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Chapter 20 Principles of Reactivity: Electron Transfer Reactions

Chapter 20 Principles of Reactivity: Electron Transfer Reactions. Oxidation-Reduction Reactions Balancing oxidation-reduction reactions Balancing Redox Equations by the Oxidation Number Change Method (Other Stuff page) Voltaic (Galvanic) Cells Cell construction Cell potential

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Chapter 20 Principles of Reactivity: Electron Transfer Reactions

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  1. Chapter 20 Principles of Reactivity: Electron Transfer Reactions

  2. Oxidation-Reduction Reactions • Balancing oxidation-reduction reactionsBalancing Redox Equations by the Oxidation Number Change Method (Other Stuff page) • Voltaic (Galvanic) Cells • Cell construction • Cell potential • Effect of concentration on cell potential • Commercial voltaic cells • Electrolytic Cells • Electrolysis • Quantitative aspects of electrolysis

  3. I. Oxidation-Reduction Reactions • Review: • 2 Na + Cl2 2 NaCl (0) (0) (+1)(–1) oxidation = increase in oxidation number (loss of electrons) reduction = decrease in oxidation number (gain of electrons) • e.g., Assign oxidation numbers to the species being oxidized and reduced in the following equation, and label the oxidizing agent and reducing agent. • NaI + 3 HOCl  NaIO3 + 3 HCl

  4. I. Oxidation-Reduction Reactions A. Balancing oxidation-reduction reactions hard way: ion-electron (half-reaction) method (text) easy way: oxidation number change method (Web - Other Stuff) 1. Determine the oxidation numbers of the species being oxidized and reduced (and make sure there are the same number of each on each side). 2. Balance the changes in oxidation numbers by multiplying each species by the appropriate coefficient (i.e., balance the electrons gained and lost). 3. Balance charges with: H+ in acidic solution OH– in basic solution 4. Balance H (and O!) with H2O.

  5. I. Oxidation-Reduction Reactions A. Balancing oxidation-reduction reactions e.g., PH3 + I2 H3PO2 + I– (acidic solution) e.g., MnO4– + H2SO3 Mn2+ + SO42– (acidic)

  6. I. Oxidation-Reduction Reactions A. Balancing oxidation-reduction reactions e.g., CrO2– + S2O82– CrO42– + SO42– (basic) e.g., Cl2 Cl– + ClO3– (basic)

  7. II. Voltaic (Galvanic) Cells • Produce electricity: • chemical energy  electrical energy • 2Na+ + 2Cl– 2Na + Cl2DG >> 0 • requires input of electrical energy (electrolysis) • 2Ag+ + Ni  2Ag + Ni2+DG < 0 • produces energy • but with Ag+ and Ni in contact, we can’t generate electricity; electrons just flow from Ni to Ag+ • have to use a voltaic (galvanic) cell…

  8. electrical device e– e– • salt bridge • KCl in gelatin • allows electrolytic conduction without mixing Ni Ag Cl– K+ Ag+ Ni2+ Ni(NO3)2(aq) AgNO3(aq) II. Voltaic (Galvanic) Cells A. Cell construction anode (–) cathode (+) – – – – – – + + + + + + Ni  Ni2+ + 2e– (oxidation) Ag++ e– Ag (reduction) net: Ni + 2Ag+ Ni2+ + 2Ag

  9. Potentiometer (voltmeter) • measures cell potential (voltage or electromotive force) • depends on: • species in redox equation • concentrations • temperature • cell potential, E • = actual cell voltage • standard cell potential, Eº • = voltage at standard state • 25ºC • 1 atm pressure • 1 M concentration II. Voltaic (Galvanic) Cells A. Cell construction Ni Ag Ni2+ Ag+ salt bridge or porous partition • Shorthand notation: • Ni | Ni2+ || Ag+ | Ag

  10. can’t measure directly net: Ni + 2Ag+ Ni2+ + 2Ag Eºcell = Eº(Ag+) – Eº(Ni2+) = 1.05 V II. Voltaic (Galvanic) Cells B. Cell potential 1. standard reduction potential, Eº a. Eº = tendency for a species to be reduced Ag+ + e– Ag Eº(Ag+) Ni2+ + 2e– Ni Eº(Ni2+) Can only measure difference in a voltaic cell: 2Ag+ + 2e– 2Ag Eº(Ag+) Ni Ni2+ + 2e– –Eº(Ni2+) (not 2 x) General: Eºcell = Eº(species reduced) – Eº(species oxidized)

  11. 1 atm H2 1 M H+ Pt electrode II. Voltaic (Galvanic) Cells B. Cell potential 1. standard reduction potential, Eº b. standard: hydrogen electrode at standard state by definition: at standard state, the reduction 2H+ + 2e– H2 has Eº = 0.000 V (exactly)

