1. Chapter 18 Electrochemistry

2. 2 GOALS Review: oxidation states oxidation/reduction oxidizing/reducing agent ch. 17 Balancing redox reactions Voltaic cells Electrochemical potentials Electrolysis …& the calculations!! …& the calculations!!

3. Why Study Electrochemistry? Batteries Batteries Corrosion Corrosion Industrial production of chemicals such as Cl 2, NaOH, F 2 and Al Industrial production of chemicals such as Cl 2, NaOH, F 2 and Al Biological redox reactions Biological redox reactions The heme group

4. Electron Transfer Reactions Electron transfer reactions are oxidation-reduction or redox reactions (i.e. changes in oxidation states). Redox reactions can result in the generation of an electric current (battery), or, may be caused by applying an electric current (electroplating). Therefore, this field of chemistry is often called ELECTROCHEMISTRY.

5. ELECTRON TRANSFER REACTIONS 0 0 1+ 1- 0 0 1+ 1- 2Na(s) + Cl 2 (g)  2NaCl(s) 1+ 0 2+ 0 1+ 0 2+ 0 2HCl(aq) + Zn(s)  ZnCl 2 (aq) + H 2 (g) 0 0 0 0 Cu(s) + 2 Ag + (aq)  Cu 2+ (aq) + 2 Ag(s) 0 0 3+ 2- 0 0 3+ 2- 2Fe(s) + xH 2 O(l) + 1½O 2 (g)  Fe 2 O 3.xH 2 O(s) 4+ 2- 0 0 CO 2 (g) + H 2 O(l) + energy  (CH 2 O) n + O 2 (g)

6. Review of Terminology for Redox Reactions ; OXIDATION—loss of electron(s) by a species; increase in oxidation number ; e- to the right of arrow. Na  Na + + e- REDUCTION—gain of electron(s); decrease in oxidation number; e- to left of arrow. ½Cl 2 (g) + e-  Cl - OXIDIZING AGENT—electron acceptor; it is reduced: ½ Cl 2 (g) + e-  Cl - REDUCING AGENT—electron donor; it is oxidized Na  Na + + e- ; OXIDATION—loss of electron(s) by a species; increase in oxidation number ; e- to the right of arrow. Na  Na + + e- REDUCTION—gain of electron(s); decrease in oxidation number; e- to left of arrow. ½Cl 2 (g) + e-  Cl - OXIDIZING AGENT—electron acceptor; it is reduced: ½ Cl 2 (g) + e-  Cl - REDUCING AGENT—electron donor; it is oxidized Na  Na + + e-

7. Electrochemical Cells Apparatus for generating an electric current through the use of a product favored reaction (spontaneous): Apparatus for generating an electric current through the use of a product favored reaction (spontaneous): voltaic or galvanic cell. (an electric current is used to bring about a nonspontaneous chemical reaction). An electrolytic cell is used to carry out electrolysis (an electric current is used to bring about a nonspontaneous chemical reaction). Batteries are voltaic cells

8. Electrochemistry Alessandro Volta, 1745-1827, Italian scientist and inventor. Luigi Galvani, 1737-1798, Italian scientist and inventor.

9. Balancing Equations for Redox Reactions Some redox reactions have equations that must be balanced by special techniques. Mn = +7 Fe = +2 Fe = +3 Mn = +2 MnO 4 - (aq) + 5 Fe 2+ (aq) + 8 H + (aq)  Mn 2+ (aq) + 5 Fe 3+ (aq) + 4 H 2 O(liq)

10. 10 Rules for Assigning Oxidation St ates rules are in order of priority 1. free elements have an oxidation state = 0 Na = 0 and Cl 2 = 0 in 2 Na(s) + Cl 2 (g) 2. monatomic ions have an oxidation state equal to their charge Na = +1 and Cl = -1 in NaCl 3. (a) the sum of the oxidation states of all the atoms in a compound is 0 Na = +1 and Cl = -1 in NaCl, (+1) + (-1) = 0

