1. ELECTROCHEMISTRY CHARGE (Q) – A property of matter which causes it to experience the electromagnetic force COULOMB (C) – The quantity of charge equal to 6.241 × 10 18 electrons ELECTROMOTIVE FORCE or POTENTIAL or VOLTAGE ( Ɛ ) – The potential difference between 2 substances, causing electrons to flow from one to the other VOLT (V) – One joule of potential energy per coulomb ELECTRIC CURRENT or AMPERAGE (I) – The rate of flow of electric charge AMPERE (A) – Flow rate of one coulomb of electric charge per second 3D-1 (of 14)

2. Spontaneous oxidation-reduction reaction: Fe (s) + Cu 2+ (aq) → Fe 2+ (aq) + Cu (s) Redox reactions can be written as the sum of 2 half-reactions Oxidation: Reduction: Fe (s) → Fe 2+ (aq) Cu 2+ (aq) → Cu (s) If the Fe (s) and Cu 2+ (aq) are separated, the electron transfer can happen through a wire 3D-2 (of 14) Iron is more reactive then copper,  iron atoms will release their valence electrons to the copper (II) ions + 2e - 2e - + Fe (s) + 2e - + Cu 2+ (aq) → Fe 2+ (aq) + Cu (s) + 2e - Fe (s) + Cu 2+ (aq) → Fe 2+ (aq) + Cu (s)

3. ANODE – The electrode where oxidation occurs CATHODE – The electrode where reduction occurs 0.78 V is called the cell potential, the cell voltage, or the cell emf 3D-3 (of 14) Cu

4. GALVANIC CELL – An electrochemical cell that produces electric current from a chemical reaction Shorthand notation: Anode | Anode Solution || Cathode Solution | Cathode Fe | Fe 2+ (1 M) || Cu 2+ (1 M) | Cu 3D-4 (of 14)

5. The ΔG of a reaction occurring in a Galvanic cell is related to Ɛ ΔG= -n F Ɛ n= number of moles of electrons transferred in the redox reaction F = the Faraday constant the charge of 1 mole of electrons, equal to 96,485 C For Galvanic cells with 1 M concentrations ΔGº= -n F Ɛ º J= (mol) (C/mol) (V) (J/C) 3D-5 (of 14)

6. Calculate ΔGº for the reaction Fe (s) + Cu 2+ (aq) → Fe 2+ (aq) + Cu (s) Ɛ º = 0.78 V ΔGº= -n F Ɛ º = -(2 mol)(96,485 C/mol)(0.78 V) = -(2 mol)(96,485 C/mol)(0.78 J/C) = -150,000 J ΔG 0 means a spontaneous process The more negative ΔG, the more spontaneous the process The more positive Ɛ, the more spontaneous the process 3D-6 (of 14)

7. REDUCTION AND OXIDATION POTENTIALS The sum of a reduction potential and an oxidation potential must equal the potential for the overall redox reaction 3D-7 (of 14) REDUCTION POTENTIAL ( Ɛ red ) – The electric potential for a reduction half- reaction OXIDATION POTENTIAL ( Ɛ ox ) – The electric potential for an oxidation half- reaction Ɛ º red and Ɛ º ox are for a standard state half-reactions The more positive the Ɛ red or Ɛ ox, the more spontaneous the half-reaction

8. The potential of an overall redox reaction in a Galvanic cell can be measured with a voltmeter Unfortunately, the potential of a half-reaction cannot be measured, so we make one up! This standard hydrogen half-reaction is assigned a potential of 0.00 V 3D-8 (of 14) H 2 (g, 1 atm)→ 2H + (aq, 1 M) + 2e - All other standard reduction potentials are measured relative to this one

9. Fe (s) → Fe 2+ (aq) + 2e - 3e - + Cr 3+ (aq) → Cr (s) USES FOR STANDARD REDUCTION POTENTIALS 1)Predicting the spontaneity of a reaction Determine if the following standard state reaction is spontaneous 3Fe (s) + 2Cr 3+ (aq) → 3Fe 2+ (aq) + 2Cr (s) Find the 2 reduction potentials that can be used to make the reaction 2e - + Fe 2+ (aq) → Fe (s) 3e - + Cr 3+ (aq) → Cr (s) Ɛ º red =-0.44 V Ɛ º red =-0.73 V Add a reduction and oxidation half-reaction to make the desired reaction 3D-9 (of 14) Ɛ º ox =0.44 V Ɛ º red =-0.73 V Ɛ º = -0.29 V Ɛ º is negative,  not spontaneous