  12. DVM 1 atm H2 Ag 1 M Ag+ 1 M H+ II. Voltaic (Galvanic) Cells B. Cell potential 1. standard reduction potential, Eº • c. standard reduction potentials • - measure others against the standard hydrogen electrode: • Find: • cathode: 2Ag+ + 2e– 2Ag • anode: H2  2H+ + 2e– • Eºcell = 0.80 V • Since Eºcell = Eº(Ag+) – Eº(H+) • 0.80 V = Eº(Ag+) – 0.00 V •  Eº(Ag+) = +0.80 V • (i.e., more easily reduced than H+)

  13. II. Voltaic (Galvanic) Cells B. Cell potential 1. standard reduction potential, Eº c. standard reduction potentials Ni | Ni2+(1 M) || H+(1 M) | H2 • Find: • cathode: 2H+ + 2e– H2 • anode: Ni  Ni2+ + 2e– • Eºcell = 0.25 V • Since Eºcell = Eº(H+) – Eº(Ni2+) • 0.25 V = 0.00 V – Eº(Ni2+) •  Eº(Ni2+) = –0.25 V • (i.e., less easily reduced than H+ or Ag+)

  14. more easily reduced  2Ag+ + 2e– 2Ag Ni Ni2+ + 2e– Ni + 2Ag+ Ni2+ + 2Ag II. Voltaic (Galvanic) Cells B. Cell potential 1. standard reduction potential, Eº d. determining cell potentials Ni | Ni2+(1 M) || Ag+(1 M) | Ag Ag+ + e– Ag Eº = +0.80 V Ni2+ + 2e– Ni Eº = –0.25 V Eºcell = Eº(Ag+) – Eº(Ni2+) = +0.80 V – (–0.25 V) = +1.05 V

  15. easiest to reduce (strongest oxidant) hardest to oxidize (weakest reductant) F2 + 2e– Ag+ + e– 2H+ + 2e– Ni2+ + 2e– Li+ + e– 2F– Ag H2 Ni Li hardest to reduce (weakest oxidant) easiest to oxidize (strongest reductant) II. Voltaic (Galvanic) Cells B. Cell potential 2. spontaneity of redox reactions Eº +2.87 V +0.80 V 0.00 V –0.25 V –3.05 V A species on the left will react spontaneously with a species on the right that is below it in the table. Or: The species with the more positiveEº will be reduced, and the species with the more negative Eº will be oxidized (Eºcell always > 0).

  16. II. Voltaic (Galvanic) Cells B. Cell potential 2. spontaneity of redox reactions Given the two half reactions below, what is the net cell reaction? What is Eº? Draw a galvanic cell using these half cells and label the anode and cathode, their charges, and the direction electrons flow in the circuit. Fe3+ + 3e– Fe Eº = –0.04 V Zn2+ + 2e– Zn Eº = –0.76 V

  17. energy e– w = # e–s  = coulombs  (= joules) joules coulomb DG = –nFE DGº = –nFEº II. Voltaic (Galvanic) Cells B. Cell potential 3. cell potential and free energy DGsys = –wsurr = coulombs  volts coulombs = nF (n = # of moles of e–s in redox reaction) (F = 96,500 coulombs/mol)  w = nFE Eºcell > 0, DGº < 0, spontaneous (voltaic) Eºcell < 0, DGº > 0, nonspontaneous (electrolytic)

  18. II. Voltaic (Galvanic) Cells B. Cell potential 3. cell potential and free energy e.g., Zn | Zn2+(1 M) || Fe3+(1 M) | Fe 2Fe3+ + 3Zn  2Fe + 3Zn2+Eºcell = 0.72 V What is DGº for the cell?

  19. E = Eº – RTlnQ nF 0.0592 n Nernst equation at 25ºC: E = Eº – logQ II. Voltaic (Galvanic) Cells C. Effect of concentration on cell potential 1. Nernst equation DG = DGº + RTlnQ –nFE = –nFEº + RTlnQ

  20. II. Voltaic (Galvanic) Cells C. Effect of concentration on cell potential 1. Nernst equation e.g., Ni | Ni2+ (0.05 M) || Ag+ (0.01 M) | Ag Ni + 2Ag+ Ni2+ + 2Ag Eº = 1.05 V

  21. e.g., AgCl(s) Ag+ + Cl–Ksp = [Ag+][Cl–] DVM 0.10 M Cl– Ag Ni 1.0 M Ni2+ [Ag+] = ? AgCl(s) II. Voltaic (Galvanic) Cells C. Effect of concentration on cell potential 2. applications a. measuring Ksp Find: E = 0.53 V; What is Ksp for AgCl? Ni + 2Ag+ Ni2+ + 2Ag Eº = 1.05 V

