1. + Electrochemistry The study of the interchange of chemical and electrical energy. AP Chemistry Ch. 17

2. + Oxidation-Reduction Reactions (REDOX) Oxidation: An increase in oxidation state (a loss of electrons). Reduction: A decrease in oxidation state (a gain of electrons). Oxidizing Agent: Electron acceptor A reactant that accepts electrons from another. Reducing Agent: Electron donor A reactant that donates electrons to another substance to reduce the oxidation state of one of its atoms.

3. + Assigning Oxidation States Rules for Assigning Oxidation States The Oxidation State of… An atom in an element is zero Na(s), O 2 (g), Hg(l) A monatomic ion is the same as its charge Na +, Cl - Fluorine is -1 in its compounds HF, PF 3 Oxygen is usually -2 in its compounds H 2 O, CO 2 Exception: Peroxides (containing O 2 2- ) in which oxygen is -1 Hydrogen is +1 in its covalent compounds H 2 O, HCl, NH 3

4. + Recall… Zinc reacts with HCl: Zn(s) + H + (aq) + Cl - (aq)  Zn 2+ (aq) + Cl - (aq) + H 2 (g) Half-Reactions

5. + Balancing a Redox Reaction that occurs in Acidic Solution Steps: Acidic 1. Write separate equations for the oxidation and reduction half- reactions. 2. For each half-reaction Balance all the elements except hydrogen and oxygen. Balance oxygen using H 2 O. Balance hydrogen using H +. Balance the charge using electrons (e - ). 3. If necessary, multiply one or both balanced half-reactions by an integer equalize the number of electrons transferred in the two half- reactions. 4. Add the half-reactions, and cancel identical species. 5. Check that the elements and charges are balanced.

6. + Practice Complete and balance the following equations that occur in acidic solutions. a.) MnO 4 - (aq) + C 2 O 4 2- (aq)  Mn 2+ (aq) + CO 2 (g) b.)Mn 2+ (aq) + NaBiO 3 (s)  Bi 3+ (aq) + MnO 4 - (aq)

7. + Balancing a Redox Reaction that occurs in Basic Solution Steps: Basic The same as acidic and then… 6. Add a number of OH - ions to both sides so that you just balance excess H + ions. 7. H + and OH - will form H 2 O on the side with excess H +. Free OH - will appear on one side of the equation. 8. Double check atoms and charges!

8. + Practice Complete and balance the following equations that occur in basic solutions. a.) NO 2 - (aq) + Al(s)  NH 3 (aq) + AlO 2 - (aq) b.)Cr(OH) 3 (s) + ClO - (s)  CrO 4 2- (aq) + Cl 2 (g)

9. + Voltaic (or galvanic) cells Voltaic (galvanic) cell: A device in which chemical energy from a spontaneous redox reaction is changed to electrical energy that can be used to do work. Uses a spontaneous redox reaction to produce a current that can be used to do work. Electrodes a conductor through which electricity enters or leaves an object, substance, or region. Zn(s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu(s) Cathode (RED CAT) Anode (AN OX)

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11. + Salt Bridge (or porous barrier): a U-tube containing an electrolyte that connects the two compartments of a galvanic cell, allowing ion flow without extensive mixing of the different solutions

12. + Analyze the Animation Additional Video (Demo) Additional Video

13. + Cell EMF Under Standard Conditions (E°) Electromotive Force (EMF) or Cell Potential (E cell ): The driving force in a galvanic cell that pulls electrons from the reducing agent in one compartment to the oxidizing agent in the other. Voltage Unit of electrical potential (V) 1 joule of work per coulomb of charge transferred

14. + Standard Reduction Potentials (E° red ) Cell potential: E° cell = E° red (cathode) - E° red (anode) For all spontaneous reactions at standard conditions, E° cell > 0. 2 H + (aq, 1 M) + 2 e -  H 2 (g, 1 atm) E° red = 0.00 V (standard hydrogen electrode)