10. USES FOR STANDARD REDUCTION POTENTIALS 2)Predicting strong oxidizing and reducing agents e - + Ag + (aq) → Ag (s) 2e - + Cu 2+ (aq) → Cu (s) 2e - + Ni 2+ (aq) → Ni (s) 3e - + Al 3+ (aq) → Al (s) Reduction Half-Reactions Stand. Reduction Potentials Ɛ º red =0.80 V Ɛ º red =0.34 V Ɛ º red =-0.23 V Ɛ º red =-1.71 V A large, positive reduction potential means the forward reaction is spontaneous (the REACTANT has a strong tendency to be REDUCED) Best oxidizing agent from the list?Ag + (aq) Good oxidizing agents?Halogens (X 2 )→ X - O2O2 → H 2 O 3D-10 (of 14)

11. USES FOR STANDARD REDUCTION POTENTIALS 2)Predicting strong oxidizing and reducing agents e - + Ag + (aq) → Ag (s) 2e - + Cu 2+ (aq) → Cu (s) 2e - + Ni 2+ (aq) → Ni (s) 3e - + Al 3+ (aq) → Al (s) Reduction Half-Reactions Stand. Reduction Potentials Ɛ º red =0.80 V Ɛ º red =0.34 V Ɛ º red =-0.23 V Ɛ º red =-1.71 V A large, negative reduction potential means the reverse reaction is spontaneous (the PRODUCT has a strong tendency to be OXIDIZED) 3D-11 (of 14) Best reducing agent from the list?Al (s) Good reducing agents? (Alkali Metals) M→ M + C→ CO 2

12. USES FOR STANDARD REDUCTION POTENTIALS 3)Predicting the potential and spontaneous reaction in a Galvanic cell 3D-12 (of 14)

13. For a Galvanic cell with silver and nickel electrodes in 1 M solutions of Ag + and Ni 2+ respectively, determine the (a) standard cell potential, (b) spontaneous reaction, and (c) anode and cathode (a)Find the 2 reduction potentials to produce the Galvanic cell e - + Ag + (aq) → Ag (s) 2e - + Ni 2+ (aq)→ Ni (s) Ɛ º red =0.80 V Ɛ º red =-0.23 V The largest positive potential is the spontaneous process, and will be the reduction the other must be reversed, and will be the oxidation e - + Ag + (aq) → Ag (s) Ni (s) → Ni 2+ (aq) + 2e - Ɛ º red =0.80 V Ɛ º ox =0.23 V Add the reduction and oxidation potentials to get the cell potential 0.80 V + 0.23 V = 1.03 V 3D-13 (of 14)

14. For a Galvanic cell with silver and nickel electrodes in 1 M solutions of Ag + and Ni 2+ respectively, determine the (a) standard cell potential, (b) spontaneous reaction, and (c) anode and cathode (b)Equalize e - s and add the reduction and oxidation half-reactions together e - + Ag + (aq) → Ag (s) Ni (s) → Ni 2+ (aq) + 2e - ( ) x 2 2e - + 2Ag + (aq)→ 2Ag (s) Ni (s) → Ni 2+ (aq) + 2e - 2Ag + (aq) + Ni (s) → 2Ag (s) + Ni 2+ (aq) 3D-14 (of 14) Nickel half-reaction is the oxidation, ∴ Ni is the anode (c) Silver half-reaction is the reduction, ∴ Ag is the cathode

16. For nonstandard cells ΔG = ΔGº + RT ln Q -n F Ɛ = -n F Ɛ º + RT ln Q Ɛ = Ɛ º – RT ln Q ____ n F THE NERNST EQUATION 3E-1 (of 11) NONSTANDARD STATE CELLS

17. Calculate the potential for the following cell at 25ºC Zn (s) | Zn 2+ (0.200 M) || Ag + (0.100 M) | Ag (s) e - + Ag + (aq) → Ag (s) 2e - + Zn 2+ (aq) → Zn (s) Ɛ º red =0.80 V Ɛ º red =-0.76 V 0.80 V + 0.76 V = 1.56 V e - + Ag + (aq) → Ag (s) Zn (s) → Zn 2+ (aq) + 2e - Ɛ º red =0.80 V Ɛ º ox =0.76 V 3E-2 (of 11)