  22. DVM 1 atm H2 Ag 1.0 M Ag+ [H+] = ? lemon juice II. Voltaic (Galvanic) Cells C. Effect of concentration on cell potential 2. applications b. measuring pH Find E = 0.94 V; What is pH? H2 + 2Ag+ 2H+ + 2Ag Eº = 0.80 V

  23. II. Voltaic (Galvanic) Cells C. Effect of concentration on cell potential 2. applications A cell was constructed using the standard hydrogen electrode ([H+] = 1.0 M) in one compartment and a lead electrode in a 0.10 M K2CrO4 solution in contact with undissolved PbCrO4 in the other. The potential of the cell was measured to be 0.51 V with the Pb electrode as the anode. Determine the Ksp of PbCrO4 from this data. (Pb2+ + 2e– Pb Eº = –0.13 V)

  24. II. Voltaic (Galvanic) Cells C. Effect of concentration on cell potential 2. applications A galvanic cell was constructed with a Cu electrode in a solution of 1.0 M Cu2+ in one compartment and a hydrogen electrode immersed in a sample of a soft drink. The cell potential was measured to be 0.523 V. What was the pH of the soft drink? (Cu2+ + 2e– Cu Eº = 0.337 V)

  25. discharge cell: Pb + PbO2 + 4H+ + 2SO42– 2PbSO4 + 2H2O E ~ 2 V (6 cells in series) charge II. Voltaic (Galvanic) Cells D. Commercial voltaic cells anode: Pb + SO42– PbSO4 + 2e– cathode: PbO2 + 4H+ + SO42– + 2e–  PbSO4 + H2O • discharge: H2SO4 consumed, H2O produced • dilutes electrolyte solution • can measure with densitometer

  26. II. Voltaic (Galvanic) Cells D. Commercial voltaic cells anode: Zn + 2OH– ZnO + H2O + 2e– cathode: 2MnO2+ H2O + 2e–  Mn2O3 + 2OH– cell: Zn + 2MnO2  ZnO + Mn2O3 E ~ 1.5 V Zinc cup anode Graphite cathode Moist paste of MnO2, KOH and H2O Porous partition

  27. Anode cap Partition Cell can Gasket Anode: Zn and KOH Cathode: Ag2O paste II. Voltaic (Galvanic) Cells D. Commercial voltaic cells anode: Zn + 2OH– ZnO + H2O + 2e– cathode: Ag2O + H2O + 2e–  2Ag + 2OH– cell: Zn + Ag2O  ZnO + 2Ag E ~ 1.5 V

  28. III. Electrolytic cells A. Electrolysis electrical energy  chemical energy e.g., NaCl(l)  Na (l) + Cl2(g) DG >> 0 cell: 2Na+ + 2Cl– 2Na + Cl2 Eº = Eº(Na+) - Eº(Cl2) = (-2.71) - (1.36) = -4.07 V DGº = +786 kJ/mol

  29. H2O more easily reduced than Na+ H2O more easily oxidized than Cl– net: (2H2O + 2e– H2 + OH–) x 2 2H2O  O2 + 4H+ + 4e– 2H2O  2H2 + O2 III. Electrolytic cells A. Electrolysis E e.g., NaCl(aq)  ? Eº cathode: 2H2O + 2e– H2 + OH– -0.83 V (reduction) Na+ + e– Na -2.71 V Eº anode: Cl2 + 2e– 2Cl– +1.36 V (oxidation) O2 + 4H+ + 4e– 2H2O +1.23 V

  30. Cu2+ more easily reduced than H2O H2O more easily oxidized than Cl– net: (Cu2+ + 2e– Cu) x 2 2H2O  O2 + 4H+ + 4e– 2Cu2+ + 2H2O  2Cu + O2 + 4H+ III. Electrolytic cells A. Electrolysis E e.g., CuCl2(aq)  ? Eº cathode: Cu2+ + 2e– Cu +0.34 V (reduction) 2H2O + 2e– H2 + OH– -0.83 V Eº anode: Cl2 + 2e– 2Cl– +1.36 V (oxidation) O2 + 4H+ + 4e– 2H2O +1.23 V

  31. IV. Electrolytic cells B. Quantitative aspects of electrolysis Units of charge: 1 faraday (F) = 1 mol e–s 1 coulomb = 1 amp ·1 sec (A·s) experimentally: 1 F = 96,500 C e.g., How many moles of Na and Cl2 are produced in the electrolysis of NaCl(l) when a current of 25 A is applied for 8.0 hours?

  32. IV. Electrolytic cells B. Quantitative aspects of electrolysis e.g., How long would it take to deposit 21.4 g of Ag from a solution of AgNO3 using a current of 10.0 A?

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