11. 11 Rules for Assigning Oxidation States 3. (b) the sum of the oxidation states of all the atoms in a polyatomic ion equals the charge on the ion N = +5 and O = -2 in NO 3 –, (+5) + 3(-2) = -1 4. (a) Group I metals have an oxidation state of +1 in all their compounds Na = +1 in NaCl (b) Group II metals have an oxidation state of +2 in all their compounds Mg = +2 in MgCl 2

12. 12 Rules for Assigning Oxidation States 5. in their compounds, nonmetals have oxidation states according to the table below (grp # - 8) nonmetals higher on the table take priority NonmetalOxidation StateExample FCF 4 H+1CH 4 O-2CO 2 Group 7ACCl 4 Group 6A-2CS 2 Group 5A-3NH 3

13. Balancing Equations Cu + Ag +  give  Cu 2+ + Ag

14. Balancing Equations Step 1:Divide the reaction into half- reactions, one for oxidation and the other for reduction. OxCu  Cu 2+ Red Ag +  Ag Step 2:Balance each for mass. Already done in this case. Step 3:Balance each half-reaction for charge by adding electrons. Ox Cu  Cu 2+ + 2e- Red Ag + + e-  Ag

15. Balancing Equations Step 4:Multiply each half-reaction by a factor so that the reducing agent supplies as many electrons as the oxidizing agent requires. Reducing agent Cu  Cu 2+ + 2e- Oxidizing agent 2 Ag + + 2 e-  2 Ag Step 5:Add half-reactions to give the overall equation. Cu + 2 Ag +  Cu 2+ + 2Ag The equation is now balanced for both charge and mass (the 2e - of the left are cancelled out with those on the right).

16. Reduction of VO 2 + with Zn

17. Balancing Equations Balance the following in acid solution— VO 2 + + Zn  VO 2+ + Zn 2+ VO 2 + + Zn  VO 2+ + Zn 2+ Step 1:Write the half-reactions OxZn  Zn 2+ RedVO 2 +  VO 2+ Step 2:Balance each half-reaction for mass. OxZn  Zn 2+ + 2e - Red VO 2 + + e-  VO 2+ Red VO 2 + + e-  VO 2+ excess of 2+ on the right Zn lost two electrons. They are written on the right. V is 5+ on the left and 4+ on the right. It gained one electron. That is written on the left side.

18. Balancing Equations Step 3:Balance half-reactions for charge Reaction is acidic, then we can use H +. Reaction is acidic, then we can use H +. OxZn  Zn 2+ + 2e- Rede- + 2 H + + VO 2 +  VO 2+ + H 2 O Step 4:Multiply by an appropriate factor. OxZn  Zn 2+ + 2e- Red 2e- + 4 H + + 2 VO 2 +  2 VO 2+ + 2 H 2 O Step 5:Add balanced half-reactions. Zn + 4 H + + 2 VO 2 +  Zn 2+ + 2 VO 2+ + 2 H 2 O

19. Tips on Balancing Equations Never add O 2, O atoms, or O 2- to balance oxygen. Never add O 2, O atoms, or O 2- to balance oxygen. Balance O with OH - or H 2 O. Never add H 2 or H atoms to balance hydrogen. Never add H 2 or H atoms to balance hydrogen. Balance H with H + /H 2 O in acid or OH - /H 2 O in base.

20. {Equations that include oxoanions like SO 4 2-, NO 3 -, ClO -, CrO 4 2-, and MnO 4 -, also fall into this category}. Tips on Balancing Equations Check your work at the end to make sure mass and charge are balanced.Check your work at the end to make sure mass and charge are balanced. PRACTICE!!!!!!!!!!!PRACTICE!!!!!!!!!!! Be sure to write the correct charges on all the ions.

21. 21 More Practice - Balance the equations below! I  (aq) + MnO 4  (aq)  I 2(aq) + MnO 2(s) in basic solution ClO 3 - + Cl -  Cl 2 (in acid) Cr 2 O 7 2- + I -  IO 3 - + Cr 3+ (in acid) MnO 4 - + H 2 SO 3  SO 4 2- + Mn 2+ (in acid) Cr(OH) 4 - + H 2 O 2  CrO 4 2- + H 2 O (in basic soln) Zn + NO 3 -  Zn(OH) 4 - + NH 3 (in basic soln) An alkaline (basic) solution of hypochlorite ions reacts with solid chromium(III) hydroxide to produce chromate and chloride ions.

