CELL POTENTIAL, Eo

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Cell diagrams. Rather than drawing an entire cell, a type of shorthand can be used.For our copper - zinc cell, it is:Zn | Zn2 (1M) || Cu2 (1M) | CuThe anode is always on the left.| = boundaries between phases||= salt bridgeAnode Left, Cathode Right (Reduction, Receiving)Electron

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CELL POTENTIAL, Eo

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1. CELL POTENTIAL, Eo This is the STANDARD CELL POTENTIAL, Eo Eo is a quantitative measure of the tendency of reactants to proceed to products when all are in their standard states at 25 °C.

2. Cell diagrams Rather than drawing an entire cell, a type of shorthand can be used. For our copper - zinc cell, it is: Zn | Zn2+ (1M) || Cu2+ (1M) | Cu The anode is always on the left. | = boundaries between phases || = salt bridge Anode Left, Cathode Right (Reduction, Receiving) Electrons flow left to right (in order of species)

3. SHE 1 M H+(aq)+ 2e- ? H2(g) (1 bar)

5. Standard Reduction Potentials

6. Problem see appendix F Arrange the following in order of increasing oxidizing strength: MnO4- in acidic media Sn2+ Co3+

7. Increasing oxidizing strength: Co3+ + e- = Co2+ 1.92 V MnO4- + 4H+ + 3e- = MnO2 + 2H2O 1.68 V MnO4- + 8H+ + 5e- = Mn2+ + 2H2O 1.51 V Sn2+ + 2e- = Sn -0.14 V

9. Zn/Cu Electrochemical Cell Anode: Zn(s) ? Zn2+(aq) + 2e- Eo = -0.76 V

10. Problem Is the following redox reaction spontaneous? Mg2+ + 2Ag(s) ? Mg(s) + 2Ag+ given: Ag+ + e- ? Ag(s) +0.80 V vs SHE Mg2+ + 2e- ? Mg(s) -2.37 V vs SHE

11. Eo and DGo Eo is related to DGo, the free energy change for the reaction. DGo = - n F Eo F = Faraday constant = 9.6485 x 104 J/V•mol n = number of moles of e-’s transferred.

12. For a product-favored reaction Galvanic cell: Chemistry ? electric current Reactants ? Products DGo < 0 and so Eo > 0 (Eo is positive) Eo and DGo

13. How do Cell Potentials Change if We are Not at Standard State? ?G = - nFE ?Go = - nFEo ?G = ?G0 + 2.303 RT log Q E = E0 - (RT/nF) ln Q aA + bB ? cC + dD At standard state temperature, Nernst equation

14. Application of cell potentials Example Determine the spontaneous direction and Ecell for the following system. Half reaction Eo Pb2+ + 2e- ? Pb -0.125 V Sn2+ + 2e- ? Sn -0.136 V

15. Calculation of cell potentials Pb2+ + 2e- Pb -0.125 V Sn2+ + 2e- Sn -0.136 V At first glance, it would appear that Pb2+ would be reduced to Pb (most positive Eo). However, we’re not at standard conditions. We need to determine the actual Ecell for each half reaction before we know what will happen.

16. Calculation of cell potentials For lead: E = -0.125 - log = -0.184 V For tin: E = -0.136 - log = -0.124 V Ecell = Ecathode - Eanode = -0.124 – (-0.184) = 0.063

17. Example of Nernst Equation

18. Relation to Equilibrium Constant Half reaction Eo V Br2 (l) + 2e- 2 Br- (aq) +1.066 v Cu2+ + 2e- Cu (s) +0.342 v Br2 (l) + Cu2+ Cu (s) + 2 Br- (aq) +0.724 v

19. Your Turn

20. Your Turn

21. Amount of Chemical Reaction when electrolysis occurs Half reaction Eo, v Cu+ + e- Cu (s) +0.153 Zn2+ + 2e- Zn (s) -0.763

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