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Electrochemistry Ch 13 pg 225 Princeton Review

Electrochemistry Ch 13 pg 225 Princeton Review. 2Mg ( s ) + O 2 ( g ) 2MgO ( s ). 2Mg 2Mg 2+ + 4e -. O 2 + 4e - 2O 2-. Electrochemical processes are oxidation-reduction reactions in which:

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Electrochemistry Ch 13 pg 225 Princeton Review

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  1. Electrochemistry Ch 13 pg 225 Princeton Review

  2. 2Mg (s) + O2 (g) 2MgO (s) 2Mg 2Mg2+ + 4e- O2 + 4e- 2O2- • Electrochemical processes are oxidation-reduction reactions in which: • the energy released by a spontaneous reaction is converted to electricity or • electrical energy is used to cause a nonspontaneous reaction to occur 0 0 2+ 2- Oxidation half-reaction (lose e-) Reduction half-reaction (gain e-) 19.1

  3. Cotton plugs SO SO 2– 2– 4 4 Flow of e- Flow of current Galvanic Cell Voltmeter e– e– Zinc anode Copper cathode Cl– K+ Salt bridge Cu2+ Zn2+ ZnSO4 solution CuSO4 solution 2e– 2e– Cu2+ Zn Zn2+ Cu Zn is oxidized to Zn2+ at anode. Cu2+ is reduced to Cu at cathode. Zn2+(aq) + 2e– Zn(s) 2e– + Cu2+(aq) Cu(s) Net reaction Zn2+(aq) + Cu(s) Zn(s) + Cu2+(aq)

  4. SO SO 2– 2– 4 4 Cu2+ Zn2+ ZnSO4 solution CuSO4 solution

  5. Cl– K+ Cotton plugs SO SO 2– 2– 4 4 Voltmeter e– e– Zinc anode Copper cathode Salt bridge Cu2+ Zn2+ ZnSO4 solution CuSO4 solution Salt bridge provides electrical neutrality by providing negative anions to equal the positive cations being created at the Zn anode during oxidation. And cations ions (K+) to replace Cu 2+ being used up at reduction.

  6. Cl– K+ Cotton plugs SO SO 2– 2– 4 4 Voltmeter e– e– Zinc anode Copper cathode Salt bridge Cu2+ Zn2+ ZnSO4 solution CuSO4 solution Zn

  7. Cl– K+ Cotton plugs SO SO 2– 2– 4 4 Voltmeter e– e– Zinc anode Copper cathode Salt bridge Cu2+ Zn2+ ZnSO4 solution CuSO4 solution 2e– Zn Zn2+ Zn is oxidized to Zn2+ at anode. Zn2+(aq) + 2e– Zn(s)

  8. Cl– K+ Cotton plugs SO SO 2– 2– 4 4 Voltmeter e– e– Zinc anode Copper cathode Salt bridge Cu2+ Zn2+ ZnSO4 solution CuSO4 solution 2e– Zn Zn2+ Cu Zn is oxidized to Zn2+ at anode. Zn2+(aq) + 2e– Zn(s)

  9. Cl– K+ Cotton plugs SO SO 2– 2– 4 4 Voltmeter e– e– Zinc anode Copper cathode Salt bridge Cu2+ Zn2+ ZnSO4 solution CuSO4 solution 2e– 2e– Cu2+ Zn Zn2+ Cu Zn is oxidized to Zn2+ at anode. Zn2+(aq) + 2e– Zn(s)

  10. Cl– K+ Cotton plugs SO SO 2– 2– 4 4 Voltmeter e– e– Zinc anode Copper cathode Salt bridge Cu2+ Zn2+ ZnSO4 solution CuSO4 solution 2e– 2e– Cu2+ Zn Zn2+ Cu Zn is oxidized to Zn2+ at anode. Cu2+ is reduced to Cu at cathode. Zn2+(aq) + 2e– Zn(s) 2e– + Cu2+(aq) Cu(s)

  11. Cotton plugs SO SO 2– 2– 4 4 Voltmeter e– e– Zinc anode Copper cathode Cl– K+ Salt bridge Cu2+ Zn2+ ZnSO4 solution CuSO4 solution 2e– 2e– Cu2+ Zn Zn2+ Cu Zn is oxidized to Zn2+ at anode. Cu2+ is reduced to Cu at cathode. Zn2+(aq) + 2e– Zn(s) 2e– + Cu2+(aq) Cu(s) Net reaction Zn2+(aq) + Cu(s) Zn(s) + Cu2+(aq)

  12. Galvanic Cells anode oxidation cathode reduction spontaneous redox reaction 19.2

  13. Current • The flow of positive charge. • Current is always in the opposite directions from electron flow.

  14. Zn (s) + Cu2+(aq) Cu (s) + Zn2+(aq) Electrolytic Cells • The difference in electrical potential between the anode and cathode is called: • cell voltage • electromotive force (emf) • cell potential • Used in electro plating Cell Diagram [Cu2+] = 1 M & [Zn2+] = 1 M Zn (s) | Zn2+ (1 M) || Cu2+ (1 M) | Cu (s) Anode OX Cathode RED 19.2

