Unit 15 Electrochemistry

1. Unit 15Electrochemistry 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

2. Oxidation Numbers Please assign oxidation numbers to the elements in each substance. HNO3 HNO2 PbSO4 (NH4)2Ce(SO4)3 Al(C2H3O2)3

3. Electrochemistry: Relationship between electricity and redox reactionsElectricity = movement of electrons • 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 Voltaic Cells

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

5. Using a Porous Disk as a Salt Bridge

6. Components of Electrochemical Cells • 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. Pneumonic Devices • 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. Standard Cell Potentials • 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 Ecell , measured in volts, V • Also called cell potential

9. Cell Potential • Cell potential is • positive for spontaneous reactions • negative for nonspontaneous reactions • Depends on • Reactions Occurring (Metals Used) • Concentrations of Solutions (Molarity) • Temperature (normally 25oC) • Eocell cell potential at standard conditions

10. EMF can then be used to determine what will happen in a chemical equation • The more positive the Eored 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 Eored table acts as an activity series for which substances act as oxidizers and reducers

11. Calculating Cell Potential • Eocell based on standard reduction potentials • Potential associated with each electrode is the ability for reduction to occur at the electrode. • Eored = standard reduction potential for metal (based on Standard Hydrogen Electrode, SHE – use Appendix E p. 1128!) • Eocell = Eored (cathode) – Eored (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 Eored for a metal, the greater the ability for reduction • Thus the metal with a larger Eored will be the better cathode in a voltaic cell

12. Example: A voltaic cell is based on the following half cells:Cd2+ + 2e- CdSn2+ + 2e- Sn • Determine voltage produced in spontaneous reaction.

13. Example: A voltaic cell is based on the following half cells:Cd2+ + 2e- CdSn2+ + 2e- Sn • Determine voltage produced in spontaneous reaction. • Eored (Cd2+ /Cd ) = -0.403 V • Eored (Sn2+ /Sn ) = -0.136 V • Cathode: Sn2+ + 2e- Sn • Anode: Cd Cd2+ + 2e- • Eocell = Eocathode – Eoanode = -0.136 V – (-0.403 V) = 0.267 V

14. Example: What would be the electrical potential for the reaction:PbO2 + Na  Pb2+ + Na+

15. Example: What would be the electrical potential for the reaction:PbO2 + Na  Pb2+ + Na+ • Na  Na+ + 1e- + 2.71 V • PbO2  Pb2+ + 1.50 V • (Na  Na+ + 1e-)x2 *DO NOT MULTIPLY Eored by 2* • 4H+ + 2e- + PbO2  Pb2+ + 2H2O • 4H+ + PbO2 + 2Na 2Na+ + Pb2+ + 2H2O Overall potential for the battery = 2.71 + 1.50 = 4.3V

16. Spontaneous Redox Reactions • 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 ∆Go = -nFEo n= number of electrons transferred F = Faraday’s constant = 96500 C/mol or J/Kmol Eo = Calculated from Eored values

17. Calculate ∆G for the following reaction: Cu2+(aq) + Fe(s)  Cu(s) + Fe2+(aq)

18. Calculate ∆G for the following reaction: Cu2+(aq) + Fe(s)  Cu(s) + Fe2+(aq) • Start with Half-Reactions • Fe  Fe2+ + 2e- • Cu  Cu2+ + 2e- • Flip the copper reaction so it matches the overall equation • Fe  Fe2+ + 2e- 0.44 V • Cu2+ + 2e-  Cu 0.34 V • Cu2+(aq) + Fe(s)  Cu(s) + Fe2+(aq) 0.78V Solve for ∆Go = -nFEo ∆Go = -(2)(96500)(0.78) = -150517 J/mol = -151 kJ/mol What is the Keq for this reaction? ∆Go = -RTlnKeq

19. Warm-Up : Do another page on the net ionic equations! The whole packet is due TODAY!The 2008 FRQ was your Homework. Please be ready to grade it and the crossword page.

20. Effect of Concentration on Cell EMF • 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 = 0 at Equilibrium • Using Nernst Will be zero Solving for the equilibrium constant

22. Example: What is the electrical potential for the cell with the following concentrations?VO21+ + Zn  Zn2+ + VO2+[VO21+] = 2.0 M [H+] = 0.50 M[VO2+] = 1.0 x 10 M [Zn2+] = 0.10 M

23. Example: What is the electrical potential for the cell with the following concentrations?VO21+ + Zn  Zn2+ + VO2+[VO21+] = 2.0 M [H+] = 0.50 M[VO2+] = 1.0 x 10-2M [Zn2+] = 0.10 M • Start with the half reactions • 2(2H+ + 1e- + VO21+ VO2++ H2O) 1.00 V • Zn  Zn2+ +2e- 0.76 V • 4H+ + 2VO21+ + Zn  Zn2+ + 2VO2+ + 2H2O 1.76 V • Write the equilibrium expression 4 x 10-5

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

25. Concentration Cells: an electrochemical cell where the same electrodes (same half-reactions) are used, but the solutions have different concentrations • 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 V

26. Use the Nernst Equation to solve for the potential • You DON’T need your calculator for this one!!! • E = 0.0592 V

27. Batteries • 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 Fuel Cells • A voltaic cell that converts chemical energy into electrical energy • Not self-contained • Produces current by combustion and electron-excitation

28. Corrosion • 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) Electrolysis • 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)

29. Electrolytic Cell

30. Calculations with Electrolytic Cells • Ampere = units to measure current in Coulombs/second • Measures the rate of electron flow • Faraday’s Constant, F = 96485 Coulomb/mol Example: Calculate the amount of time required to produce 1000 g of magnesium metal by electrolysis of moltenMgCl2using a current of 50 A.

31. Calculations with Electrolytic Cells • Ampere = units to measure current in Coulombs/second • Measures the rate of electron flow • Faraday’s Constant, F = 96485 Coulomb/mol Example: Calculate the amount of time required to produce 1000 g of magnesium metal by electrolysis of molten MgCl2 using a current of 50 A. • Use a t-table! • Time = 158823 sec = 2647 min = 44 hours = 1.84 days

32. Example: a Cr3+ (aq) solution is electrolyzed using a current of 7.60 A. What mass of Cr (s) is plated out after 2.00 days?

33. Example: a Cr3+ (aq) solution is electrolyzed using a current of 7.60 A. What mass of Cr (s) is plated out after 2.00 days? • 3 e + Cr3+ Cr • Use a t-table! Mass of Cr = 2.36 x 102 g

34. Electrolysis can be used to electroplate metals • 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 NO MORE AP CHEM NOTES! GOOD LUCK ON YOUR EXAM!