1. Warm up 3/17/15 Balance the half reaction

2. Ch. 17 Electrochemistry

3. The study of the interchange of chemical and electrical energy. Electrochemistry

4. ● Oxidation-reduction reaction involves a transfer of electrons from the reducing agent to the oxidizing agent. ● Oxidation involves a loss of electrons (an increase in oxidation number), reduction involves a gain of electrons (a decrease in oxidation number). 17.1 Galvanic Cells

5. Fe 2+ is oxidized MnO 4 - is reduced e-s are trasferred from Fe 2+ (the reducing agent) to MnO 4 - (the oxidizing agent) MnO 4 - and Fe 2+

6. Half reaction You know how to figure out? MnO 4 - and Fe 2+

7. When MnO 4 - and Fe 2+ are present in the same solution, the electrons are transferred directly when the reactants collide. How can we harness this energy? Separate the oxidizing agent from the reducing agent, making the electron transfer to occur through a wire.

8. The current produced in the wire by the electron flow Electrons should flow through the wire from Fe 2+ to MnO 4 -. However there is no flow of electrons apparent.

9. Example for Redox Fe 2+ +MnO 4 -  Mn 2+ + Fe 3+ OxidationReduction Gain OLose O Lose e-Gain e- OxidizedReduced e- on the right sidee- on the left side Reducing agentOxidizing agent Increase Oxidation # Decrease oxidation #

10. The solutions must be connected so that ions can flow to keep the net charge in each compartment zero. Need a salt bridge (a U-tube filled with an electrolyte) or a porous disk in a tube

11. c. Salt bridge/porous disk i. Contain strong electrolyte (KNO 3, KCl, etc) ii. to keep flow of ions from one half-cell to the other so net charge is _____. iii. “anions to anode, cations to cathode” zero

12. Either devices allows ions to flow without extensive mixing of solutions. Electrons flow through the wire from reducing agent to oxidizing agent Ions flow keep the net charge zero http://www.youtube. com/watch?v=oqUL R_u96L4 Process of cell

13. ● A device in chemical energy is changed to electrical energy. (The opposite process is called electrolysis) ● The reaction occurs at the interface between the electrode and the solution where the electron transfer occurs. Galvanic cell

14. Zinc metal in Copper (II) sulfate solution i.Net ionic equation  ii.Zn (s)  Zn 2+ (aq) + 2e - Reaction iii.Cu 2+ (aq) +2e -  Cu (s)Reaction Oxidation Reduction Zn (s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu (s)

15. v. Electron transfer Electrons are lost from Zn, then gained by Cu 2+ vi. If you put Copper metal in zinc sulfate solution, what will happen? Nothing will happen. Zinc metal in Copper (II) sulfate solution Why is Zinc oxidized? Zinc lost e-

16. Electrode – a conductor (metal) plate used to make contact with the nonmetallic part (electrolyte solution).

17. Two types of electrodes i. Cathode – Reduction (______ electrons) ii. Anode – Oxidation (_______ electrons) *Cathode starts with consonant, so Reduction *Anode starts with vowel, so Oxidation gain lose

18. Wire carries the electrons in the external circuit From the Zn rod to the Cu rod

19. Electrons travel…. From the Anode ( ) charge to the Cathode ( ) charge - +

20. Warm up 3/20/15 Indicate the oxidizing agent, the reducing agent, the species being oxidized, and the species being reduced. 2AgNO 3 (aq)+Cu(s)  Cu(NO 3 ) 2 (aq)+2Ag(s)

21. Using the reaction Zn(s)+Cu 2+ (aq) → Zn 2+ (aq) +Cu(s) i. Electrons are produced at the zinc rod. Zn (s) → Zn 2+ (aq) + 2e - ii. Why? Zinc loses electrons. Zinc is more reactive than Copper iii. It is oxidized  Zn rod is anode vowels

22. Zn(s)+Cu 2+ (aq) → Zn 2+ (aq) +Cu(s) iv. The electrons leave the ____ anode and pass through the external circuit to the ____ rod. Zn Cu

23. Zn(s)+Cu 2+ (aq) → Zn 2+ (aq) +Cu(s) v. Electrons enter the ____ rod and interact with copper ions in solution. Therefore, the following reduction half- reaction occurs. Cu 2+ (aq) + 2e - → Cu (s) vi. It is reduced  Cu rod is Cu Cathode consonant

24. Zn(s)+Cu 2+ (aq) → Zn 2+ (aq) +Cu(s) vi. To complete the circuit, both positive and negative ions move through the aqueous solutions via the salt bridge. Oxidation Reduction

25. ● A galvanic cell pulls electrons through a wire from a reducing agent to an oxidizing agent. ● The “pull” or driving force on the electrons is called the cell potential (E cell ) or the electromotive force (emf) of the cell. Cell Potential

