1. Determine the oxidation number for each atom in the following molecules 1.H 2 S 2.P 2 O 5 3.S 8 4.SCl 2 5.Na 2 SO 3 6. SO 4 -2 7. NaH 8.Cr 2 O 7 -2 9.SnBr 4 10. Ba(OH) 2 A

2. Intersection 14 12/05/06 Electrochemistry 19.9-19.13 p 941-955 A

3. December in Studio SMTuWThFS 12/5 Exam 3 12/6 Studio 12/8 Polymers; check out 12/11 Poster session, paper due 12/12 final IS 12/13 In- class assignment 12/17 Review session 7- 9 pm 12/19 Final exam 8-10 am A

4. Watershed Poster Session Monday, December 11 in USB 2165 Board (4 ft x 4ft), easel, pins Set up by 1:10 and 3:10 One person stationed at poster; others evaluate Rubric available Paper due same time A

5. Last In-Class Assignment Wednesday, December 13 th in studio Available on-line Read papers before coming to class; bring them with you. May make any notes you like on the papers Goal: to evaluate scientific method and data A

6. Outline Ed’s demos Balancing Redox Reactions Electrochemistry –Electrochemical cells and Standard Hydrogen Electrodes –Nernst –Quantifying current A

7. Ed’s Demos A

8. Oxidation States of Vanadium: Reduction of V 5+ to V +2 Reaction 1 –Zn (s) + 2 VO 3 - (aq) + 8 H 3 O + (aq) ↔ 2 VO 2 + (aq) + Zn +2 (aq) + 12 H 2 O (l) Reaction 2 –Zn (s) + 2 VO 2 + (aq) + 8 H 3 O + (aq) ↔ 2 V 3+ (aq) + Zn +2 (aq) + 6 H 2 O (l) Reaction 3 –Zn (s) + 2 V 3+ (aq) ↔ 2 V 2+ + Zn +2 (aq) V +5 (aq) → V +4 (aq)yellow to green V +4 (aq) → V +3 (aq)green to blue V +3 (aq) → V +2 (aq)blue to violet A

9. Oxidation States of Manganese: Mn +7, Mn +6, Mn +4, and Mn +2 +7 (purple) to +2 (colorless) –2 MnO 4 - (aq) + H + (aq) + 5 HSO 3 - (aq) ↔ 2 Mn +2 (aq) + 5 SO 4 -2 (aq) + 3 H 2 O(l) + 7 (purple) to +4 (brown) –OH - + 2 MnO 4 - (aq) + 3 HSO 3 - (aq) ↔ 2 MnO 2 (s) + 3 SO 4 -2 (aq) + 2 H 2 O(l) + 7 (purple) to + 6 (green) –2 MnO 4 - (aq) + 3 OH - + HSO 3 - (aq) ↔ 2 MnO 4 -2 (aq) + SO 4 -2 (aq) + 2 H 2 O(l) A

10. Thinking back…. What happened when Na (s) was added to water? Na (s) + H 2 O (l)  Na + (aq) + H 2(g) + OH - (aq)  Determine the oxidation state of each reactant and product What was oxidized? What was reduced? A

11. Balancing Redox Reactions When KMnO 4 (potassium permanganate) is mixed with Na 2 C 2 O 4 (sodium oxalate) under acidic conditions, Mn +2 (aq) ions and CO 2 (g) form. The unbalanced chemical equation is: KMnO 4(aq) + Na 2 C 2 O 4(aq)  Mn +2 (aq) + CO 2(g) + K + (aq) + Na + (aq) K + and Na + are spectator ions, so we can ignore them at this point. MnO 4 - (aq) + C 2 O 4 -2 (aq)  Mn +2 (aq) + CO 2(g) M

12. Half-Reactions Reduction reaction Oxidation reaction MnO 4 - (aq) + C 2 O 4 -2 (aq)  Mn +2 (aq) + CO 2(g) M

13. Reduction reaction Step 1: Balance all elements other than oxygen and hydrogen. Step 2: Balance the oxygens by adding water. Step 3: Balance the hydrogens using H + Step 4: Balance the electrons Mn +7 on reactants side Mn +2 on products side Step 5: Check charge balance and elemental balance MnO 4 -  Mn +2 M