22. VOLTAIC CELLS The Zn|Zn 2+ and Cu|Cu 2+ Cell Zn(s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu(s) - use a chemical rxn to produce an electric current.

23. Oxidation: Zn(s)  Zn 2+ (aq) + 2e- Reduction: Cu 2+ (aq) + 2e-  Cu(s)-------------------------------------------------------- Cu 2+ (aq) + Zn(s)  Zn 2+ (aq) + Cu(s) Electrons are transferred from Zn to Cu 2+, but there is no useful electric current. With time, Cu plates out onto Zn metal strip, and Zn strip “disappears.” A CHEMICAL CHANGE PRODUCES AN ELECTRIC CURRENT

24. To obtain a useful current, we separate the oxidizing and reducing agents so that electron transfer occurs through an external wire. To obtain a useful current, we separate the oxidizing and reducing agents so that electron transfer occurs through an external wire. This is accomplished in a GALVANIC or VOLTAIC cell. A group of such cells is called a battery. Zn(s)  Zn 2+ (aq) + 2e- A CHEMICAL CHANGE PRODUCES AN ELECTRIC CURRENT

25. Electrons travel through external wire. Salt bridge allows anions and cations to move between electrode compartments. Salt bridge allows anions and cations to move between electrode compartments. Electrons travel through external wire. Salt bridge allows anions and cations to move between electrode compartments. Salt bridge allows anions and cations to move between electrode compartments. Zn  Zn 2+ + 2e- Cu 2+ + 2e-  Cu  Anions Cations  OxidationAnodeNegativeOxidationAnodeNegative ReductionCathodePositiveReductionCathodePositive

26. CELL POTENTIAL, E Electrons are “driven” from anode to cathode by an electromotive force or emf. Electrons are “driven” from anode to cathode by an electromotive force or emf. For Zn/Cu cell, this is indicated by a voltage of 1.10 V at 25 ˚C and when [Zn 2+ ] = [Cu 2+ ] = 1.0 M. For Zn/Cu cell, this is indicated by a voltage of 1.10 V at 25 ˚C and when [Zn 2+ ] = [Cu 2+ ] = 1.0 M. Zn and Zn 2+, anode Cu and Cu 2+, cathode 1.10 V 1.0 M

27. Need Calculate Cell Voltage. Balanced half-reactions can be added together to get the overall, balanced equation. Zn(s)  Zn 2+ (aq) + 2e- Cu 2+ (aq) + 2e-  Cu(s) -------------------------------------------- Cu 2+ (aq) + Zn(s)  Zn 2+ (aq) + Cu(s) Zn(s)  Zn 2+ (aq) + 2e- Cu 2+ (aq) + 2e-  Cu(s) -------------------------------------------- Cu 2+ (aq) + Zn(s)  Zn 2+ (aq) + Cu(s) If we know E o for each half-reaction, we add them to get E o for overall reaction. -need E o for Zn & Cu half-cells

28. 28 Standard Reduction Potential We can measure ALL other half-cell potentials relative to another half-reaction. We select as a standard half-reaction, the reduction of H + to H 2 under standard conditions (1atm, 1 M @ 25 o C) and which we assign a potential difference = 0 V Standard Hydrogen Electrode, SHE

29. Zn/Zn 2+ half-cell hooked up to a SHE. E o for the cell = +0.76 V Zn/Zn 2+ half-cell hooked up to a SHE. E o for the cell = +0.76 V Negative electrode Supplier of electrons Acceptor of electrons Positive electrode 2 H + + 2e-  H 2 ReductionCathode Zn  Zn 2+ + 2e- OxidationAnode Zn is a better reducing agent than H 2

30. Acceptor of electrons Supplier of electrons Cu 2+ + 2e-  Cu ReductionCathode H 2  2 H + + 2e- OxidationAnode Positive Negative Cu/Cu 2+ half-cell hooked up to a SHE. E o for the cell = +0.34 V Cu/Cu 2+ half-cell hooked up to a SHE. E o for the cell = +0.34 V H 2 is now a better reducing agent than Cu!

