1. 1 Electrochemistry Chapter 20

2. 2 Half-reaction method…remember? Example Example Al (s) + Cu 2+ (aq)  Al 3+ (aq) + Cu (s) Oxidation: Al (s)  Al 3+ (aq) + 3e - Oxidation: Al (s)  Al 3+ (aq) + 3e - Reduction: 2e - + Cu 2+ (aq)  Cu (s) Reduction: 2e - + Cu 2+ (aq)  Cu (s) –Use lowest common multiple to make both equivalent in number of electrons Oxidation  multiply by 2 Oxidation  multiply by 2 –2Al (s)  2Al 3+ (aq) + 6e - Reduction  multiply by 3 Reduction  multiply by 3 –6e - + 3Cu 2+ (aq)  3Cu (s) –Collate (electrons cross out) Net reaction: Net reaction: 2Al (s) + 3Cu 2+ (aq)  2Al 3+ (aq) + 3Cu (s) DOES EVERYTHING BALANCE? (Make sure to balance after every step!)

3. 3 In acidic milieu Fe 2+ (aq) + MnO 4 - (aq)  Fe 3+ (aq) + Mn 2+ (aq) Oxidation: Oxidation: Fe 2+ (aq)  Fe 3+ (aq) + e - Reduction: Reduction: 8H + (aq) + MnO 4 - (aq)  4H 2 O (l) + Mn 2+ (aq) –What did I do in the above half-rxn? –Is it fully balanced? 5e - + 8H + (aq) + MnO 4 - (aq)  4H 2 O (l) + Mn 2+ (aq) Balance both half-reactions: Balance both half-reactions: 5Fe 2+ (aq)  5Fe 3+ (aq) + 5e - (multiply by 5; why?) 5e - + 8H + (aq) + MnO 4 - (aq)  4H 2 O (l) + Mn 2+ (aq) Collate Collate Net rxn: Net rxn: 5Fe 2+ (aq) + 8H + (aq) + MnO 4 - (aq)  5Fe 3+ (aq) +4H 2 O (l) + Mn 2+ (aq)

4. 4 Solve VO 2 + (aq) + Zn (s)  VO 2+ (aq) + Zn 2+ (aq)

5. 5 Answer Oxidation: Oxidation: Zn (s)  Zn 2+ (aq) + 2e - Reduction: Reduction: e - + 2H + (aq) + VO 2 + (aq)  VO 2+ (aq) + H 2 O (l) Balancing both half-reactions: Balancing both half-reactions: Zn (s)  Zn 2+ (aq) + 2e - 2e - + 4H + (aq) + 2VO 2 + (aq)  2VO 2+ (aq) + 2H 2 O (l) Collate Collate Net reaction: Net reaction: Zn (s) + 4H + (aq) + 2VO 2 + (aq)  2VO 2+ (aq) + 2H 2 O (l) + Zn 2+ (aq)

6. 6 In basic milieu I - (aq) + MnO 4 - (aq)  I 2(aq) + MnO 2(s) Oxidation: Oxidation: I - (aq)  I 2(aq) + e - 2I - (aq)  I 2(aq) + 2e - Reduction: Reduction: MnO 4 - (aq)  2OH - (aq) + MnO 2(s) 2H + (aq) + 2OH - (aq) + MnO 4 - (aq)  2OH - (aq) + MnO 2(s) 2H 2 O (l) + MnO 4 - (aq)  2OH - (aq) + MnO 2(s) 2H 2 O (l) + MnO 4 - (aq)  4OH - (aq) + MnO 2(s) 3e - + 2H 2 O (l) + MnO 4 - (aq)  4OH - (aq) + MnO 2(s) Balance both half-reactions: Balance both half-reactions: 6I - (aq)  3I 2(aq) + 6e - 6e - + 4H 2 O (l) + 2MnO 4 - (aq)  8OH - (aq) + 2MnO 2(s) Collate Collate Net rxn: Net rxn: 6I - (aq) + 4H 2 O (l) + 2MnO 4 - (aq)  3I 2(aq) + 8OH - (aq) + 2MnO 2(s)

7. 7 Solve Al (s) + H 2 O (l)  Al(OH) 4 - (aq) + H 2(g) Al (s) + H 2 O (l)  Al(OH) 4 - (aq) + H 2(g)

