1. CHE1102, Chapter 19 Learn, 1 Chapter 19 Electrochemistry Lecture Presentation

2. CHE1102, Chapter 19 Learn, 2 Electrochemistry - The relationship between chemical processes and electricity In electrochemical reactions, electrons are transferred from one species to another.

3. CHE1102, Chapter 19 Learn, 3 Oxidation Numbers In order to keep track of what loses electrons and what gains them, we assign oxidation numbers.

4. CHE1102, Chapter 19 Learn, 4 Oxidation and Reduction A species is oxidized when it loses electrons. –Here, zinc loses two electrons to go from neutral zinc metal to the Zn 2+ ion.

5. CHE1102, Chapter 19 Learn, 5 Oxidation and Reduction A species is reduced when it gains electrons. –Here, each of the H + gains an electron, and they combine to form H 2.

6. CHE1102, Chapter 19 Learn, 6 Oxidation and Reduction What is reduced is the oxidizing agent. –H + oxidizes Zn by taking electrons from it. What is oxidized is the reducing agent. –Zn reduces H + by giving it electrons.

7. CHE1102, Chapter 19 Learn, 7 Synopsis of Assigning Oxidation Numbers (as a Reminder) 1.Elements = 0 2.Monatomic ion = charge 3.F: –1 4.O: –2 (unless peroxide = –1) 5.H: +1 (unless a metal hydride = –1) 6.The sum of the oxidation numbers equals the overall charge (0 in a compound).

8. CHE1102, Chapter 19 Learn, 8 Half-Reactions The oxidation and reduction are written and balanced separately. We will use them to balance a redox reaction. For example, when Sn 2+ and Fe 3+ react,

9. CHE1102, Chapter 19 Learn, 9 Balancing Redox Equations: The Half-Reactions Method (a Synopsis) 1)Make two half-reactions (oxidation and reduction). 2)Balance atoms other than O and H. Then, balance O and H using H 2 O/H +. 3)Add electrons to balance charges. 4)Multiply by common factor to make electrons in half- reactions equal. 5)Add the half-reactions. 6)Simplify by dividing by common factor or converting H + to OH – if basic. 7)Double-check atoms and charges balance!

10. CHE1102, Chapter 19 Learn, 10 Electrolytic process – an electrochemical process that requires the continual, external input of electrical energy to drive a nonspontaneous redox reaction Two fundamental types of electrochemical processes 2. Voltaic (or galvanic) processes – (spontaneous) 1. Electrolytic processes – (nonspontaneous)

11. CHE1102, Chapter 19 Learn, 11 electrolytic cells – the containers in which the electrolysis occurs electrodes – the actual surfaces where oxidation and reduction physically occur cathode – electrode where the process of reduction occurs anode – electrode where the process of oxidation occurs

12. CHE1102, Chapter 19 Learn, 12 B-36 Peacemaker gross weight = 410,000 lbs each B-36 had 22,000 lbs Mg

13. CHE1102, Chapter 19 Learn, 13

14. CHE1102, Chapter 19 Learn, 14 positively charged ++++++ negatively charged ------ voltage source anode cathode cell contains molten MgCl 2 ( )…. (which is Mg 2+ and 2 Cl - ) Mg 2+ migrates toward the cathode Cl - migrates toward the anode

15. CHE1102, Chapter 19 Learn, 15 voltage source positively charged ++++++ negatively charged ------ anode cathode (reduction) Cl - cathode: Mg 2+ Mg + 2 e - anode: Cl 2 2 + 2 e - (oxidation) Mg 2+ 2 e - + + 2 Cl - Mg + Cl 2 + 2 e -

16. CHE1102, Chapter 19 Learn, 16 Michael Faraday 1791 – 1867 “The amount of chemical change that occurs during electrolysis is directly proportional to the amount of electricity that passes through the cell”

17. CHE1102, Chapter 19 Learn, 17 To deposit one mole of metallic magnesium requires two moles of electrons. The half-reaction for an oxidation or reduction relates the amount of chemical substance consumed or produced to the amount of electrons that the electric current must supply. We need to be able to relate current to electrons produced. Mg 2+ Mg + 2 e -

18. CHE1102, Chapter 19 Learn, 18 Coulomb, C – SI unit of charge C = (amps) (sec) 1 mole e - = 96485 C = 1 F ampere, amp – SI unit of electrical current The amount of charge that passes by a given point in a wire when an electric current of one ampere flows for one second.

19. CHE1102, Chapter 19 Learn, 19 How many grams of elemental magnesium, Mg (s) can be produced by the electrolysis of molten MgCl 2 ( ) with an electrical current of 6.00 amps for 2222 seconds ?

