1. Electrochemistry The first of the BIG FOUR

2. Introduction of Terms  Electrochemistry- using chemical changes to produce an electric current or using electric current to produce a chemical change  Current- movement of electrons or a transfer of electrons  Conductor- contains charge carriers 1.Metallic conductor- wire where the metal ions float around the electrons 2.Electro - ions present from ionization or dissociation of a dissolved electrolyte.

3. Voltaic Cells A. Definition: An electrochemical cell in which a spontaneous reaction generates an electric current. B. This type of cell is spontaneous. C. Consists of two half cells; a portion of the cell contains one half of the reaction D. Oxidation - ANODE “an ox” Reduction - CATHODE “red cat”

4. Illustration Galvanic Cell Animation

5. Notation for Voltaic Cells A. Anode || Cathode (the double line represents the salt bridge B. Reactants, products || reactants, products C. | - represents a phase boundary D. Examples: 1.Zn(s) | Zn 2+ (aq) || Cu 2+ (aq)| Cu(s) 2.Zn(s) | Zn 2 +(aq)|| H + (aq) | H 2 (g) | Pt 3.Tl + Sn 2+ → Tl 3+ + Sn 4.Zn + Fe 3+ → Zn 2+ + Fe 2+ E. A gas is involved, Pt is usually the electrode 1.|| Cl 2 (g) | Cl - | Pt

6. Notation for Voltaic Cells 2 H + (aq) + 2e -  H 2(g) Zn (s)  Zn 2+ + 2e - Tl (s) │ Tl + (aq) ║ Sn 2+ │ Sn (s) Zn (s) │ Zn 2+ (aq) ║ Fe 3+ (aq), Fe 2+ (aq) │ Pt

7. Electromotive Force (EMF) A. Definition- The maximum potential difference between the electrodes of a voltaic cell. B. W max = ∆G = -n F E cell where F is a faraday = 9.6485 x 10 4 C ∆G is Gibb’s Free Energy (thermodynamics) C. The maximum cell potential is directly related to the free energy difference between the reactants and the products in the cell. D. Example 1: Cu 2+ (aq) + Fe (s) → Cu (s) + Fe 2+ Answer: -1.5 x 10 5 J

8. Example 2  The emf of a voltaic cell with the reaction: Hg 2 2+ (aq) + H 2 ↔ 2 Hg(l) + 2 H + (aq) is 0.650 V. Calculate the maximum electrical work of this cell when 0.500 g of H 2 is consumed. Answer: -2 x 96, 480 x 0.650 = -1.25 x 10 5 J.500g (1/2.02)(-1.25 x 10 5 J/mol)= -3.09 x 10 4 J

9. Standard Cell EMF’s A. A cell emf is a measure of the driving force of the cell reaction. B. Anode + Cathode 1.Ox potential- a measure in V of the tendency for a species to lose electrons in the oxidation half reaction 2.Red. Potential- a measure in V of the tendency for a species to gain electrons in the reduction half reaction.

10. Calculating E cell 1. Equation: E cell = E ox + E red 2. Turning E red into a E ox a)Reverse the half cell reaction (to a ox. Rxn) i.Sn 2+ + 2e - → Sn Sn → Sn 2+ + 2e - b)Reverse the sign on the potential value i.+0.14 V

12. Strengths of Oxidizing & Reducing Agents 1. Oxidizing Agents – these are the substances reduced, so the strongest oxidizing agents have the highest positive E red. (F 2 is the strongest oxidizing agent while Li + is the weakest oxidizing agent). 2. Reducing Agents- these are the substances being oxidized so they appear on the right side of the yield sign. The strongest reducing agents correspond with the highest negative E red. (Li is the strongest reducing agent and F - is the weakest.

13. Examples 3. Fe 3+, Cl 2, H 2 O 2 4. Cu, H 2, Al

14. Nernst Equation 1. Calculate cell potentials when conditions are not standard 2. Calculate cell potentials of half reactions 3. Predict whether the cell voltage will decrease or increase depending on Q.

15. The Equation Controls the voltage of the cell Q and E are inversely related Q>1 : then the log value is positive and E decreases Q<1 : then the log value is negative and E increases

16. Electrolytic Cells A. Definition: An electrochemical cell in which an electric current drives an otherwise nonspontaneous reaction. B. Electrolysis- the process by which a direct electrical current causes a chemical reaction to occur.