8. 8 Answer Oxidation: Oxidation: Al (s) + 4OH - (aq)  Al(OH) 4 - (aq) + 3e - Reduction: Reduction: 2e - + 2H 2 O (l)  2OH - (aq) + H 2(g) Balance each half-reaction: Balance each half-reaction: 2Al (s) + 8OH - (aq)  2Al(OH) 4 - (aq) + 6e - 6e - + 6H 2 O (l)  6OH - (aq) + 3H 2(g) Collate Collate Net-reaction: Net-reaction: 2Al (s) + 2OH - (aq) + 6H 2 O (l)  2Al(OH) 4 - (aq) +3H 2(g)

9. 9 Electricity Movt of electrons Movt of electrons Movt of electrons through wire connecting 2 half-reactions  electrochemical cell Movt of electrons through wire connecting 2 half-reactions  electrochemical cell Also called voltaic or galvanic cell Also called voltaic or galvanic cell Cell produces current from spontaneous rxn Cell produces current from spontaneous rxn –Example: copper in solution of AgNO 3 is spontaneous On the other hand, an electrolytic cell uses electrical current to drive a non-spontaneous chemical rxn On the other hand, an electrolytic cell uses electrical current to drive a non-spontaneous chemical rxn

10. 10 Voltaic cell Solid Zn in zinc ion solution = half-cell Solid Zn in zinc ion solution = half-cell Likewise, Cu/Cu-ion solution Likewise, Cu/Cu-ion solution Wire attached to each solid Wire attached to each solid Salt bridge = Salt bridge = 1. contains electrolytes, 1. contains electrolytes, 2. connects 2 half-cells, 2. connects 2 half-cells, 3. anions flow to neutralize accumulated cations at anode and cations flow to neutralize accumulated anions at cathode (completes circuit) 3. anions flow to neutralize accumulated cations at anode and cations flow to neutralize accumulated anions at cathode (completes circuit) “An Ox” = anode oxidation “An Ox” = anode oxidation Has negative charge because releases electrons Has negative charge because releases electrons “Red Cat” = reduction cathode “Red Cat” = reduction cathode Has positive charge because takes up electrons Has positive charge because takes up electrons

11. 11 Electrical current Measured in amperes (A) Measured in amperes (A) 1 A = 1 C/s 1 A = 1 C/s Coulomb = unit of electric charge Coulomb = unit of electric charge e - = 1.602 x 10 -19 C e - = 1.602 x 10 -19 C 1 A = 6.242 x 10 18 e - /s 1 A = 6.242 x 10 18 e - /s Electric current driven by difference in potential energy per unit of charge: J/C Electric current driven by difference in potential energy per unit of charge: J/C Potential difference (electromotive force or emf) = volt (V) Potential difference (electromotive force or emf) = volt (V) Where 1 V = 1 J/C Where 1 V = 1 J/C

12. 12 More… In the voltaic cell, potential difference (emf) between cathode and anode is referred to as In the voltaic cell, potential difference (emf) between cathode and anode is referred to as –Cell potential (E cell ) Under standard conditions (1 M, 1 atm, 25°C), cell potential is Under standard conditions (1 M, 1 atm, 25°C), cell potential is Standard cell potential = E ° cell Standard cell potential = E ° cell Cell potential = measure of overall tendency of redox rxn to occur spontaneously Cell potential = measure of overall tendency of redox rxn to occur spontaneously Thus, the higher the E ° cell, the greater the spontaneity Thus, the higher the E ° cell, the greater the spontaneity

13. 13 Electrochemical notation Cu (s) |Cu 2+ (aq) ||Zn 2+ (aq) |Zn (s) Cu (s) |Cu 2+ (aq) ||Zn 2+ (aq) |Zn (s) Notation describes voltaic cell Notation describes voltaic cell An ox on left An ox on left Red cat on right Red cat on right Separated by double vertical line (salt bridge) Separated by double vertical line (salt bridge) Single vertical line separates diff phases Single vertical line separates diff phases

14. 14 Electrochemical notation Some redox rxns reactants & products in same phase Some redox rxns reactants & products in same phase Mn doesn’t precipitate out  uses Pt at cathode Mn doesn’t precipitate out  uses Pt at cathode –Pt is inert, but provides area for electron gain/loss Fe (s) |Fe 2+ (aq) ||MnO 4 - (aq), H + (aq), Mn 2+ (aq) |Pt (s) Fe (s) |Fe 2+ (aq) ||MnO 4 - (aq), H + (aq), Mn 2+ (aq) |Pt (s) Write out net reaction Write out net reaction