18. e - + Ag + (aq) → Ag (s) Zn (s) → Zn 2+ (aq) + 2e - ( ) x 2 Calculate the potential for the following cell at 25ºC Zn (s) | Zn 2+ (0.200 M) || Ag + (0.100 M) | Ag (s) 2Ag + (aq) + Zn (s) → 2Ag (s) + Zn 2+ (aq) Ɛ = Ɛ º – RT ln Q ____ n F = 1.56 V – (8.314 CV/K)(298.2 K) ln 0.200 ____________________________ ________ (2 mol)(96,485 C/mol) 0.100 2 = 1.56 V – 0.0385 V= 1.52 V 3E-3 (of 11) Q = [Zn 2+ ] _________ [Ag + ] 2

19. For a nickel-cadmium cell with solutions of 0.00100 M nickel (II) sulfate and 0.10 M cadmium sulfate, determine the (a) standard cell potential, (b) spontaneous reaction, (c) anode and cathode (d) cell potential at 25ºC 2e - + Ni 2+ (aq) → Ni (s) 2e - + Cd 2+ (aq) → Cd (s) Ɛ º red =-0.23 V Ɛ º red =-0.40 V -0.23 V + 0.40 V = 0.17 V 2e - + Ni 2+ (aq) → Ni (s) Cd (s) → Cd 2+ (aq) + 2e - Ɛ º red =-0.23 V Ɛ º ox = 0.40 V 3E-4 (of 11)

20. For a nickel-cadmium cell with solutions of 0.00100 M nickel (II) sulfate and 0.10 M cadmium sulfate, determine the (a) standard cell potential, (b) spontaneous reaction, (c) anode and cathode (d) cell potential at 25ºC 2e - + Ni 2+ (aq)→ Ni(s) Cd (s) → Cd 2+ (aq) + 2e - Ni 2+ (aq) + Cd (s) → Ni (s) + Cd 2+ (aq) Ni – cathode Cd – anode Ɛ = Ɛ º – RT ln Q ____ n F = 0.17 V – (8.314 CV/K)(298.2 K) ln 0.10 ____________________________ __________ (2 mol)(96,485 C/mol) 0.00100 = 0.17 V – 0.059 V = 0.11 V 3E-5 (of 11) Q = [Cd 2+ ] _________ [Ni 2+ ]

21. 3E-6 (of 11) For a nickel-cadmium cell with solutions of 0.00100 M nickel (II) sulfate and 0.10 M cadmium sulfate, determine the (e) the cell potential at 25ºC once 90.0% of the nickel (II) ions have reacted away Ni 2+ (aq) + Cd (s) → Ni (s) + Cd 2+ (aq) 0.10090 Initial M’s Change in M’s Final M’s 0.001000.10 - 0.00090+ 0.00090 0.00010 Ɛ = Ɛ º – RT ln Q ____ n F = 0.17 V – (8.314 CV/K)(298.2 K) ln 0.10090 ____________________________ __________ (2 mol)(96,485 C/mol) 0.00010 = 0.17 V – 0.089 V = 0.08 V Q = [Cd 2+ ] _________ [Ni 2+ ]

22. Ɛ = Ɛ º – RT ln Q ____ n F For a reaction at equilibrium ΔG = 0 ∴ Ɛ = 0 0 = Ɛ º – RT ln K eq ____ n F RT ln K eq = Ɛ º ____ n F K eq = e Ɛ ºn F/ RT 3E-7 (of 11)

23. Find the equilibrium constant at 25ºC for Fe (s) + I 2 (s) → Fe 2+ (aq) + 2I - (aq) 2e - + I 2 (s) → 2I - (aq) 2e - + Fe 2+ (aq) → Fe (s) Ɛ º red =0.54 V Ɛ º red =-0.44 V 0.54 V + 0.44 V = 0.98 V 2e - + I 2 (s) → 2I - (aq) Fe (s) → Fe 2+ (aq) + 2e - Ɛ º red =0.54 V Ɛ º ox = 0.44 V 3E-8 (of 11) I 2 (s) + Fe (s) → 2I - (aq) + Fe 2+ (aq)