31. Zn/Cu Electrochemical Cell oxid: Zn(s)  Zn 2+ (aq) + 2e- E o = +0.76 V oxid: Zn(s)  Zn 2+ (aq) + 2e- E o = +0.76 V red: Cu 2+ (aq) + 2e-  Cu(s) E o = +0.34 V red: Cu 2+ (aq) + 2e-  Cu(s) E o = +0.34 V--------------------------------------------------------------- Cu 2+ (aq) + Zn(s)  Zn 2+ (aq) + Cu(s) Cu 2+ (aq) + Zn(s)  Zn 2+ (aq) + Cu(s) E o (calc’d) = +1.10 V Cathode; positive; sink for electrons Anode; negative; source of electrons+

32. Uses of E o Values Organize half-reactions by relative ability to act as oxidizing/reducing agents. Half-rxns are written as reduction rxns!! Cu 2+ (aq) + 2e-  Cu(s)E o = +0.34 V Zn 2+ (aq) + 2e-  Zn(s) E o = –0.76 V When a reaction is reversed, the sign of E˚ is reversed!

34. 34 reducing agents oxidizing agents

35. Using Standard Potentials, E o Which is the best reducing agent: Hg (+0.79 V), Al (-1.66 V), or Sn (-0.14 V)? Which is the best oxidizing agent: O 2 (1.23 V); H 2 O 2 (1.77 V) or Cl 2 (1.36 V)? H 2 O 2 (1.77 V) Al (-1.66 V)

36. Using Standard Potentials, E o Which element/ion is the best reducing agent? Fe 3+ + e-  Fe 2+ (+0.77 V) I 2 + 2e-  2I - (+0.54 V) Sn 4+ + 2e-  Sn 2+ (+0.15 V) Which substance is the best oxidizing agent? Cr 2 O 7 2- + 6e- + 14H +  2Cr 3+ + 7H 2 O(+1.33 V) O 2 + 4e- + 4H +  2H 2 O(+1.23 V) Fe 3+ + e-  Fe 2+ (+0.77 V) Cr 2 O 7 2- Sn 2+

37. Standard Redox Potentials, E o Any substance on the right will reduce any substance HIGHER than it on the LEFT. Any substance on the right will reduce any substance HIGHER than it on the LEFT. Zn can reduce H + and Cu 2+. Zn can reduce H + and Cu 2+. H 2 can reduce Cu 2+ but not Zn 2+ H 2 can reduce Cu 2+ but not Zn 2+ Cu cannot reduce H + or Zn 2+. Cu cannot reduce H + or Zn 2+.

38. Standard Redox Potentials, E o Cu 2+ + 2e- --> Cu +0.34 + 2 H + 2e- --> H 2 0.00 Zn 2+ + 2e- --> Zn -0.76 Northwest-southeast rule: product-favored reactions occur between reducing agent at southeast corner (ANODE) & reducing agent at southeast corner (ANODE) & oxidizing agent at northwest corner (CATHODE) oxidizing agent at northwest corner (CATHODE) Any substance on the right will reduce any substance higher than it on the left. Ox. agent Red. agent

39. Standard Redox Potentials, E o Cu 2+ + 2e- --> Cu +0.34 V 2+ Ni + 2e- --> Ni -0.25 V Zn 2+ + 2e- --> Zn -0.76 V Northwest-southeast rule: product-favored reactions occur between reducing agent at southeast corner (ANODE) & reducing agent at southeast corner (ANODE) & oxidizing agent at northwest corner (CATHODE) oxidizing agent at northwest corner (CATHODE) Zn will reduce Ni 2+, Cu 2+ ; Ni will reduce Cu 2+. Ox. agent Red. agent