15. 15 Answer Fe (s) |Fe 2+ (aq) ||MnO 4 - (aq), H + (aq), Mn 2+ (aq) |Pt (s) Fe (s) |Fe 2+ (aq) ||MnO 4 - (aq), H + (aq), Mn 2+ (aq) |Pt (s) Oxidation: Oxidation: Fe (s)  Fe 2+ (aq) + 2e - Reduction: Reduction: 5e - + MnO 4 - (aq) + 8H + (aq)  Mn 2+ (aq) + 4H 2 O (l) Net-reaction: Net-reaction: 5Fe (s) + 2MnO 4 - (aq) + 16H + (aq)  5Fe 2+ (aq) + 2Mn 2+ (aq) + 8H 2 O (l)

16. 16 Standard reduction potentials One half-cell must have a potential of zero to serve as reference One half-cell must have a potential of zero to serve as reference –Standard hydrogen electrode (SHE) half- cell Comprises Pt electrode in 1 M HCl w/ H 2 bubbling at 1 atm: Comprises Pt electrode in 1 M HCl w/ H 2 bubbling at 1 atm: 2H + (aq) + 2e -  H 2(g) ; E ° red = 0.00 V 2H + (aq) + 2e -  H 2(g) ; E ° red = 0.00 V

17. 17 Example Throw zinc into 1M HCl Throw zinc into 1M HCl Zn (s) |Zn 2+ (aq) ||2H + (aq) |H 2(g) Zn (s) |Zn 2+ (aq) ||2H + (aq) |H 2(g) E ° cell = E ° ox + E ° red = 0.76 V E ° cell = E ° ox + E ° red = 0.76 V If E ° red = 0.00 V (as the reference) If E ° red = 0.00 V (as the reference) Then E ° ox = 0.76 V (= oxid of Zn half-rxn) Then E ° ox = 0.76 V (= oxid of Zn half-rxn) Reduction of Zn-ion Reduction of Zn-ion –Is = -0.76 V (non-spontaneous)

18. 18 Problem Cr (s) |Cr 3+ (aq) ||Cl - (aq) |Cl 2(g) Cr (s) |Cr 3+ (aq) ||Cl - (aq) |Cl 2(g) What is the std cell pot (E ° cell ) given oxid of Cr = 0.73 V and Cl red = 1.36V? What is the std cell pot (E ° cell ) given oxid of Cr = 0.73 V and Cl red = 1.36V? Hint: standard electrode potentials are intensive properties; e.g., like density Hint: standard electrode potentials are intensive properties; e.g., like density –Stoichiometry irrelevant!

19. 19 Solution E ° cell = E ° ox + E ° red = 0.73V + 1.36V E ° cell = E ° ox + E ° red = 0.73V + 1.36V = 2.09V

20. 20 Appendix M, pages A-33-35 Standard reduction potentials in aqueous solution @ 25°C Standard reduction potentials in aqueous solution @ 25°C Also, pg. 967, Table 20.1 (gives increasing strengths of ox/red agents) Also, pg. 967, Table 20.1 (gives increasing strengths of ox/red agents) –Let’s take a look at it Does increasing strengths of ox/red agents make sense? Does increasing strengths of ox/red agents make sense? What happens to oxidizing agent, reducing agent? What happens to oxidizing agent, reducing agent?

21. 21 Problem Calculate the standard cell potential for the following: Calculate the standard cell potential for the following: Al (s) + NO 3 - (aq) + 4H + (aq)  Al 3+ (aq) + NO (g) + 2H 2 O (l)

22. 22 Answer Oxidation Oxidation Al (s)  Al 3+ (aq) + 3e - ; E ° ox = 1.66V Reduction Reduction NO 3 - (aq) + 4H + (aq) + 3e -  NO (g) + 2H 2 O (l) ; E ° red = 0.96V E ° cell = E ° ox + E ° red =1.66V + 0.96V = 2.62V E ° cell = E ° ox + E ° red =1.66V + 0.96V = 2.62V

23. 23 Predicting the spontaneous direction of a redox rxn Generally, any reduction half-rxn is spontaneous when paired w/reverse of half-rxn below it in table of standard reduction potentials Generally, any reduction half-rxn is spontaneous when paired w/reverse of half-rxn below it in table of standard reduction potentials Let’s look at table Let’s look at table Predict the exact value and spontaneity for the following: Predict the exact value and spontaneity for the following: Fe (s) + Mg 2+ (aq)  Fe 2+ (aq) + Mg (s)