24. Find the equilibrium constant at 25ºC for Fe (s) + I 2 (s) → Fe 2+ (aq) + 2I - (aq) K eq = e Ɛ ºn F/ RT = e [(0.98 V)(2 mol)(96,485 C/mol)] / [(8.314 CV/K)(298.2 K)] = 1.3 x 10 33 3E-9 (of 11)

25. Find the solubility product constant for mercury (I) chloride at 25ºC Hg 2 Cl 2 (s) ⇄ Hg 2 2+ (aq) + 2Cl - (aq) 2e - + Hg 2 Cl 2 (s) → 2Hg (l) + 2Cl - (aq) 2e - + Hg 2 2+ (aq) → 2Hg (l) Ɛ º red =0.33 V Ɛ º red =0.80 V 2e - + Hg 2 Cl 2 (s) → 2Hg (l) + 2Cl - (aq) 2Hg (l) → Hg 2 2+ (aq) + 2e - Ɛ º red = 0.33 V Ɛ º ox = -0.80 V Hg 2 Cl 2 (s) ⇄ Hg 2 2+ (aq) + 2Cl - (aq) 0.33 V – 0.80 V = -0.47 V 3E-10 (of 11)

26. K sp = e Ɛ ºn F/ RT = e [(-0.47 V)(2 mol)(96,485 C/mol)] / [(8.314 CV/K)(298.2 K)] = 1.3 x 10 -16 Find the solubility product constant for mercury (I) chloride at 25ºC for Hg 2 Cl 2 (s) ⇄ Hg 2 2+ (aq) + 2Cl - (aq) 3E-11 (of 11)

28. BATTERIES BATTERY – One or more electrochemical cells that produce electricity from a chemical reaction 3F-1 (of 16)

29. Dry Cell Graphite rod (cathode) Paste of MnO 2, NH 4 Cl, and H 2 O Zinc casing (anode) Anode: Cathode: Zn (s) → Zn 2+ (aq) + 2e - 2e - + 2MnO 2 (s) + 8NH 4 + (aq) → 2Mn 3+ (aq) + 4H 2 O (l) + 8NH 3 (aq) Potential or Voltage: 1.5 VCurrent or Amperage: Depends on battery size 3F-2 (of 16)

30. Dry Cell The acidic content tends to corrode the Zn Fast usage :NH 3 insulates the cathode, reducing the voltage With rest : Zn 2+ migrates to center, forming Zn(NH 3 ) 4 2+ to bind the NH 3 Graphite rod (cathode) Paste of MnO 2, NH 4 Cl, and H 2 O Zinc casing (anode) 3F-3 (of 16)

31. Alkaline Battery Graphite rod (cathode) Paste of MnO 2, KOH, and H 2 O Powdered zinc (anode) Anode: Cathode: Zn (s) + OH - (aq) → ZnO (s) + H 2 O (l) + 2e - 2e - + 2MnO 2 (s) + H 2 O (l) → 2Mn 2 O 3 (s) + 2OH - (aq) Zn resists corrosion in a basic solution 3F-4 (of 16)

32. Lithium-Ion Battery Lithium cobalt oxide (anode) Graphite (cathode) Anode: Cathode: LiCoO 2 (s) → CoO 2 (s) + Li + (org) + e - e - + Li + (org) + 6C (s) → LiC 6 (s) Because both products stick to the electrodes, by applying an external source of electricity the reverse reaction will occur, reforming the reactants This is called RECHARGING Separator 3F-5 (of 16)

33. Lead Storage Battery Lead (anode) Lead + lead (IV) oxide (cathode) 4 M sulfuric acid Anode: Cathode: Pb (s) + SO 4 2- (aq) → PbSO 4 (s) + 2e - 2e - + 2PbO 2 (s) + 4H + (aq) + SO 4 2- (aq)→ PbSO 4 (s) + 2H 2 O (l) 3F-6 (of 16)

34. Lead Storage Battery Potential or Voltage: 2.1 V x 6 cells = 12.6 V _______ cell Because products stick to the electrodes, this battery is rechargeable Lead (anode) Lead + lead (IV) oxide (cathode) 4 M sulfuric acid 3F-7 (of 16)

