 31
  32. 32. Problem: Determine ▲G for the reaction: Alo(s) | Al3+(aq) | | Cu2+(aq) | Cuo (s)oxi: Al → Al3+ + 3 e- X2 1.66 Vred: Cu2+ + 2 e- → Cu X 3 0.34 V 2 Al + 3 Cu2+ → 2 Al3++ 3 Cu 2.00 V = Eo▲G = - (n F Eocell) = - ( 6 mol e- X 96500 C/mol e- X 2.00 V ) = - 1,158,000 J or = - 1,158 kJ 32
  33. 33. Problem:2) Give the cell diagram and determine ▲G for thereaction: Zn(s) + Cl2(g) ZnCl2(aq).Answers:Cell diagram: Zn(s) | Zn2+(aq) | | Cl2(g) , Cl-(aq) | Pt (s)Ox: Zn → Zn2+ + 2e- 0.76 VRed: Cl2 + 2e- → 2 Cl- 1.36 VZn + Cl2 → Zn2+ + 2 Cl- 2.12 V = Eo▲G = - (n F Eocell) = - ( 2 mol e- X 96500 C/mol e- X 2.12 V ) 33 = - 409,160 J or - 409.2 kJ
  34. 34. EXERCISES: 34
  35. 35. I. Give the following:a) cell diagram, c) Eo for the net reaction andb) anode and cathode reactions, d) ▲G for the following reactions 1. Ag+(aq) + Mgo(s) → Mg2+(aq) + Ago(s) 2. I2(s) + Cl-(aq) → Cl2(g) + I-(aq) 3. Br2o(aq) + Fe2+(aq) → FeBr3(aq) 4. Pb2+(aq) is displaced from solution by Al(s) 5. MgBr2(aq) is produced from Mg(s) and Br2(l) 6. Cl2(g) is reduced to Cl-(aq) and Fe(s) is oxidized to Fe2+(aq) 35
  36. 36. Sketch or draw and label the voltaic cell from the spontaneous reaction of Cu2+ and Sn+2 solutions. Indicate the following:the a) anode, b) cathode, c) oxidation reaction, d) reduction reaction, e) final equation, and f) direction of electron flow. 36
  37. 37. Arrange the following metals in the order of decreasing oxidizing strength: Ni, Au, Ag, Al, and Cr. 37
  38. 38. 38
  39. 39. CORROSION OF IRON 39
  40. 40. CORROSION OF IRON No indication of corrosion  Basic or alkaline solutions  (NaOH, KOH, Mg(OH)2) With indication of corrosion  Neutral solutions  (including water, Na2SO4, K3PO4)  Acidic solutions  (HCl, H2SO4, NH4Cl) 40
  41. 41. Table 21.2 Selected Standard Electrode Potentials (298K) Half-Reaction E0(V) F2(g) + 2e- 2F-(aq) +2.87 Cl2(g) + 2e- 2Cl-(aq) +1.36 MnO2(g) + 4H+(aq) + 2e- Mn2+(aq) + 2H2O(l) +1.23 NO3-(aq) + 4H+(aq) + 3e- NO(g) + 2H2O(l) +0.96 strength of reducing agent Ag+(aq) + e- Ag(s) +0.80strength of oxidizing agent Fe3+(g) + e- Fe2+(aq) +0.77 O2(g) + 2H2O(l) + 4e- 4OH-(aq) +0.40 Cu2+(aq) + 2e- Cu(s) +0.34 2H+(aq) + 2e- H2(g) 0.00 N2(g) + 5H+(aq) + 4e- N2H5+(aq) -0.23 Fe2+(aq) + 2e- Fe(s) -0.44 Cr3+(aq) + 3e- Cr(s) -0.74 Zn2+(aq) + 2e- Zn(s) -0.76 Mg2+(aq) + 2e- Mg(s) -2.37 41
  42. 42. Corrosion of the nail occurs at strained regions(more anodic). Contact with zinc protects the nailfrom corrosion. Zinc is oxidized instead of the iron(forming the faint white precipitate of zincferricyanide). Copper does not protect the nailfrom corrosion. 42
  43. 43. The use of sacrificial anodes to prevent iron corrosion. 43
  44. 44. General characteristics of voltaic and electrolytic cells. VOLTAIC CELL ELECTROLYTIC CELL System does work on its Energy is released from Surroundings(power supply) Energy is absorbed to drive a spontaneous redox reaction surroundings nonspontaneous redox reaction do work on system(cell) Oxidation half-reaction Oxidation half-reaction X X+ + e - A- A + e- Reduction half-reaction Reduction half-reaction Y++ e- Y B+ + e - B Overall (cell) reaction Overall (cell) reaction X + Y+ X+ + Y; ∆G < 0 A- + B+ A + B; ∆G > 0 44
  45. 45. Comparison of Voltaic and Electrolytic Cells ElectrodeCell Type ∆G Ecell Name Process SignVoltaic <0 >0 Anode Oxidation -Voltaic <0 >0 Cathode Reduction +Electrolytic >0 <0 Anode Oxidation +Electrolytic >0 <0 Cathode Reduction - 45
  46. 46. Electrolysis of PbBr2(s)Oxi: 2 Br- → Br2(g) + 2 e-Red: Pb2+ + 2 e- → Pb(s)Final: Pb2+ + 2 Br- → Pb(s) + Br2(g) 46
  47. 47. Electrolysis of Dilute H2SO4 Electrolysis of water H SO serves as a catalyst 2 4 As the electrolysis proceeds, the H2SO4 will become more and more concentrated as the water is used up. 47
  48. 48. The electrolysis of water.Oxidation half-reaction Reduction half-reaction2H2O(l) 4H+(aq) + O2(g) + 4e- 2H2O(l) + 4e- 2H2(g) + 2OH-(aq) Overall (cell) reaction 48 2H2O(l) 2H2(g) + O2(g)