24. 24 Answers Fe (s) + Mg 2+ (aq)  Fe 2+ (aq) + Mg (s) Oxidation Oxidation Fe (s)  Fe 2+ (aq) + 2e - ; E ° ox = 0.45V Reduction Reduction Mg 2+ (aq) + 2e -  Mg (s) ; E ° red = -2.37V E ° cell = E ° ox + E ° red =0.45V + -2.37V = E ° cell = E ° ox + E ° red =0.45V + -2.37V =-1.92Vnonspontaneous

25. 25 Will metal X dissolve in acid? Metals whose reduction half-rxns lie below reduction of proton to hydrogen gas will dissolve in acids Metals whose reduction half-rxns lie below reduction of proton to hydrogen gas will dissolve in acids Why? Why? –Just look at the table! Nitric acid is exception Nitric acid is exception –Let’s take a look

26. 26 E ° cell,  G °, K What must the values for E ° cell,  G °, & K be in order to have a spontaneous rxn? What must the values for E ° cell,  G °, & K be in order to have a spontaneous rxn?  G ° <0  G ° <0 E ° cell >0 E ° cell >0 K>1 K>1 –Product-favored

27. 27 Relationship between  G ° & E ° cell Faraday’s Constant (F) = 96,485 C/mol e - Faraday’s Constant (F) = 96,485 C/mol e -  G ° =-n e - FE ° cell  G ° =-n e - FE ° cell Problem: Problem: Calculate  G ° for Calculate  G ° for I 2(s) + 2Br - (aq)  2I - (aq) + Br 2(l) Is it spontaneous?

28. 28 Solution: it’s nonspontaneous! I 2(s) + 2Br - (aq)  2I - (aq) + Br 2(l) Oxidation Oxidation 2Br - (aq)  Br 2(l) + 2e - ; E ° ox = -1.09V Reduction Reduction I 2(s) + 2e -  2I - (aq) ; E ° red = 0.54V E ° cell = -1.09V + 0.54V = -0.55V E ° cell = -1.09V + 0.54V = -0.55V

29. 29 Problem 2Na (s) + 2H 2 O (l)  H 2(g) + 2OH - (aq) + 2Na + (aq) Is it spontaneous?

30. 30 Solution: it’s spontaneous! Oxidation Oxidation 2Na (s)  2Na + (aq) + 2e - ; E ° ox = 2.71V Reduction Reduction 2H 2 O (l) + 2e -  H 2(g) + 2OH - (aq) E ° red = -0.83V E ° cell = 2.71V + -0.83V = 1.88V E ° cell = 2.71V + -0.83V = 1.88V

31. 31 Relationship between E ° cell & K

32. 32 Problem Calculate K for Calculate K for 2Cu (s) + 2H + (aq)  Cu 2+ (aq) + H 2(g)

33. 33 Solution: is it product-favored? Oxidation Oxidation 2Cu (s)  Cu 2+ (aq) + 2e - ; E ° ox = -0.34V Reduction Reduction 2H + (aq) + 2e -  H 2(g) ; E ° red = 0.00V E ° cell = -0.34V

34. 34 Cell potential & concentration: Nernst Equation Concentration ≠ 1M Concentration ≠ 1M –Non-standard conditions Under standard conditions, Q = 1 Under standard conditions, Q = 1  E cell = E ° cell  E cell = E ° cell

35. 35 Problem Compute the cell potential, given Compute the cell potential, given Cu (s)  Cu 2+ (aq, 0.010 M) + 2e - MnO 4 - (aq, 2.0 M) + 4H + (aq, 1.0M) + 3e -  MnO 2(s) + 2H 2 O (l)

36. 36 Solution Balance the equation! Balance the equation! Oxidation Oxidation 3Cu (s)  3Cu 2+ (aq) + 6e - ; E ° ox = -0.34V Reduction Reduction 2MnO 4 - (aq) + 8H + (aq) + 6e -  2MnO 2(s) + 4H 2 O (l) ; E ° red = 1.68V

37. 37 To summarize If Q<1, rxn goes to products If Q<1, rxn goes to products –E cell > E ° cell If Q>1, rxn goes to reactants If Q>1, rxn goes to reactants –E cell < E ° cell If Q = K, @ eq., If Q = K, @ eq., –E ° cell = 0 (& E cell = 0) Explains why all batteries die Explains why all batteries die