26. ● The unit of electrical potential is the volt (abbreviated V), which is defined as 1 joule of work per coulomb of charge transferred. ● E° = cell potential measured at “standard” conditions: 1 atm and 25°C (unlike STP for gases) Cell Potential

27. ● The reaction in a galvanic cell can be broken down into two half reactions. ● We can obtain the cell potential by summing the half-cell potentials. 17.2 Standard Reduction Potentials

28. The anode contains a zinc metal electrode with Zn 2+ and SO 4 2-. The anode reaction is the oxidation half reaction: Zn → Zn 2+ + 2e - 2H + (aq) + Zn(s) → Zn 2+ (aq) + H 2 (g)

29. ● The anode compartment will contain 1 M Zn 2+. ● The cathode reaction of this cell is the reduction half-reaction: 2H + + 2e - → H 2 2H + (aq) + Zn(s) → Zn 2+ (aq) + H 2 (g)

30. ● The cathode consists of an inert conductor platinum electrode in contact with 1 M H + ions ● Produce hydrogen gas at 1 atm. ● That electrode is called the standard hydrogen electrode. 2H + (aq) + Zn(s) → Zn 2+ (aq) + H 2 (g)

31. No way to measure the potentials of the individual electrode [H + ] = 1 M and P H2 = 1 atm (potencial=0V) Zn → Zn 2+ + 2e - (0.76V) E cell = E H+ → H2 + E Zn → Zn2+ 0.76V = 0V + 0.76 V Standard potential for the half reaction 2H + + 2e - → H 2 = 0.00 V (Reference)

32. Figure 17.6 A Galvanic Cell involving the Half-Reactions Zn(s) + Cu 2+ (aq) → Zn 2+ (aq) + Cu (s) Anode : Zn → Zn 2+ + 2e - Cathode: Cu 2+ + 2e - → Cu E cell = E Zn → Zn2+ + E Cu2+ → Cu 1.10 V = 0.76 V + 0.34 V (measured) (known) (calculated)

33. The half-reaction potentials based on the assignment of zero volts to the process 2H + + 2e → H 2 (under standard conditions). Standard Reduction Potentials: ● Only reduction (gain electrons) ● All solutes at 1 M ● All gases at 1 atm

34. Calculating Cell Potential of a Redox Reaction: Tips a. Identify and write each half- reaction. b. Compare the reaction to the Reduction Potential Chart. c. One with higher E value will be E Red. d. One with lower E value, you need to reverse the reaction, then change the E to the opposite sign. Hint Lower E 0, reverse equation and change sign

35. Practice: Mg and Al a. Identify and write each half-reaction. One with higher E value will be E Red. d. One with lower E value, you need to reverse the reaction, then change the E to the opposite sign. Lower E 0, reverse equation and change sign Mg 2+ (aq) + 2e -  Mg(s) E= -2.37V Al 3+ (aq) + 3e -  Al (s) E= -1.66V Mg(s)  Mg 2+ (aq) + 2e - E= +2.37V

36. Calculating Cell Potential of a Redox Reaction e. If needed, multiply one (or both) rxns by an integer so the number of electrons are equal on opposite sides. f. Do NOT multiply the E value by the integer here! E values are intensive (doesn’t depend on the amount – it will always stay the same, similar to temperature).

37. Practice: Mg and Al e. If needed, multiply one (or both) rxns by an integer so the number of electrons are equal on opposite sides. Mg(s)  Mg 2+ (aq) + 2e - E= +2.37 Al 3+ (aq) + 3e -  Al (s) E= -1.66V x2 x3 2Al 3+ (aq) + 6e -  2Al (s) E= -1.66V 3Mg(s)  3Mg 2+ (aq) + 6e - E= +2.37

38. Calculating Cell Potential of a Redox Reaction g. Add the two E values to obtain E cell (cell potential of the reaction). h. + Positive Voltage = spontaneous  good cell E 0 = +0.71V 2Al 3+ (aq) + 6e -  2Al (s) E= -1.66V 3Mg(s)  3Mg 2+ (aq) + 6e - E= +2.37 2Al 3+ (aq) + 3Mg(s)  2Al (s) + 3Mg 2+ (aq) E= +0.71V

39. More practices

40. e. Does the reaction Mg 2+ (aq) + 2e -  Mg(s) occur by itself? Find out the E 0 value of Mg E 0 = -2.37V 2 H + (aq) + 2 e -  H 2 (g) (reference electrode) 0.00 Fe 2+ + 2e -  Fe -0.44 Al 3+ (aq) + 3e - → Al (s) -1.66 Mg 2+ (aq) + 2e -  Mg(s) -2.37 nonspontaneous, b/c E is negative.