40. Using Standard Potentials, E o In which direction do the following reactions go? In which direction do the following reactions go? Cu(s) + 2 Ag + (aq)  Cu 2+ (aq) + 2 Ag(s) +0.46 V Cu(s) + 2 Ag + (aq)  Cu 2+ (aq) + 2 Ag(s) +0.46 V Cu 2+ (aq) + Zn(s)  Cu(s) + Zn 2+ (aq) +1.10 V Cu 2+ (aq) + Zn(s)  Cu(s) + Zn 2+ (aq) +1.10 V Go to the right as written Go to the right as written Fe 2+ (aq) +  Fe 2+ (aq) + Cd(s)  Fe(s) + Cd 2+ (aq) Goes LEFT, opposite to direction written Goes LEFT, opposite to direction written What is E o net for this reverse reaction? -0.04 V +0.04 V

41. Cd  Cd 2+ + 2e- or Cd 2+ + 2e-  Cd (-0.40 V) Fe  Fe 2+ + 2e- or Fe 2+ + 2e-  Fe (-0.44 V) E o for a Voltaic Cell Which way does the reaction proceed? In which direction is it spontaneous? Fe(s) + Cd 2+ (aq)  Cd(s) + Fe 2+ (aq)

42. From the table, Fe is a better reducing agent than Cd (-0.44 V; anode) Fe is a better reducing agent than Cd (-0.44 V; anode) Cd 2+ is a better oxidizing agent than Fe 2+ (-0.40 V; cathode)Cd 2+ is a better oxidizing agent than Fe 2+ (-0.40 V; cathode) E o for a Voltaic Cell E o = E˚ cathode - E˚ anode (reverse the smaller, more negative, then add) negative, then add) cathode: cathode: Cd 2+ (aq) + 2e-  Cd(s) -0.40 V (red) - anode: Fe(s)  Fe 2+ (aq) + 2e- +0.44 V (oxid) Overall : Fe(s) + Cd 2+ (aq)  Cd(s) + Fe 2+ (aq) +0.04 V

43. More 0n Cell Voltage More 0n Cell Voltage When two half-rxns (written as reduction) are joined in an electrochemical cell, the one with the larger half-cell potential occurs in the forward direction, and the one with the smaller potential occurs in the reverse direction. Cd 2+ + 2e-  Cd (-0.40 V) larger Fe 2+ + 2e-  Fe (-0.44 V) smaller Fe 2+ + 2e-  Fe (-0.44 V) smaller (reverse this rxn) Cd 2+ + 2e-  Cd (-0.40 V) larger Cd 2+ + 2e-  Cd (-0.40 V) larger Fe  Fe 2+ + 2e- (+0.44 V) Cd 2+ + Fe  Cd + Fe 2+ (+0.04 V) overall: Cd 2+ + Fe  Cd + Fe 2+ (+0.04 V)

44. More 0n Cell Voltage More 0n Cell Voltage When two half-rxns (written as reduction) are joined in an electrochemical cell, the one with the larger half-cell potential occurs in the forward direction, and the one with the smaller potential occurs in the reverse direction. Ni 2+ + 2e-  Ni (-0.23 V) larger Mn 2+ + 2e-  Mn (-1.18 V) smaller Mn 2+ + 2e-  Mn (-1.18 V) smaller (reverse this rxn) Ni 2+ + 2e-  Ni (-0.23 V) Ni 2+ + 2e-  Ni (-0.23 V) Mn  Mn 2+ + 2e- (+1.18 V) Ni 2+ + Mn  Ni + Mn 2+ (+0.95 V) overall: Ni 2+ + Mn  Ni + Mn 2+ (+0.95 V)

45. More 0n Cell Voltage More 0n Cell Voltage Assume I - ion can reduce water.  2 H 2 O + 2e-  H 2 + 2 OH - Cathode  2 I -  I 2 + 2e- Anode -------------------------------------------------  2 I - + 2 H 2 O  I 2 + 2 OH - + H 2  2 H 2 O + 2e-  H 2 + 2 OH - Cathode  2 I -  I 2 + 2e- Anode -------------------------------------------------  2 I - + 2 H 2 O  I 2 + 2 OH - + H 2 Assuming reaction occurs as written, E˚ net = E˚ cathode - E˚ anode (from values in table) = (-0.828 V) - (+0.535 V) = -1.363 V Minus E net ˚ means net rxn. occurs in the opposite direction (favors I - + H 2 O). I 2 + 2 OH - + H 2  2I - + H 2 O (+1.363 V)!!!