38. 38 Concentration cells Voltaic cells can be constructed from two similar half-rxns where difference in concentration drives current flow Voltaic cells can be constructed from two similar half-rxns where difference in concentration drives current flow Cu (s) + Cu 2+ (aq, 2.0M)  Cu 2+ (aq, 0.010M) + Cu (s) –E ° cell = 0 since both half-rxns are the same However, using Nernst equation, different concentrations yield 0.068V However, using Nernst equation, different concentrations yield 0.068V –Let’s take a look Flow is from lower Cu-ion concentration half-cell to higher one Flow is from lower Cu-ion concentration half-cell to higher one –Down the concentration gradient The electrons will flow to the concentrated cell where they dilute the Cu-ion concentration The electrons will flow to the concentrated cell where they dilute the Cu-ion concentration Results in  Cu-ion concentration in dilute cell &  Cu-ion concentration in concentrated cell Results in  Cu-ion concentration in dilute cell &  Cu-ion concentration in concentrated cell

39. 39 Batteries Dry-cell batteries Dry-cell batteries –Don’t contain large amounts of water Anode Anode Zn (s)  Zn 2+ (aq) + 2e - Cathode Cathode 2MnO 2(s) + 2NH 4 + (aq) + 2e -  Mn 2 O 3(s) + 2NH 3(g) + H 2 O (l) –Cathode is carbon-rod immersed in moist (acidic) paste of MnO 2 that houses NH 4 Cl 1.5 V 1.5 V

40. 40 Batteries More common dry-cell type More common dry-cell type –Alkaline battery Anode Anode Zn (s) + 2OH - (aq)  Zn(OH) 2(s) + 2e - Zn (s) + 2OH - (aq)  Zn(OH) 2(s) + 2e - Cathode Cathode 2MnO 2(s) + 2H 2 O (l) + 2e -  2MnO(OH) (s) + 2OH - (aq) 2MnO 2(s) + 2H 2 O (l) + 2e -  2MnO(OH) (s) + 2OH - (aq) Longer shelf-life, “live” longer Longer shelf-life, “live” longer Cathode in basic paste Cathode in basic paste

41. 41 Car Batteries Lead-acid storage batteries Lead-acid storage batteries 6 electrochemical cells (2V) in series 6 electrochemical cells (2V) in series Anode Anode Pb (s) + HSO 4 - (aq)  PbSO 4(s) + H + (aq) + 2e - Cathode Cathode PbO 2(s) + HSO 4 - (aq) + 3H + (aq) + 2e -  PbSO 4(s) + 2H 2 O (l) In 30% soln of sulfuric acid In 30% soln of sulfuric acid If dead due to excess PbSO 4 covering electrode surfaces If dead due to excess PbSO 4 covering electrode surfaces Re-charge (reverse rxn)  converts PbSO 4 to Pb and PbO 2 Re-charge (reverse rxn)  converts PbSO 4 to Pb and PbO 2

42. 42 Rechargeable batteries Ni-Cd Ni-Cd Anode Anode Cd (s) + 2OH - (aq)  Cd(OH) 2(s) + 2e - Cathode Cathode 2NiO(OH) (s) + 2H 2 O (l) + 2e -  2Ni(OH) 2(s) + 2OH - (aq) KOH, usually, used KOH, usually, used 1.30 V 1.30 V Reverse rxn recharges battery Reverse rxn recharges battery Excess recharging  electrolysis of water Excess recharging  electrolysis of water EXPLOSION!!! EXPLOSION!!! Muhahahaha! Muhahahaha!

43. 43

44. 44 Rechargeable batteries Since Cd is toxic Since Cd is toxic –Developed safer alternative Ni-MH Ni-MH Hybrid car batteries: high energy density Hybrid car batteries: high energy density Same cathode rxn as previous Same cathode rxn as previous Anode Anode MH (s) + OH - (aq)  M (s) + H 2 O (l) + e - Commonly, M = AB 5, where A is rare earth mixture of La, Ce, Nd, Pr, and B is Ni, Co, Mn, and/or Mn Commonly, M = AB 5, where A is rare earth mixture of La, Ce, Nd, Pr, and B is Ni, Co, Mn, and/or Mn Very few use AB 2, where A = Ti and/or V Very few use AB 2, where A = Ti and/or V

45. 45

46. 46 Rechargeable batteries Anode made of graphite w/incorporated Li-ions between carbon layers Anode made of graphite w/incorporated Li-ions between carbon layers Ions spontaneously migrate to cathode Ions spontaneously migrate to cathode Cathode = LiCoO 2 or LiMn 2 O 4 Cathode = LiCoO 2 or LiMn 2 O 4 Transition metal reduced Transition metal reduced Used in laptop computers, cell phones, digital cameras Used in laptop computers, cell phones, digital cameras Light weight and high E density Light weight and high E density