35. Hydrogen Fuel Cell Platinum Catalyst Polymer Electrolyte Membrane Anode: Pt catalyst splits hydrogen atoms into hydrogen ions and electrons Electrolyte: PEM allows hydrogen ions to pass through to the cathode Cathode: Oxygen and electrons combine with hydrogen ions to make water 3F-8 (of 16)

36. ELECTROLYTIC CELL – An electrochemical cell that uses electricity to produce a chemical reaction H2OH2O Anode: Oxidation H 2 O (l)→ O 2 (g) +1 -2 Cathode: Reduction H 2 O (l) → H 2 (g) 2+ 4e - + 4H + (aq) 2e - + + OH - (aq) 3F-9 (of 16) 22

37. Electricity throughAnodeCathode KF (l) NaCl (l) NaCl (aq) KF (aq) CuBr 2 (aq) HCl (aq) HNO 3 (aq) H 2 SO 4 (aq) Na 2 SO 4 (aq) AgNO 3 (aq) F 2 (g) Cl 2 (g) F 2 (g) Br 2 (l) Cl 2 (g) O 2 (g) K (s) Na (s) H 2 (g) Cu (s) H 2 (g) Ag (s) H 2 (g) N in HNO 3 cannot be oxidized 3F-10 (of 16), so O in H 2 O will be oxidized Na + (aq) + e - → Na (s) 2H 2 O (l) + 2e - → H 2 (g) + 2OH - (aq) Ɛ º red =-2.71 V Ɛ º red =-0.83 V

38. Ag Ag + Electrolytic cells are used for (1) producing elements (Na, Cl 2, etc.) (2) purification of metals from ore (3) electroplating metals (Au, Ag, Pt, etc.) Anode: Oxidation Ag (s) → Ag + (aq) Cathode: Reduction Ag + (aq) → Ag (s) + e - e - + 3F-11 (of 16) Ag +

39. FARADAY’S LAWS OF ELECTROLYSIS (1)Passing the same quantity of electricity through a cell always leads to the same amount of chemical change (2)It takes 96,485 C of electricity to deposit or liberate 1 mole of a substance that gains or loses 1 mole of e - s during the cell reaction 96,485 C= 1 mole e - = 1 Faraday 3F-12 (of 16)

40. Calculate the mass of copper deposited by a current of 7.89 amperes flowing for 1.20 x 10 3 seconds, if the cathode reaction is Cu 2+ (aq) + 2e - → Cu (s) 7.89 Ax 1 C ______ 1 As x 1.20 x 10 3 s x 1 F ___________ 96,485 C x 1 mol e - __________ 1 F x 1 mol Cu ___________ 2 mol e - x 63.55 g Cu ______________ 1 mol Cu = 3.12 g Cu 3F-13 (of 16)

41. Calculate the mass of aluminum deposited by a current of 5.00 amperes flowing for 10.0 minutes through an aluminum nitrate solution. Al 3+ (aq) + 3e - → Al (s) 5.00 Ax 1 C ______ 1 As x 10.0 min x 1 F ___________ 96,485 C x 1 mol e - __________ 1 F x 1 mol Al ___________ 3 mol e - x 26.98 g Al _____________ 1 mol Al = 0.280 g Al x 60 s ________ 1 min 3F-14 (of 16)

42. Calculate the current needed to plate 0.150 grams of zinc onto an electrode in 60.0 seconds from a zinc acetate solution. Zn 2+ (aq) + 2e - → Zn (s) 0.150 g Znx 2 mol e - ___________ 1 mol Zn x 1 F __________ 1 mol e - x 96,485 C ___________ 1 F x 1 As ______ 1 C x 1 ________ 60.0 s = 7.38 A x mol Zn _____________ 65.38 g Zn 3F-15 (of 16)

43. Calculate the time, in minutes, needed to deposit 0.400 grams of chromium from a chromium (III) nitrate solution with a current of 10.0 amperes. Cr 3+ (aq) + 3e - → Cr (s) 0.400 g Crx 3 mol e - ___________ 1 mol Cr x 1 F __________ 1 mol e - x 96,485 C ___________ 1 F x 1 As ______ 1 C x 1 ________ 10.0 A = 3.71 min x mol Cr _____________ 52.00 g Cr x 1 min _______ 60 s 3F-16 (of 16)