41. Practice: i. Mg and Cl 2 a. Identify and write each half-reaction. One with higher E value will be E Red. c. One with lower E value, you need to reverse the reaction, then change the E to the opposite sign. Lower E 0, reverse equation and change sign Mg 2+ (aq) + 2e -  Mg(s) E= -2.37V Cl 2 (g) + 2e -  2Cl - (aq) E= +1.36V Mg(s)  Mg 2+ (aq) + 2e - E= +2.37V

42. Practice: i. Mg and Cl 2 e. If needed, multiply one (or both) rxns by an integer so the number of electrons are equal on opposite sides. Mg(s)  Mg 2+ (aq) + 2e - E= +2.37 Cl 2 (g) + 2e -  2Cl - (aq) E= +1.36V Cl 2 (g) + Mg(s)  2Cl - (aq) + Mg 2+ (aq) E= +3.73V

43. Practice: ii. Cu and Ag + a. Identify and write each half-reaction. One with higher E value will be E Red. c. One with lower E value, you need to reverse the reaction, then change the E to the opposite sign. Lower E 0, reverse equation and change sign Cu 2+ (aq) + 2e -  Cu(s) E= +0.337V Ag + (aq) + e -  Ag (s) E= +0.80V Cu(s)  Cu 2+ (aq) + 2e - E= -0.337V

44. f. Practice: i. Mg and Cl 2 b. If needed, multiply one (or both) rxns by an integer so the number of electrons are equal on opposite sides. 2Ag + (aq) + Cu(s)  2Ag (s) + Cu 2+ (aq) E= +0.46V Ag + (aq) + e -  Ag (s) E= +0.80V Cu(s)  Cu 2+ (aq) + 2e - E= -0.337V x2 2Ag + (aq) + 2e -  2Ag (s) E= +0.80V

45. ● Anode components are listed on the left ● Cathode components are listed on the right Mg(s) │ Mg 2+ (aq) ││ Al 3+ (aq) │ Al(s) ● The anode is listed on the far left ● The cathode is listed on the far right ● All components involved in the redox reaction are ions Line Notation

46. ● A nonreacting (inert) conductor must be used (usually platinum) Pt(s)|CIO 3 - (aq), CIO 4 - (aq),H + (aq)||H + (aq), MnO 4 - (aq), Mn 2+ (aq)|Pt(s) Line Notation

47. Warm up 3/23/15 Cathode/Anode Cell potential Diagram w/solution and electrode Line notation

48. Reaction 2 must be reversed. 17.2 Description of a Galvanic Cell

49. Ag + gains e- and Fe 2+ release e-s. The e-s will flow from Fe 2+ to Ag + Ag + (cathode, reduction) Fe 2+ (anode, oxidation) Electrode for cathode  silver metal Electrode for anode  Pt metal (Fe 2+, Fe 3+ ) 17.2 Description of a Galvanic Cell

50. Line notation Pt(s)│Fe 2+ (aq), Fe 3+ (aq)││Ag + (aq)│Ag(s)

51. Do only #33 on the expt. Chem. Choice I (prelab and post lab questions) Due Mon. 3/30 Ch. 17 quiz – Tue. 3/31 Ch. 17 test - Thu. 4/2 Expt #15 info (Dry lab)

52. ● A cell will always run spontaneously in the direction that produces a positive cell potential (cathode) ● Leads to a positive cell potential E o cell = E o (cathode) - E o (anode) Or E o cell = E o (cathode) + (- E o (anode)) Complete Description of a Galvanic Cell

53. ● The physical setup of the cell ● Anode: Pure metallic iron (Fe) and 1.0 M Fe 2+ Fe → Fe 2+ + 2e - An oxidation reaction 1

54. ● Cathode: 1.0 M MnO 4 -, 1.0 M H +, 1.0 M Mn 2+ MnO 4 - + 5e - + 8H + → Mn 2+ + 4H 2 O ● A reduction reaction ● Neither MnO 4 - nor Mn 2+ ions can serve as the electrode  a nonreacting conductor (platimun) 1

55. ● the electrons flow from Fe to MnO 4 - or from the anode to the cathode 2. The direction of electron flow

56. The cell potential (always positive for a galvanic cell) E o cell = E o (cathode) - E o (anode) and the balanced cell reaction the direction of electron flow  obtained by inspecting the half- reactions and using the direction that gives a positive E cell Galvanic cell contains:

57. ● Designation of the anode and cathode. ● The nature of each electrode and the ions present in each compartment. ● A chemically inert conductor is required if none of the substances participating in the half-reaction is a conducting solid.

58. ● Try #4-8 at home Worksheet problems?