  49. 49. Electrolysis of KI(aq) 49
  50. 50. Electrolysis of KI(aq)Oxi: 2 I - → I2 + 2 e -Red: 2 H + + 2 e - → H2Final: 2 H + + 2 I - → H 2 + I2 50
  51. 51. Figure 21.11 A concentration cell based on the Cu/Cu2+ half-reaction.Oxidation half-reaction Reduction half-reactionCu(s) Cu2+(aq, 0.1M) + 2e- Cu2+(aq, 1.0M) + 2e- Cu(s) Overall (cell) reaction Cu2+(aq,1.0M) Cu2+(aq, 0.1M) 51
  52. 52. Quantitative Aspects ofElectrolysis Faraday’s Law: the mass of product formed (or reactant consumed) at an electrode is proportional to both the amount of electricity transferred at the electrode and the molar mass of the substance in question. 52
  53. 53. E.g.1) cathode reaction for CuSO4 solution Cu2+ + 2 e- → Cuo2) cathode reaction for the electrolysis of molten NaCl Na+ + 1 e- → Nao  According to Faraday’s law: a) The mass of Na produced is proportional to its atomic mass divided by one, and the mass of Cu produced is proportional to its atomic mass divided by two. b) 1 mole of Cu ion reacts with 2 F of electricity to form 1 mole of Cu atom, and 1 mole of Na ion reacts with 1 F of electricity to form 1 mole of Na atom. 53
  54. 54.  Cu+2 + 2 e- → Cuo 2 F = 1 mole Cu2+ Na+ + 1 e- → Na 1 F = 1 mole Na+ Al3+ + 3 e- → Alo 3 F = 1 mole Al+3 54
  55. 55. In an electrolysis experiment, wegenerally measure the current (inamperes) that passes through anelectrolytic cell in a given period of time. 1C=1AX1sthat is, a coulomb is the quantity of electrical charge passing any point in the circuit in 1 second when the current is 1 ampere. 55
  56. 56. A summary diagram for the stoichiometry of electrolysis. MASS (g) MASS (g)of substance of substance oxidized or M(g/mol) oxidized or reduced reduced AMOUNT (MOL) AMOUNT (MOL) AMOUNT (MOL) AMOUNT (MOL) of substance of substance Faraday of electrons of electrons oxidized or oxidized or constant transferred transferred reduced reduced (C/mol e-) balanced CHARGE (C) CHARGE (C) half-reaction time(s) CURRENT (A) CURRENT (A) 56
  57. 57. Problems:1. A current of 0.452 A is passed through an electrolytic cell containing molten CaCl2 for 1.50 hours. Write the electrode reactions and calculate the quantity of products (in grams) formed at the electrodes. Anode: 2 Cl- → Cl2 + 2 e- Cathode: Ca2+ + 2 e- → CaFinal: Ca2+ + 2 Cl- → Ca + Cl2 57
  58. 58. Answers: ?C = 0.452 A X 1.50 hr X 3600s X 1C 1 hr 1AXs = 2.44 X 103 C? g Ca = 2.44 X 103 C X 1 F X 1 mol Ca X 40.08 g Ca 96,500 C 2F 1 mol Ca = 0.507 g Ca? g Cl2 = 2.44X103 C X 1F X 1 mol Cl2 X 70.90 g Cl2 96,500 C 2F 1 mol Cl2 = 0.896 g Cl2 58
  59. 59. Problems:2. If you wish to silver plate a spoon will you make it the cathode or the anode? How much Ag will be deposited by 0.3 F of electric charge?Ans.The object to be plated is always made the cathode. Ag+ + 1 e- → Ago x = (0.3 F) (1 mol Ag+ / 1 F) (107.87 g Ag / 1 mol Ag+) x = 32.4 g of Ag 59
  60. 60. Problems:3. How many moles of electrons are needed to deposit 0.45 g of aluminum from fused AlCl3?Ans. Al3+ + 3 e- → Alo x = (0.45 g) (1 mol) (3 mol e-) 27 g Al 1 mol x = 0.05 mol e- 60
  61. 61. Problems:4. Calculate the time required for a current of 1.8 Amperes to deposit 12.5 grams of copper from CuSO4 solution.Ans. Cu+2 + 2 e- → Cuo time = (g) (n) (F) (A) (M) = (12.5 g) (2 e-) (96,500 C) (1.8 A) (64 g/mole) = 20,942 secs. Or 5.82 hours 61
  62. 62. Problems:5. A constant current is passed through an electrolytic cell containing molten MgCl2 for 18 hours. If 4.8 X 105 grams of Cl2 are obtained, what is the current in Amperes?Ans. = 2.0 X 104 A 62
  63. 63. 63
  64. 64. 64
  65. 65. 65
  66. 66. 66
  67. 67. The corrosion of iron. 67
  68. 68. Enhanced corrosion at sea. 68
  69. 69. The effect of metal-metal contact on the corrosion of iron.faster corrosion cathodic protection 69

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