  15. Reduction Potentials • reduction potential (E0): is the voltage associated with a reduction reaction • The more positive E0 the greater the tendency for the substance to be reduced • On the AP test you will be given a chart of reduction potentials. You can reverse them and change the sign on the voltage to get oxidation potentials. 19.3

  16. The half-cell reactions are reversible • The sign of E0changes when the reaction is reversed • Changing the stoichiometric coefficients of a half-cell reaction does not change the value of E0 19.3

  17. Using the Table TOP: F2 + 2e-  2F- = E° = +2.87 Large reduction potential = more likely to be reduced and thus a strong oxidizing agent. Bottom: Li+ + e-  Li = E° = -3.05 Li Li+ + e- = E° = +3.05 Large oxidation potential = more likely to lose an electron and become oxidized thus it’s a good reducing agent.

  18. 0 =-nFEcell Spontaneity of Redox Reactions Redox rxns will occur spontaneously if its cell potential has a positive value. Or if its free energy (G) is negative. DG0 19.4

  19. Example Look at the spontaneous rxn: Zn + Ag+ Zn2+ + Ag E0 = + 1.56 Zn  Zn2+ + 2e- (LEO) E0 = -0.76 reverse = +0.76 Ag+ + e- Ag (GER) E0 = + 0.80 E = Eox + Ered = .76 + .80 = + 1.56 E is positive so G is negative and the rxn is spontaneous

  20. Example with coefficients • The number of electrons lost must equal the number of electrons gained. In order to insure this we must multiply the half reaction by integers to balance. This does not change the Cell potential E°.

  21. Example Fe3+ + Cu  Cu2+ + Fe 2+ ½ rxns GERFe3+ + e-  Fe 2+ E° = 0.77 V LEO Cu  Cu2+ + 2 e- E° = -0.34 V • The reduction must occue 2x for every 1 OX rxn • Note the voltage of the OX is reversed because our chart is of reduction potentials E° cell = - 0. 34 V + 0.77 E° = 0.43 V

  22. Under standard conditions the maximum cell potential (E°) is directly related to the free energy (G) difference between the reactants and the products. This equation allows us to test for ∆G° in the lab. ∆G° = - nFE° G = Gibbs free energy kJ/mol n = moles of e- exchanged F = Faraday’s constant 96,500 coulombs/mole (how much charge is produced for every 1 mole of e- ) E° = Standard reduction potential

  23. Example Calculate ∆G° for the reaction Fe + Cu2+ Cu + Fe 2+ • Write ½ rxns to ID red.ox atoms Fe +  Fe 2+ 0 +2 LEO (rev potential) Cu2+ Cu +2 0 GER • Use chart to find reduction potentials Fe +  Fe 2+ E = -0.44 = +0.44 v Cu2+ Cu E = +0.34 v

  24. E cell = 0.44+0.34 = 0.78 F = 96,500 n = 2 mole (2 e- per atom) 3. Plug in G = -2(96,500)(0.78) = - 1.5 x 105 J 4. Conclude G = negative so rxn is spontaneous E cell = positive spontaneous ** if E is negative rxn will not occur and it is not spontaneous

  25. Nernst equation Gives the relationship between cell potential and concentrations of cell components. E = E°cell - (0.0591/n) log Q

  26. Q = reaction quotient Q = concentration of products [ ] raised to its coefficient divided by [ ] reactants raised to its coefficient

  27. Example Calculate E for the following reaction given [Al3+] = 1.5M [Mn2+] = 0.5 M Ecell = 0.48 2 Al + 3 Mn2+ 2Al 3+ + 3 Mn Q = 1.52 = 18 0.53 n = 2 Al  2Al 3+3 Mn2+ 3 Mn 0 +3 = 3(2) = 6 +2 0 = 2(3)= 6 n = 6

  28. Cont. Plug it in: E = 0.48 – (0.0591/6) log 18 = 0.47 V

  29. Electrolysis is the process in which electrical energy is used to cause a nonspontaneous chemical reaction to occur. 19.8

  30. Electrolysis of Water 19.8

  31. Electrolysis and Mass Changes charge (C) = current (A) x time (s) 1 mole e- = 96,500 C 19.8

  32. How much Ca will be produced in an electrolytic cell of molten CaCl2 if a current of 0.452 A is passed through the cell for 1.5 hours? 2 mole e- = 1 mole Ca mol Ca = 0.452 x 1.5 hr x 3600 C s 2Cl- (l) Cl2 (g) + 2e- hr s Ca2+(l) + 2e- Ca (s) 1 mol Ca 1 mol e- x x 96,500 C 2 mol e- Ca2+ (l) + 2Cl- (l) Ca (s) + Cl2 (g) Anode: Cathode: = 0.0126 mol Ca = 0.50 g Ca 19.8

  33. 2+ 2+ 2+ Hg2 /Ag2Hg3 0.85 V Sn /Ag3Sn -0.05 V Sn /Ag3Sn -0.05 V Chemistry In Action: Dental Filling Discomfort

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