14. Oxidation reaction C 2 O 4 -2  CO 2 M

15. Combine Half Reactions 5 e - + 8H + + MnO 4 -  Mn +2 + 4 H 2 O C 2 O 4 -2  2 CO 2 + 2e - We are assuming the reaction takes place under acidic conditions! M

16. Balancing in Base 5 e - + 8H + + MnO 4 -  Mn +2 + 4 H 2 O C 2 O 4 -2  2 CO 2 + 2e - Change H + to water by adding OH - to each side M

17. M

18. Electrochemical Cells A

19. Definitions Electrochemical cell: A combination of anode, cathode, and other materials arranged so that a product-favored redox reaction can cause a current to flow or an electric current can cause a reactant- favored redox reaction to occur Voltaic cell (battery): An electrochemical cell or group of cells in which a product-favored redox reaction is used to produce an electric current. Galvanic cell: A cell in which an irreversible chemical reaction produces electrical current Electrolytic cell: electrochemical reactions are produced by applying electrical energy A

20. A Copper-Zinc battery – What Matters? Consider reduction potentials: Cu +2 + 2e - → Cu(s)0.3419 V Zn +2 + 2e - → Zn(s)-0.7618 V Place Zn electrode in copper sulfate solution – What happens? Cu +2 + 2e- → Cu(s)0.3419 V Zn(s) → Zn +2 + 2e-0.7618 V Cu +2 + Zn(s) → Zn +2 + Cu(s)1.1 V E > 0, spontaneous Note, no need for electron to flow external to cell for reaction to occur!! Copper is plated on Zn electrode A

21. A Copper-Zinc battery – What Matters? Consider reduction potentials: Cu +2 + 2e- → Cu(s)0.3419 V Zn +2 + 2e- → Zn(s)-0.7618 V Place Cu electrode in zinc sulfate solution – What happens? Cu(s) → Cu +2 + 2e- -0.3419 V Zn +2 + 2e- → Zn(s)-0.7618 V Zn +2 + Cu(s) → Cu +2 + Zn(s)-1.1 V E < 0, not spontaneous No reaction occurs !! Zn doesn’t plate on copper electrode?! A

22. Fig. 19-3, p.918 A

23. What are the ½ reactions? What is the overall reaction? Identify the oxidation, reduction, anode, and cathode A

24. Fig. 19-7, p.922 SHE: Standard Hydrogen Electrode 2 H 3 O + (aq, 1.00 M) + 2e - H 2(g, 1 atm) + 2H 2 O (l) E o = 0V Standard conditions: 1M, 1atm, 25 o C A

25. Measuring Relative Potentials Table of Standard Reduction Potentials A

26. Standard Reduction Potentials What is the standard potential of a Au +3 /Au/Mg +2 /Mg cell? A

27. The half-reaction with the more positive standard reduction potential occurs at the cathode as reduction. The half-reaction with the more negative standard reduction potential occurs at the anode as oxidation. A

28. Picture from: www.corrosion-doctors.org/ Biographies/images/ www.corrosion-doctors.org/ Biographies/images/ Nernst M

29. Is potential always the same? Standard conditions: 1 atm, 25 o C, 1 M What will influence the potential of a cell? M

30. Mathematical Relationships: Nernst The Nernst Equation: E o = standard potential of the cell R = Universal gas constant = 8.3145 J/mol*K T = temperature in Kelvin n = number of electrons transferred F = Faraday’s constant = 96,483.4 C/mol Q = reaction quotient (concentration of anode divided by the concentration of the cathode) E = E o - RT ln Q nF Cu +2 + Zn(s) → Zn +2 + Cu(s) Q = M

31. Applying the Nernst Equation This cell is operating at 25 o C with 1.00x10 -5 M Zn 2+ and 0.100M Cu 2+ ? Predict if the voltage will be higher or lower than the standard potential E = E o - RT ln Q nF Cu +2 + Zn(s) → Zn +2 + Cu(s) M