20. CHE1102, Chapter 19 Learn, 20 How many grams of elemental magnesium, Mg (s) can be produced by the electrolysis of molten MgCl 2 ( ) with an electrical current of 6.00 amps for 2222 seconds ? 1. Write the ½ reaction of interest 2. If possible, get Coulombs, C Mg 2+ Mg + 2 e - C = (amps) (sec) = (6.00)(2222) =13332

21. CHE1102, Chapter 19 Learn, 21 How many grams of elemental magnesium, Mg (s) can be produced by the electrolysis of molten MgCl 2 ( ) with an electrical current of 6.00 amps for 2222 seconds ? 3. Determine moles of electrons 4. Determine grams of Mg 13332 C x = 0.13818 mol e - 1 mol e - 96485 C 0.13818 mol e - x x = 1.680 g Mg 1 mol Mg 2 mol e - 24.31 g Mg 1 mol Mg

22. CHE1102, Chapter 19 Learn, 22 What current, in amps, is required to produce 45.5 g of elemental chlorine, Cl 2 (g) from a melt of NaCl ( ) if electrolyzed for 2 hours, 15 minutes and 25 seconds ?

23. CHE1102, Chapter 19 Learn, 23 What current, in amps, is required to produce 45.5 g of elemental chlorine, Cl 2 (g) from a melt of NaCl ( ) if electrolyzed for 2 hours, 15 minutes and 25 seconds ? 2. Determine mols of e - 45.5 g Cl 2 x x = 1.2835 mol e - 1 mol Cl 2 2 mol e - 1 mol Cl 2 70.90 g Cl 2 1. Write the ½ reaction of interest Cl - Cl 2 2 + 2 e -

24. CHE1102, Chapter 19 Learn, 24 What current, in amps, is required to produce 45.5 g of elemental chlorine, Cl 2 (g) from a melt of NaCl ( ) if electrolyzed for 2 hours, 15 minutes and 25 seconds ? 4. Calculate the current 1.2835 mol e - x = 123840 C 3. Convert electrons to Coulombs 1 mol e - 96485 C Amps = = 15.24 A 8125 s 123840 C

25. CHE1102, Chapter 19 Learn, 25 Voltaic process – an electrochemical cell which generates an electrical current from a spontaneous chemical redox process Two fundamental types of electrochemical processes 2. Voltaic (or galvanic) processes – (spontaneous) 1. Electrolytic processes – (nonspontaneous)

26. CHE1102, Chapter 19 Learn, 26 John Daniell 1790 – 1845 Daniell and Faraday

27. CHE1102, Chapter 19 Learn, 27 Zn + Cu 2+ Zn 2+ + Cu (spontaneous) If a piece of zinc metal is placed into an aqueous solution of copper sulfate, a spontaneous reaction occurs. The reaction is exothermic, but no usable energy can be harnessed because all the energy is dispersed as heat.

28. CHE1102, Chapter 19 Learn, 28 Separating the zinc and copper into different “half- cells” allows for transmission of an electrical current through an external wire – a more useful application Cu 2+ + 2 e - Cu Zn Zn 2+ + 2 e - Zn + Cu 2+ Zn 2+ + Cu

29. CHE1102, Chapter 19 Learn, 29 Anode (oxidation)Zn(s) Zn 2+ (aq) + 2e - Cathode (reduction)Cu2 + (aq) + 2e - Cu(s) Electrons become available as Zn is oxidized at the anode. They flow through the external circuit to the cathode, where they are consumed as Cu 2+ is reduced. The Zn electrode loses mass, and the concentration of the Zn 2+ solution increases as the cell operates; the Cu electrode gains mass, and the Cu 2+ solution becomes less concentrated.

30. CHE1102, Chapter 19 Learn, 30 salt bridge – any medium that allows the flow of ions but prevents the direct mixing of solutions – contains an electrolyte solution whose ions will not react with other ions in the cells Cu 2+ + 2 e - Cu Zn Zn 2+ + 2 e - For a voltaic cell to work, the solutions in the two half- cells must remain electrically neutral.

31. CHE1102, Chapter 19 Learn, 31

32. CHE1102, Chapter 19 Learn, 32 Voltage (or Potential) – the force pushing the electrons Alessandro Volta 1745 – 1827

33. CHE1102, Chapter 19 Learn, 33 The driving force behind a transfer of electrons from one atom to another is a difference in potential energy. Measured in volts, the “potential difference” between the two electrodes of a voltaic cell provides the driving force that pushes electrons through the external circuit. Also called the electromotive force (emf) or cell potential. For any cell reaction that proceeds spontaneously, such as that in a voltaic cell, the cell potential will be positive.