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16. + Calculating Cell EMF Determine the E° cell for the reaction Zn(s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu(s) Electrons lost must equal electrons gained! Determine the E° cell for the reaction Cu(s) + Fe 3+ (aq)  Cu 2+ (aq) + Fe 2+ (aq)

17. + Example Consider a galvanic cell based on the following reactions. Give the balanced cell reaction and calculate the standard emf (E°) for the cell. a.Al 3+ (aq) + Mg(s)  Al(s) + Mg 2+ (aq) b.MnO 4 - (aq) + H + (aq) + ClO 3 - (aq)  ClO 4 - (aq) + Mn 2+ (aq) + H 2 O(l)

18. + Complete Description of a Galvanic Cell Consider a galvanic cell based on the following half- reactions: Fe 2+ + 2 e -  Fe E° = -0.44 V MnO 4 - + 5 e - + 8 H +  Mn 2+ + 4 H 2 OE° = 1.51 V In a working galvanic cell, one of these reactions must run in reverse. Which one? What is the balanced cell reaction? What is the E° cell ?

19. + Example Continued… On the diagram below, draw what ions must be present in each solution and what metal is each electrode made out of. A chemically inert conductor (like platinum) is required if none of the substances participating in the half-reaction is a conducting solid.

20. + Example Problem Describe completely (cell reaction, cell potential, and physical setup of cell) the galvanic cell based on the following half-reactions under standard conditions. Ag + + e -  AgE° = 0.80 V Fe 3+ + e -  Fe 2+ E° = 0.77 V

21. + Strengths of Oxidizing and Reducing Agents The more positive the E° red value for a half-reaction, the greater the tendency for the reactant of the half-reaction to be reduced and, therefore, to oxidize another species. Rank the following species from the strongest to the weakest reducing agent: I - (aq), Fe(s), Al(s).

22. + Free Energy and Redox Reactions E° = E° red (reduction process) - E° red (oxidation process) What is the relationship between a spontaneous process and the E° cell ?

23. + Example Problem Using standard reduction potentials, determine whether the following reaction is spontaneous under standard conditions. Cu(s) + 2 H + (aq)  Cu 2+ (aq) + H 2 (g)

24. + Cell Potential, Electrical Work, and Free Energy The work that can be accomplished when electrons are transferred through a wire depends on the “push” (the thermodynamic driving force) behind the electrons. This driving force (the emf) is defined in terms of a potential difference (in volts) between two points in the circuit. Emf = potential difference (V) = work (J) / charge (C) E° = -W/q (work is viewed from the point of view of the system…thus work flows out of the system and work has a – sign.) W max = -q E° max q = nF q is charge ; n is number of electrons; F is Faraday’s constant

25. + Cell Potential, Electrical Work, and Free Energy Faraday’s Constant (F): quantity of electrical charge in 1 mole of electrons. 1 F = 96,485 C/mol e - = 96,485 J/V∙mol EMF and Δ G W max = Δ G = -nFE°

26. + Example Problem Using standard reduction potentials, calculate Δ G for the reaction Cu 2+ (aq) + Fe(s)  Cu(s) + Fe 2+ (aq) Is this reaction spontaneous?

27. + Example Problem Using standard reduction potentials, calculate Δ G and the equilibrium constant, K, at 25°C for the reaction 4 Ag(s) + O 2 (g) + 4 H + (aq)  4 Ag + (aq) + 2 H 2 O(l)

28. + Cell EMF Under Nonstandard Conditions Standard Conditions – 1.0 M solutions Nonstandard Conditions: Cu + 2 Ce 4+  Cu 2+ + 2 Ce 3+ E°cell = 1.36 V What would happen to E cell if we increase [Ce 4+ ]? Concentration Cells Ag + + e -  AgE°cell = 0.80 V

29. + The Nernst Equation **Excluded in AP Curriculum** Qualitative Understanding ONLY! E cell = E° cell – (RT/nF)lnQ = E° cell – (2.303RT/nF)logQ at 25°C E cell = E° cell – (0.0592/n)logQ