46. 46 Calculate E  cell for the reaction at 25  C Al (s) + NO 3 − (aq) + 4 H + (aq)  Al 3+ (aq) + NO (g) + 2 H 2 O (l) (This is the reaction of Al with nitric acid) Separate the reaction into the oxidation and reduction half-reactions Find the E  for each half- reaction and sum to get E  cell ox:Al (s)  Al 3+ (aq) + 3 e − red: NO 3 − (aq) + 4 H + (aq) + 3 e −  NO (g) + 2 H 2 O (l) E° red = +0.96 V E  ox = −E  red = +1.66 V E  ox = −(E  red ) = +1.66 V E  red = +0.96 V E  cell = (+1.66 V) + (+0.96 V) = +2.62 V

47. 47 For the reaction Al (s) + NO 3 − (aq) + 4 H + (aq)  Al 3+ (aq) + NO (g) + 2 H 2 O (l) ox:Al (s)  Al 3+ (aq) + 3 e − red: NO 3 − (aq) + 4 H + (aq) + 3 e −  NO (g) + 2 H 2 O (l) E  ox = −E  red = +1.66 V E  red = +0.96 V E  cell = (+1.66 V) + (+0.96 V) = +2.62 V The symbol of the cell is Al (s) | Al 3+ (aq) || NO 3 − (aq), 4 H + (aq), NO (g) |Pt This is the symbol for the salt bridge This is the symbol for the electrode-solxn contact

48. E at Nonstandard Conditions The NERNST EQUATION E = potential under nonstandard conditions n = no. of electrons exchanged ln = “natural log” If [P] and [R] = 1 mol/L, then E = E˚ If [R] > [P], then E is LARGER than E˚ If [R] < [P], then E is smaller than E˚

49. Zn(s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu(s) Zn is anode (-0.76 V) in 0.40 M Zn 2+ (aq) and Cu is cathode (0.34 V) in 4.8  10 -3 M Cu 2+ (aq). Eº cell = Eº cathode – Eº anode = (0.34 V) – (–0.76 V) = 0.34 + 0.76 = +1.10 V Substituting Calculate the cell potential. Solution: ( need standard cell potential, E o cell )

50. E o cell = 1.10 V Zn(s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu(s) Zn is anode (-0.76 V) and Cu is cathode (0.34 V) E o cell = 1.10 V n = 2 (Cu 2+ + 2e- = Cu 0 ) Solution: = 1.04 V E = 1.10 V – (0.01285)(4.42) = 1.04 V

51. 2Fe 3+ (aq) + 3Mg(s)  2Fe(s) + 3Mg 2+ (aq) Fe 3+ = 1.0  10 -3 M ; Mg 2+ = 2.5 M Calculate the cell potential. Eº cell = Eº cathode – Eº anode = (-0.036 V) – (–2.37 V) = -0.036 + 2.37 = +2.33 V Substituting Solution: ( need standard cell potential, E o cell ) What is n???

52. 2Fe 3+ (aq) + 3Mg(s)  2Fe(s) + 3Mg 2+ (aq) Fe 3+ = 1.0  10 -3 M ; Mg 2+ = 2.5 M Calculate the cell potential. Eº cell = +2.33 V; n = 6 Substituting Solution: ( need standard cell potential, E o cell ) Ans = 2.26 V

53. BATTERIES Primary, Secondary, and Fuel Cells

54. Dry Cell Battery Anode (-) Zn  Zn 2+ + 2e- Cathode (+) 2 NH 4 + + 2e-  2 NH 3 + H 2 Primary battery — uses redox reactions that cannot be restored by recharge.