47. 47 Fuel cell Reactants flow through battery Reactants flow through battery –Undergo redox rxn Generate electricity Generate electricity Hydrogen-oxygen fuel cell Hydrogen-oxygen fuel cell Anode Anode 2H 2(g) + 4OH - (aq)  4H 2 O (l) + 4e - Cathode Cathode O 2(g) + 2H 2 O (l) + 4e -  4OH - (aq) Used in space-shuttle program Used in space-shuttle program –And Arnold’s Hummah

48. 48 Electrolysis Electrical current used to drive nonspontaneous redox rxn Electrical current used to drive nonspontaneous redox rxn In electrolytic cells In electrolytic cells Used in Used in –Electrolysis of water –Metal plating: silver coated on metal, jewelry, etc.

49. 49 Electrolytic cells: using electricity to run a rxn Anode is “+”  gives electrons, connected to positive terminal of power source Anode is “+”  gives electrons, connected to positive terminal of power source Cathode is “-”  takes electrons, connected to negative terminal of power source Cathode is “-”  takes electrons, connected to negative terminal of power source Opposite scheme of voltaic cell! Opposite scheme of voltaic cell!

50. 50 Predicting the products of electrolysis 1. Pure molten salts –Anion oxidized/cation reduced Obtain 2Na (s) and Cl 2(g) from electrolysis of NaCl Obtain 2Na (s) and Cl 2(g) from electrolysis of NaCl 2. Mixture of cations or anions –K + /Na + and Cl - /Br - present Look at page 967 & compare half-cell potentials Look at page 967 & compare half-cell potentials –Cation/anion preferably reduced that has least negative, or most positive, half-cell potential

51. 51 Example Predict the half-rxn occurring at the anode and the cathode for electrolysis of Predict the half-rxn occurring at the anode and the cathode for electrolysis of AlBr 3 & MgBr 2 Oxidation Oxidation Br - (l)  Br 2(g) + 2e - ; E ° ox = - 1.09V Bromide will be oxidized at the anode Reduction Reduction Al 3+ (l) + 3e -  Al (s) ; E ° red = - 1.66V Mg 2+ (l) + 2e -  Mg (s) ; E ° red = -2.37V Reduction of Al will occur at the cathode since its potential is greater than Mg’s

52. 52 Predicting the products of electrolysis 3. aqueous solns: same as #2 –Water redox might occur simultaneously 2H 2 O (l)  O 2(g) + 4H + (aq) + 4e - 2H 2 O (l) + 2e-  H 2(g) + 2OH - (aq) E ° ox = -0.82 V & E ° red = -0.41 V  E ° cell = -1.23 V

53. 53 Problem Given oxidation of I - = -0.54 V & reduction of Li + = -3.04 V, which, if any, gases would be formed and where; i.e., at cathode/anode? Given oxidation of I - = -0.54 V & reduction of Li + = -3.04 V, which, if any, gases would be formed and where; i.e., at cathode/anode?

54. 54 Solution Oxidation Oxidation 2I - (aq)  2I 2(aq) + 2e - ; E ° ox = -0.54V 2H 2 O (l)  O 2(g) + 4H + (aq) + 4e - ; E ° ox = -0.82V I - will be oxidized at the anode Reduction Reduction 2Li + (aq) + 2e -  2Li (s) ; E ° red = -3.04 V 2H 2 O (l) + 2e -  H 2(g) + 2OH - (aq) ; E ° red = -0.41 V Water will be reduced at the cathode

55. 55 Stoichiometry of electryolysis Can use e - stoichiometric relations to predict moles and/or grams of substances Can use e - stoichiometric relations to predict moles and/or grams of substances Remember, unit of current = ampere = A = 1 C (magnitude of current)/s (time of current flow) Remember, unit of current = ampere = A = 1 C (magnitude of current)/s (time of current flow) Also, F = 96,485 C/mole e - Also, F = 96,485 C/mole e -

56. 56 Problem Gold can be plated out of a soln containing the Au 3+ according to Gold can be plated out of a soln containing the Au 3+ according to Au 3+ (aq) + 3e -  Au (s) What mass of gold (in grams) will be plated by the flow of 5.5 A of current for 25 mins? What mass of gold (in grams) will be plated by the flow of 5.5 A of current for 25 mins?

57. 57 Solution