1. Chapter 20 Problem Set: p. 890-898 3, 11, 13, 17, 23, 27, 31, 35, 43, 45, 47, 53, 61, 63, 69, 72, 75, 79, 85, 87, 95

3. The energy released in a spontaneous redox reaction that is used to perform electrical work is harnessed in Voltaic Cells Also called Galvanic Cells Electrons transfer through an external pathway rather than directly between reactants Electrodes: metals used in the circuit

4. http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/galvan5.swf

6. Anode: electrode at which oxidation occurs Cathode: electrode at which reduction occurs Salt Bridge: a connection between two half-cells that allows for the flow of ions Composed of a strong electrolyte Must be a solution of a compound that will not produce a precipitate with any other ion in solution Salt anions migrate toward the anode; cations migrate toward the cathode **Electrons flow from the anode to the cathode**

7. Keep your vowels together and your consonants together Anode = oxidation Cathode = reduction An ox and Red cats Anode is oxidation, Cathode is reduction Electrons flow from A to C (in alphabetical order)

8. Used to determine which metal will be the cathode/anode for the reaction to be spontaneous Cell EMF: Electromotive Force Caused by a difference in potential energy between the different electrodes. Allows electrons to be pushed Denoted E cell, measured in volts, V Also called cell potential

9. Cell potential is positive for spontaneous reactions negative for nonspontaneous reactions Depends on 1.Reactions Occurring (Metals Used) 2.Concentrations of Solutions (Molarity) 3.Temperature (normally 25 o C) E o cell cell potential at standard conditions

10. The more positive the E o 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 The half reaction with the smallest reduction potential is most easily reversed as an oxidation The E o red table acts as an activity series for which substances act as oxidizers and reducers

11. E o cell based on standard reduction potentials Potential associated with each electrode is the ability for reduction to occur at the electrode. E o red = standard reduction potential for metal (based on Standard Hydrogen Electrode, SHE – use Appendix E p. 1128!) E o cell = E o red (cathode) – E o red (anode) If the stoichiometry of the reaction changes DO NOT multiply the value by the the cell potential for that metal The more positive the E o red for a metal, the greater the ability for reduction Thus the metal with a larger E o red will be the better cathode in a voltaic cell

12. Determine voltage produced in spontaneous reaction.

13. E o red (Cd 2+ /Cd ) = -0.403 V E o red (Sn 2+ /Sn ) = -0.136 V Cathode: Sn 2+ + 2e -  Sn Anode: Cd  Cd 2+ + 2e - E o cell = E o cathode – E o anode = -0.136 V – (-0.403 V) = 0.267 V

15. Na  Na + + 1e - + 2.71 V PbO 2  Pb 2+ + 1.50 V (Na  Na + + 1e - )x2 *DO NOT MULTIPLY E o red by 2* 4H + + 2e - + PbO 2  Pb 2+ + 2H 2 O 4H + + PbO 2 + 2Na  2Na + + Pb 2+ + 2H 2 O Overall potential for the battery = 2.71 + 1.50 = 4.3V

16. Reduction Potentials can be used to calculate energies of reactions when using redox processes Remember: E = +Spontaneous E = - Nonspontaneous E can be converted into G for Gibbs Free Energy or other thermochemical quantities ∆G o = -n F E o n= number of electrons transferred F = Faraday’s constant = 96500 C/mol or J/Kmol E o = Calculated from E o red values

18. Start with Half-Reactions Fe  Fe 2+ + 2e - Cu  Cu 2+ + 2e - Flip the copper reaction so it matches the overall equation Fe  Fe 2+ + 2e - 0.44 V Cu 2+ + 2e -  Cu0.34 V Cu 2+ (aq) + Fe(s)  Cu(s) + Fe 2+ (aq)0.78V Solve for ∆G o = -n F E o ∆G o = -(2)(96500)(0.78) = -150517 J/mol = -151 kJ/mol What is the K eq for this reaction? ∆G o = -RTlnK eq

20. As voltage is produced, concentrations vary. Reactants are consumed and products are produced. When E = 0, the cell is dead, the concentrations cease to change due to equilibrium being reached. For different concentrations, the Nernst equation is used:

21. To calculate voltage, use the Nernst Equation Cell EMF and Equilibrium ∆G = 0, ∆E = 0at Equilibrium Using Nernst Will be zero Solving for the equilibrium constant

23. Start with the half reactions 2(2H + + 1e - + VO 2 1+  VO 2+ + H 2 O ) 1.00 V Zn  Zn 2+ +2e - 0.76 V 4H + + 2VO 2 1+ + Zn  Zn 2+ + 2VO 2+ + 2H 2 O 1.76 V Write the equilibrium expression 4 x 10 -5

24. =1.76 – (-0.13) E = 1.89 V The voltage went up! Use LeChatelier’s Principle to explain why

25. What will the voltmeter read for the following electrochemical cell? Anode Reaction – Oxidation - dilute solution becomes more concentrated. Ag(s) → Ag + (0.1 M) + e- +0.8 V Cathode Reaction – Reduction- concentrated solution becomes more dilute. Ag + (1.0 M) + e- → Ag (s) -0.8 V Net Cell Reaction Ag+ (1.0 M) → Ag+ (0.1 M) 0.0V VV

26. You DON’T need your calculator for this one!!! E = 0.0592 V

27. Self-contained electrochemical power source consisting of 1 or more galvanic cells. Voltages are additive when batteries are connected due to a continuation of the flow of electrons A voltaic cell that converts chemical energy into electrical energy Not self-contained Produces current by combustion and electron- excitation

28. A spontaneous redox reaction in which a metal is attacked by some substance in the environment and converted to an unwanted compound Is often prevented, or reversed, by use of a sacrificial anode (a more active metal) The use of electricity to cause nonspontaneous reactions to occur by driving the reaction in the opposite direction (basis of rechargeable batteries Take place in electrolytic cells Cathode is connected to – terminal to accept e -, anode is attached to + terminal to donate e - (opposite of voltage)

30. Ampere = units to measure current in Coulombs/second Measures the rate of electron flow Faraday’s Constant, F = 96485 Coulomb/mol

31. Ampere = units to measure current in Coulombs/second Measures the rate of electron flow Faraday’s Constant, F = 96485 Coulomb/mol Use a t-table! Time = 158823 sec = 2647 min = 44 hours = 1.84 days

33. 3 e + Cr 3+  Cr Use a t-table! Mass of Cr = 2.36 x 10 2 g

34. Thin coating is plated onto an existing metal to help protect it Metal to be plated is attached to the cathode. Metal coating is produced by the metal attached to anode and a “like” solution