55. Nearly same reactions as in common dry cell, but under basic conditions. Alkaline Battery Anode (-): Zn + 2 OH -  ZnO + H 2 O + 2e- Cathode (+): 2 MnO 2 + H 2 O + 2e-  Mn 2 O 3 + 2 OH -

56. Lead Storage Battery Secondary battery Secondary battery Uses redox reactions that can be reversed. Uses redox reactions that can be reversed. Can be restored by recharging Can be restored by recharging

57. Lead Storage Battery Anode (-) E o = +0.36 V Pb + HSO 4 -  PbSO 4 + H + + 2e- Cathode (+) E o = +1.68 V PbO 2 + HSO 4 - + 3 H + + 2e-  PbSO 4 + 2 H 2 O

58. Ni-Cad Battery Anode (-) Cd + 2 OH -  Cd(OH) 2 + 2e- Cathode (+) NiO(OH) + H 2 O + e-  Ni(OH) 2 + OH -

59. Fuel Cells: H 2 as a Fuel Fuel cell - reactants are supplied continuously from an external source.Fuel cell - reactants are supplied continuously from an external source. Cars can use electricity generated by H 2 /O 2 fuel cells.Cars can use electricity generated by H 2 /O 2 fuel cells. H 2 carried in tanks or generated from hydrocarbons.H 2 carried in tanks or generated from hydrocarbons.

60. Storing H 2 as a Fuel One way to store H 2 is to adsorb the gas onto a metal or metal alloy.

61. Hydrogen—Air (O 2 ) Fuel Cell Cathode: O 2 (g) + 2H 2 O(liq) + 4e-  4OH - (aq) ---------------------------------- Net: O 2 (g) + 2H 2 (g)  2H 2 O(liq) Anode: 2H 2 (g)  4H + (aq) + 4e-

62. Electrolysis Using electrical energy to produce chemical change.  Sn 2+ (aq) + 2 Cl - (aq)  Sn(s) + Cl 2 (g) Electrolysis of water; electroplating; refining metals; production of chemicals.

63. Electrolysis Electrolysis Electric Energy  Chemical Change Electrolysis of molten NaCl. Electrolysis of molten NaCl. Here a battery “pumps” electrons from Cl - to Na +. Here a battery “pumps” electrons from Cl - to Na +. NOTE: Polarity of electrodes is reversed from batteries. NOTE: Polarity of electrodes is reversed from batteries.

64. Electrolysis of Molten NaCl Anode (+) 2Cl - (l)  Cl 2 (g) + 2e- (-1.36 V) Cathode (-) Na + (l) + e-  Na (-2.71 V) E o for cell (in melted NaCl) = E˚ c + E˚ a = - 2.71 V + (-1.36 V) = - 2.71 V + (-1.36 V) = - 4.07 V (in melted NaCl) rxn is nonspontaneous = - 4.07 V (in melted NaCl) rxn is nonspontaneous External electrical energy needed because E o is (-).

65. Electrolysis of Aqueous NaOH Anode (+) E° = -0.40 V 4 OH -  O 2 (g) + 2 H 2 O + 4e- Cathode (-) 4 H 2 O + 4e-  2 H 2 + 4 OH - E o for cell = -1.23 V Electric Energy  Chemical Change AnodeCathode H 2 O is more easily reduced than Na + !! (E o -2.71 V) NaOH + H 2 O  Na + (aq) + OH - (aq)

66. Electrolysis of Aqueous NaCl Anode (+) 2 Cl -  Cl 2 (g) + 2e - E° =-1.36 V Cl 2 (g) + 2e - E° =-1.36 V Cathode (-) 2 H 2 O + 2e-  H 2 + 2 OH - H 2 + 2 OH - E o for cell = -2.19 V Note that H 2 O is more Easily reduced than Na + (E° = -2.71 V) 2H 2 O(l)  O 2 (g) + 4H + + 4e - O 2 (g) + 4H + + 4e - Also, Cl - is oxidized in preference to H 2 O because of kinetics (overvoltage) E° red < -1.23 V, may be down to ~ -2.00 V Also, Cl - is oxidized in preference to H 2 O because of kinetics (overvoltage) E° red < -1.23 V, may be down to ~ -2.00 V NaCl + H 2 O  Na + (aq) + Cl - (aq)

67. E o and Thermodynamics E o is related to ∆G o, the free energy change for the reaction. E o is related to ∆G o, the free energy change for the reaction. ∆G˚ proportional to –nE˚ ∆G˚ proportional to –nE˚ ∆G o = -nFE o where F = Faraday constant = 9.6485 x 10 4 J/Vmol of e - = 9.6485 x 10 4 J/Vmol of e - (or 9.6485  10 4 coulombs/mol) and n is the number of moles of electrons transferred.