32. E o = standard potential of the cell R = Universal gas constant = 8.3145 J/mol*K T = temperature in Kelvin n = number of electrons transferred F = Faraday’s constant = 96,483.4 C/mol Q = reaction quotient (concentration of anode divided by the concentration of the cathode) E = E o - RT ln Q nF Zn +2 + 2e - -> Zn -0.76 V Cu +2 + 2e - -> Cu 0.34 V 25 o C + 273 = K n = 1.00x10 -5 M Zn 2+ and 0.100M Cu 2 Cu +2 + Zn(s) → Zn +2 + Cu(s) Q = [Zn +2 ]/[Cu +2 ] M

33. Were your predictions correct? M

34. Picture from: musee-ampere.univ-lyon1.fr/ musee-ampere.univ-lyon1.fr/ Ampere A

35. The Units of Electrochemistry Coulomb –1 coulomb equals 2.998 x 10 9 electrostatic units (eu) –eu is amount of charge needed to repel an identical charge 1 cm away with unit force –Charge on one electron is -1.602 x 10 -19 coulomb Problem:An aluminum ion has a +3 charge. What is this value in coulombs? magnitude of charge is same at that of e-, opposite sign 3 x 1.602 x10 -19 = 4.806 x 10 -19 coulomb Key Point: electrons or ions charges can be measured in coloumbs A

36. The Units of Electrochemistry Ampere –Amount of current flowing when 1 coulomb passes a given point in 1 second –Units of Amperes are Coulombs per second –Current (I) x time (C/s x s) gives an amount of charge. Problem:How much current is flowing in a wire in which 5.0 x 10 16 electrons are flowing per second? The charge transferred each second = (5.0 x 10 16 electrons/sec) x (1.602 x 10 -19 coulomb/electron) = 8.0 x 10 -3 coulombs/sec = amps A

37. The Units of Electrochemistry Ampere –Amount of current flowing when 1 coulomb passes a given point in 1 second –Units of Amperes are Coulombs per second –Current (I) x time (C/s x s) gives an amount of charge. –We can express electron or ION flow in amps! Problem:If 1 mol Al +3 ions passes a given point in one hour, what is the current flow? 1 hour = 3600 seconds; Al +3 charge is 4.806 x 10 -19 coulomb 1 mol Al +3 ions6.022 x 10 23 Al +3 ions4.806 x 10 -19 coulomb1 hour Hour1 mol Al +3 ions1 Al +3 ion3600 sec = 80 C/s = 80 A A

39. Electrolysis The process of using electrical current to drive a redox reaction that is not spontaneous. Balance the reaction in acid. At the anode (where oxidation occurs): H 2 O (l) → O 2 (g) At the cathode (where reduction occurs): H 3 O + (aq) → 2 H 2 (g) Overall reaction: M

40. Overall reaction: 2 H2O (l) → 2 H2(g) + O2 (g) p.951 Overall reaction: 2 H 2 O (l) → 2 H 2 (g) + O 2 (g) M

41. Standard Reduction Potentials What is the standard potential for the electrolysis of water? M

42. Electrolysis Reactions How many coulombs of charge are required to electrolyze 1 mol of water? Note that 6 water are consumed at the anode, but 4 are produced at the cathode. At the anode (where oxidation occurs): 6 H 2 O (l) → O 2 (g) + 4 H 3 O + (aq) + 4e - At the cathode (where reduction occurs): 4 H 3 O + (aq) + 4 e - → 2 H 2 (g) + 4 H 2 O (l) The net amount of water consumed per every 4 electrons is 2 molecules. 1.0 mol H 2 O4 mol e-96,487 coulombs 2 mol H 2 O1 mol e- = 190000 C M

43. Electrolysis Reactions A power supply puts out a maximum current of 10 amps. How long will it take to electrolyze 18 g (1 mole) water? 190,000 coulombs sechour 10 coulombs3600 secs = 5.4 hours M

44. Electrolysis What similarities do you find between voltaic cells and electrolysis? What differences? Overvoltage: However, the potential that must be applied to electrolysis cell is always greater than that calculated from standard reduction potentials. This excess potential is called an overvoltage. This additional voltage is needed to overcome limitations in electron transfer rate at the interface between electrode and solution. M