17. 1. Electrolysis of MOLTEN Salts (Down’s Cell) AnodeCathode Withdrawing electrons so it’s positive ReductionOxidation The battery acts as an electron pump, pulling electrons from the anode and pushing them to the cathode. Adding electrons so it’s negative 2 Cl - → Cl 2 + 2 e - 2 Na + + 2 e - → 2 Na Overall cell reaction: 2Na + + 2Cl - → 2Na + Cl 2 Overall cell reaction: 2Na + + 2Cl - → 2Na + Cl 2

18. Examples: a) KCl Anode: 2 Cl - → Cl 2 + 2 e - Cathode: 2 e - + 2 K + → 2 K  KOH Anode: 4 OH - → O 2 + 2H 2 O + 4 e - Cathode: 4 K + + 4 e - → 4 K

19. 2. Electrolysis of Aqueous Solutions a. There are two reactions possible at each electrode for an aqueous solution.  Water can be oxidized to O 2 at the anode: Anode: 2 H 2 O → O 2 + 4 H + + 4 e - ♦ Water can be reduced to H 2 at the cathode: Cathode: 2 H 2 O + 2 e - → H 2 + 2 OH -

20. b. Chloride Solutions At the anode there are 2 possible half reactions: 2 Cl - → Cl 2 + 2 e - -1.36 V 2 H 2 O → O 2 + 4 H + + 4 e - - 1.23 V At the cathode there are 2 possible half reactions: 2 Na+ + 2 e- → 2 Na -2.71 V 2 H 2 O + 2 e - → H 2 + 2 OH - -0.83 V Overall Balanced Reaction: 2 H 2 O + 2 Cl - → Cl 2 + H 2 + 2 OH - Since O 2 has the more positive potential, we would expect to see O 2 produced at the anode. This does not happen. There is a much higher voltage needed to produce the O 2, in excess of the expected value (overvoltage). The overvoltage is greater than the production of Cl 2 so the chlorine is produced at the anode.

21. c. Sulfuric acid solutions/ SO 4 2- ions Possible Anode Reactions: 2 SO 4 2- → S 2 O 8 2- + 2 e - -2.71 V 2 H 2 O → O 2 + 4 H + + 4 e - -1.23 V Possible Cathode Reactions: 2 H + + 2 e - → H 2 2 H 2 O + 2 e - → H 2 + 2 OH - Overall Balanced Reaction: 2 H 2 O → O 2 + 2 H 2 In this case, the sulfate is not as easily oxidized as chloride in the previous example, so this time the water is oxidized as expected.

22. Examples 1. Half reactions of CuSO 4 (aq) Anode: 2 SO 4 2- → S 2 O 8 2- + 2 e - -2.71 V 2 H 2 O → O 2 + 4 H + + 4 e - -1.23 V Cathode: Cu 2+ + 2e - → Cu+0.34 V 2 H 2 O + 2 e - → H 2 + 2 OH - -0.83V Overall Balanced Reaction: 2 Cu 2+ + 2 H 2 O → 2 Cu + O 2 + 4H + (acidic)

23. Examples 2. Aqueous AgNO 3 (NO 3 - is not oxidized) Anode: 2 H 2 O → O 2 + 4 H + + 4 e - -1.23 V Cathode: Ag + + e - → Ag+0.80 V 2 H 2 O + 2 e - → H 2 + 2 OH - -0.83V Overall Balanced Reaction: 4 Ag + + 2 H 2 O → Ag + O 2 + 4 H + (acidic)

24. *NOTE: It is difficult to reduce elements in group IA (alkali metals) in the presence of water (II A too!). You have the formation of H 2 instead.

25. Answers to Homework  Anode: 2 Cl -  Cl 2 + 2e -  Cathode: 2e - + Ca 2+  Ca  Anode: 4 OH - → O 2 + 2H 2 O + 4 e -  Cathode: 4e - + 4 Cs +  4 Cs  Anode: 2 H 2 O → O 2 + 4 H + + 4 e -  Cathode: 2 H 2 O + 2 e - → H 2 + 2 OH -  Overall Balanced Reaction: 2 H 2 O → O 2 + 2 H 2

26. Answers to Homework  Anode: 2 Br -  Br 2 + 2e -  Cathode: 2 H 2 O + 2 e - → H 2 + 2 OH -  Overall Reaction: 2 Br - + 2 H 2 O  Br 2 + H 2 + 2 OH -

27. Stoichiometry of Electrolysis A. Electrical Units 1.ampere- base unit of current (rate of flow) 2.Coulomb – SI unit of electrical charge (ampsec): a current of 1 amp flowing for 1 second. 3.Faraday- 1 mole of electrons = 6.02x10 23 e - 1 F = 96,500 C (96,480 book #) = 96,500 ampsec

28. Examples  1.  2.  3.

29. Example problems: 1. How many grams of copper is plated out when a current of 10.0 amps is passed for 30.0 minutes? 2.How long must a current of 5.00 A be applied to a solution of Ag + to produce 10.5 g silver metal?