68. Electrolysis of Aqueous CuCl 2 Anode (+) 2 Cl -  Cl 2 (g) + 2e- Cathode (-) Cu 2+ + 2e-  Cu E o for cell = -1.02 V Note that Cu is more easily reduced than either H 2 O or Na + (check redox potentials). CuCl 2 + H 2 O  Cu 2+ (aq) + 2Cl - (aq)

69. Calculate  G o for the reaction, Zn 2+ (aq) + Ni(s)  Zn(s) + Ni 2+ (aq) Solution: use  G o = -nFE° no. of electrons, n = 2 F = 9.6485  10 4 C Zn 2+ (aq) + 2e-  Zn(s), -0.763 V Ni(s)  Ni 2+ (aq) + 2e-, +0.25 V cathode anode need E o cell

70. E o cell = E cathode – (E anode ) E o cell = -0.763 – (-0.25 V) = -0.51 V  G o = (-2  96485 J/V  -0.51 V)  1 kJ/1000 J  1 kJ/1000 J  98 kJ ∆Go > 0, reaction is nonspontaneous Reagents are favored

71. E o and ∆G o ∆G o = - n F E o For a product-favored reaction Reactants  Products Reactants  Products ∆G o 0 E o is positive For a reactant-favored reaction Reactants  Products Reactants  Products ∆G o > 0 and so E o 0 and so E o < 0 E o is negative

72. 72 E° cell,  G° and K!! for a spontaneous reaction  G° < 0 (negative) E° > 0 (positive) K > 1 (large) When E cell = 0 (no net rxn), reactants and products are at equilibrium……and Q = K

73. Quantitative Aspects of Electrochemistry Consider electrolysis of aqueous silver ion. Ag + (aq) + e-  Ag(s) 1 mol e-  1 mol Ag If we could measure the moles of e-, we could know the quantity of Ag formed. But do we measure moles of e-?

74. But, how is charge related to moles of electrons? = 96,500 C/mol e- = 1 Faraday Michael Faraday 1791-1867 Quantitative Aspects of Electrochemistry

75. 1.50 amps flow thru a Ag + (aq) solution for 15.0 min. What mass of Ag metal is deposited? What mass of Ag metal is deposited?Solution (a)Calc. charge Charge (C) = current (A) x time (t) Charge (C) = current (A) x time (t) = (1.5 amps)(15.0 min)(60 s/min) = 1350 C = (1.5 amps)(15.0 min)(60 s/min) = 1350 C

76. Quantitative Aspects of Electrochemistry Solution (a)Charge = 1350 C (b)Calculate moles of e- used 1.50 amps flow thru a Ag + (aq) solution for 15.0 min. What mass of Ag metal is deposited? Ag + + e-  Ag(s) (c)Calc. quantity of Ag

77. Quantitative Aspects of Electrochemistry The anode reaction in a lead storage battery is Pb(s) + HSO 4 - (aq)  PbSO 4 (s) + H + (aq) + 2e- If a battery delivers 1.50 amp, and there is 454 g of Pb, how long will the battery last? Solution a) 454 g Pb = 2.19 mol Pb b) Calculate moles of e- each Pb atom is loosing 2e − each Pb atom is loosing 2e − c)Calculate charge 4.38 mol e- 96,500 C/mol e- = 423,000 C 4.38 mol e- 96,500 C/mol e- = 423,000 C

78. Quantitative Aspects of Electrochemistry The anode reaction in a lead storage battery is Pb(s) + HSO 4 - (aq)  PbSO 4 (s) + H + (aq) + 2e- If a battery delivers 1.50 amp, and you have 454 g of Pb, how long will the battery last? Solution a)454 g Pb = 2.19 mol Pb b)Mol of e- = 4.38 mol c)Charge = 423,000 C About 78 hours d)Calculate time