30. + Example Problem Describe the cell based on the following half-reactions: VO 2 + + 2 H + + e -  VO 2+ + H 2 OE° = 1.00 V Zn 2+ + 2 e -  ZnE° = -0.76 V where: T = 25°C [VO 2 + ] = 2.0 M [H + ] = 0.50 M [VO 2+ ] = 0.010 M [Zn 2+ ] = 0.10 M

31. + Example Problem Calculate the emf at 298K for the cell when [Cr 2 O 7 2- ] = 2.0 M, [H + ] = 1.0 M, [I - ] = 1.0 M, and [Cr 3+ ] = 1.0 x 10 -5 M. Cr 2 O 7 2- (aq) + 14 H + (aq) + 6 I - (aq)  2 Cr 3+ (aq) + 3 I 2 (s) + 7 H 2 O(l)

32. + Concentration Cell: A galvanic cell in which both compartments contain the same components, but at different concentrations. Although the standard emf for this cell is zero, the cell operates under nonstandard conditions because the concentration of Ni 2+ (aq) is different in the two compartments. Ni(s)  Ni 2+ (aq) + 2 e - E° red = -28.0 Vanode Ni 2+ (aq) + 2 e -  Ni(s)E° red = -28.0 Vcathode Calculate E cell

33. + Batteries and Fuel Cells Lead-Acid Battery (12-V car battery) Alkaline Battery Nickel-Cadmium Batteries

34. + Corrosion Corrosion of Iron Preventing the Corrosion of Iron

35. + Electrolysis Electrolysis: A process that involves forcing a current through a cell to cause a nonspontaneous chemical reaction to occur. Electrolytic Cell: A cell that uses electrical energy to produce a chemical change that would otherwise not occur spontaneously. Stoichiometry of electrolytic processes Coulombs = Amperes (C/s) x seconds q = I x t 1 mole of electrons carries a charge of 1 faraday (96,485 C)

36. + Example Problems Calculate the number of grams of aluminum produced in 1.00 hour by the electrolysis of molten AlCl 3 if the electrical current is 1.00 A. How long must a current of 5.00 A be applied to a solution of Ag + to produce 10.5 g silver metal?

37. + Example Problem A solution in an electrolytic cell contains the ions Cu 2+, Ag +, and Zn 2+. If the voltage is initially low and gradually turned up, in which order will the metals be plated out onto the cathode? The standard reduction potentials: Ag + + e -  AgE° = 0.80 V Cu 2+ + 2 e -  CuE° = 0.34 V Zn 2+ + 2 e -  ZnE° = -0.76 V

38. + Example Problem Calculate the number of kilowatt-hours of electricity required to produce 1.0 x 10 3 kg of aluminum by electrolysis of Al 3+ if the applied voltage is 4.50 V. 1 W = 1 J/s 1 kWh = 3.6 x 10 6 J

39. + Example Problem An unknown metal M is electrolyzed. It took 74.1 s for a current of 2.00amp to plate out 0.107 g of the metal from the solution containing M(NO 3 ) 3. Identify the metal.

40. + Additional Practice Textbook problems: Beginning on page 830 # 13, 15, 16, 25, 27, 29, 31, 33, 43, 49, 53 (conceptually only), 65, 73, 81, and 87 Unit 8 Review Packet

41. + Final Exam Analysis 25 minutes individual review…..for half credit (based on reflection)! Learning is a PROCESS! For credit you must correct your answer and provide a brief reflection of why your answer choice was incorrect and how you corrected it. (3 to 4 sentences per question) Question Number (MC or FRQ) Incorrect Answer Corrected Answer REFLECTION of LEARNING

42. + RAFT – Final Project R – Role(Recruiter/Motivator) A- Audience(Future AP Chemistry Students) F – Format (Your CHOICE!... Get CREATIVE) T –Topic (Your CHOICE!... Any of the 8 Units) See Rubric Due May 21 st for Presenting!