34. CHE1102, Chapter 19 Learn, 34 Zn + Cu 2+ Zn 2+ + Cu standard half-cell potential – voltage required to undergo a ½ reaction redox process E° red – voltage of “reduction ½ reaction” process E° ox – voltage of “oxidation ½ reaction” process standard conditions – 25  C and 1 atm and all solutions are 1 M

35. CHE1102, Chapter 19 Learn, 35

36. CHE1102, Chapter 19 Learn, 36 Cu 2+ + 2 e - Cu Zn Zn 2+ + 2 e - Zn + Cu 2+ Zn 2+ + Cu 2 e - + Zn + Cu 2+ Zn 2+ + Cu + 2 e - E° red = E° ox = + 0.34 V + 0.76 V Overall Cell Potential, E° cell – sum of the ½ cell reaction potentials E° cell = (E° red + E° ox )

37. CHE1102, Chapter 19 Learn, 37 Zn + Cu 2+ Zn 2+ + Cu Cu 2+ + 2 e - Cu E° red = + 0.34 V Zn Zn 2+ + 2 e - E° ox = + 0.76 V E° cell = (E° red + E° ox ) E° cell = 0.34 V + 0.76 V E° cell = + 1.10 V E° cell is often called “standard overall cell potential” or “electromotive force” or “EMF”

38. CHE1102, Chapter 19 Learn, 38 E° cell is the thermodynamic driving force for electrochemical processes When E° cell is positive (E° cell = + # V), electrochemical process is spontaneous When E° cell is negative (E° cell = – # V), electrochemical process is nonspontaneous

39. CHE1102, Chapter 19 Learn, 39 1. Determine the E° cell for the reaction and 2. Can a Voltaic cell be constructed using the reaction ? Fe 2+ + Al Fe + Al 3+ Voltage is an Intensive property 3(Fe 2+ + 2 e- Fe) 2(Al Al 3+ + 3 e - ) E° red = - 0.44 V E° ox = + 1.66 V

40. CHE1102, Chapter 19 Learn, 40 Relationship between E° cell and ΔG° rxn ΔG° rxn = – (n) (F) (E° cell ) ΔG° rxn is the standard Gibbs free energy n = smallest integer number of electrons transferred in the overall, balanced redox rxn F = Faraday constant = 96,485 C E° cell = the standard overall cell potential or EMF

41. CHE1102, Chapter 19 Learn, 41 3 Fe 2+ + 2 Al 3 Fe + 2 Al 3+ Determine ΔG° rxn for the following reaction E° cell = +1.22 V 6 electrons are transferred in this process 3(Fe 2+ + 2 e- Fe) 2(Al Al 3+ + 3 e - ) E° red = - 0.44 V E° ox = + 1.66 V ΔG° rxn = – (n) (F) (E° cell ) = – (6) (96,500 J/V-mol) (1.22 V) = – 706.38 kJ/mol

42. CHE1102, Chapter 19 Learn, 42 ΔG° rxn = – RT (ln K) R = 8.314 J/mole∙K T = Temperature in Kelvin K = the equilibrium constant at standard conditions Fe 3+ + Pt Fe 2+ + Pt 2+ Determine E° cell, ΔG° rxn and K for the following rxn

43. CHE1102, Chapter 19 Learn, 43 Fe 3+ + e - Fe 2+ 2 e - + 2 Fe 3+ + Pt 2 Fe 2+ + Pt 2+ + 2 e - E° red = + 0.77 V Pt Pt 2+ + 2 e - E° ox = - 1.19 V Fe 3+ + Pt Fe 2+ + Pt 2+ [] 2 Fe 3+ + 2 e - 2 Fe 2+ E° red = + 0.77 V E° cell = - 0.42 V 2 This electrochemical process is nonspontaneous

44. CHE1102, Chapter 19 Learn, 44 ΔG° rxn = – (n) (F) (E° cell ) = – (2) (96,500 J/V-mol) (-0.42 V) = + 81.06 kJ/mol ΔG° rxn = – RT (ln K) 81.06 kJ/mol = – (8.314 J/mol-K)(298.15 K)(ln K) – 32.70 = ln K K = e -32.70 = 6.289 x 10 -15

45. CHE1102, Chapter 19 Learn, 45 Some Applications of Cells Electrochemistry can be applied as follows:  Batteries: a portable, self-contained electrochemical power source that consists of one or more voltaic cells.  Batteries can be primary cells (cannot be recharged when “dead”—the reaction is complete) or secondary cells (can be recharged).  Prevention of corrosion (“rust-proofing”)  Electrolysis

46. CHE1102, Chapter 19 Learn, 46 Some Batteries Lead–Acid BatteryAlkaline Battery

47. CHE1102, Chapter 19 Learn, 47 Corrosion Corrosion is oxidation. Its common name is rusting.

48. CHE1102, Chapter 19 Learn, 48 Preventing Corrosion Corrosion is prevented by coating iron with a metal that is more readily oxidized. Cathodic protection occurs when zinc is more easily oxidized, so that metal is sacrificed to keep the iron from rusting.

49. CHE1102, Chapter 19 Learn, 49 Preventing Corrosion Another method to prevent corrosion is used for underground pipes. A sacrificial anode is attached to the pipe. The anode is oxidized before the pipe.

50. CHE1102, Chapter 19 Learn, 50