1. The Nernst Equation Galvanic and Electrolytic Cells 1.Galvanic cells and Electrolysis Cells: in an electrolysis cell, the cell reaction runs in the non--spontaneous direction electrolysis of water 2.Thermodynamics of the relationship of  E and  G, electrical work  G = - n F  E 3.Concentration dependence of the cell potential, the Nernst Equation, the pH meter, the glass electrode, the calomel electrode 4.Adding Half-Cell potentials (problem 12.23a) 5.Disproportionation reactions (problem 12.23b)

2. Galvanic Cells and Electrolysis Cells Cu(s) + 2 Ag + (aq) Cu 2+ (aq) + 2 Ag(s) Ag(s) Cu(s) Cu 2+ (aq) salt bridge Ag + (aq) Galvanic cell: cell reaction runs in the spontaneous direction Electroytic cell: cell reaction runs in the non-spontaneous direction, driven by an applied electrical potential (a voltage) -that is greater than the spontaneous potential galvanic: E o = 0.+4554 V electrolytic: E (applied) > E o positive negative cathode if galvanic anode if electrolytic anode if galvanic cathode if electrolytic

3. Cathode and Anode in Galvanic Cells and Electrolysis Cells Cu(s) (negative) Cu 2+ (aq) salt bridge Ag + (aq) By definition, the cathode is where reduction occurs and the anode is where oxidation occurs. In the galvanic cell, the copper electrode is the anode, silver is the cathode. In the electrolytic cell, the silver electrode is the anode, copper is the cathode. Note that the sign of the potential doesn't change - copper is always negative, silver is always positive. But the designations "cathode" and "anode" change between galvanic and electrolytic cells. Ag(s) (positive)

4. The Galvanic Cell Overall cell reaction: 2 H 2 (g) + O 2 (g) 2 H 2 O (aq)  E o =+1.229 V 2H 3 O 3+ (aq) + 2e - H 2 (g) + 2H 2 O voltmeter salt bridge This galvanic cell “burns” H 2 (g) and O 2 (g) to form water: - + O 2 (g) + 4 H 3 O + (aq) + 4 e - 6 H 2 O (l) Pt(s) H 3 O + (aq) O 2 (g) anode Pt(s) H 3 O + (aq) H 2 (g) cathode

5. The Electrolysis Cell Reaction in the electrolysis cell: 2 H 2 (g) + O 2 (g) 2 H 2 O (aq) O 2 (g) + 4 H 3 O + (aq) + 4 e - 6 H 2 O (l) 2H 3 O 3+ (aq) + 2e - H 2 (g) + 2H 2 O power supply: E >  E o cell salt bridge The electrolysis cell runs in the non-spontaneous direction, driven by an external voltage source: - + Pt(s) H 3 O + (aq) O 2 (g) cathode Pt(s) H 3 O + (aq) H 2 (g) anode

6. Faraday's Laws of Electrolysis 3 Cu(s) + 2 Au 3+ (aq) 3 Cu 2+ (aq) + 2 Au(s) An electrical current of 1 ampere equals one coulomb per second - Q (coulombs) = I (amperes) * t (time) 1 mole of electrons has a charge of -96,485 C (the Faraday) moles of e- = (coulombs passed through cell) / 96485 = (I*t ) / 96,485 How much gold is deposited if 100 A is passed through this electrolysis cell for an hour? (Ans. 1.24 moles or 245 g) 1.Mass is proportional to electric charge passed through the cell. 2.Equivalent masses of different substances require equal amounts of electric charge passed through the cell.

7. Thermodynamics of Electrochemical Cells  G =  H - T  S(const T)  =  E + p  V - T  S(const P)  = (q rev + w) + p  V - q rev (reversible, const T) w = - p  V + w el,rev expansion work electrical work w el = - Q  Ework = charge * (potential difference) = - n F  E  G = w el,rev  = - n F  E(const T, P, reversible) Joules Coulombs Volts

8. Concentration Effects and The Nernst Equation  G =  G o + RT ln Q - this allows you to calculate  G at non-standard concentrations and partial pressures.  E cell =  E o cell - (RT/nF) ln e Q - the Nernst equation. It allows you to calculate  G at non-standard conditions  G = - n F  E cell - this is always true; the cell potential is a very fundamental quantity, just like  G  G o is the standard  G - all solute concentrations must be 1 M and partial pressures must be 1 atm.

9. Measuring Equilibrium Constants  E cell =  E o cell - (RT/nF) ln e Q - The Nernst equation relates  E cell to  E o cell  G = - n F  E cell - this is always true; the cell potential is a very fundamental quantity, just like  G  G o = - RT ln e K- K can be calculated from  G o, which for many compounds can be obtained from thermodynamic tables.  E o = (RT/nF) ln e K- For reactions in solution, K and can be measured from a standard cell potential. Go,Go,

10. The Nernst Equation  E cell =  E o cell - (0.0592/n) log 10 Qat T = 25 o C PbO 2 (s) + SO 4 2- (aq) + 4 H 3 O + (aq) + 2 e - PbSO 4 (s) + 6 H 2 O(l) E o =+1.685 V Pt(s) SO 4 2- (aq) H 3 O + (aq) slurry of PbO 2 (s) + PbSO 4 (s) E = E o - (.0592/2) log 10 [H 3 O + ] 4 [SO 4 2- ] 1 E changes by 0.118 V for each order of magnitude change in [H 3 O + ] or by 0.0296 V for each order of magnitude change in [SO 4 2- ]: pH = 0E = +1.685 V pH = 4E = +1.211 V

11. pH measurement Standard Hydrogen Electrode: 2 H 3 O + (aq, 1M) + 2 e - H 2 (g) + 2 H 2 O(l) Pt(s) H 3 O + (aq) salt bridge H 2 (g) Pt(s) sat’d KCl(aq) paste of Hg(l) and Hg 2 Cl 2 (s) standard hydrogen electrode calomel reference electrode Calomel Electrode: Hg 2 Cl 2 (s) + 2 e - 2Hg(l) + 2 Cl - (sat’d) E = +0.241 V - (.0592/2) log 10 [H 3 O + ] 2 = +0.241 +.0592 * pH E° = +0.241 V E° = +0.000 V KCl(s)

12. The Glass Electrode and the pH Meter The glass electrode is non-metallic electrode composed of a special glass that is porous to H 3 O +. A diffusion potential develops across the membrane in response to a pH gradient. The potential varies by 59.6 mV for each unit change in pH. E cell = E o cell - 0.0592 V 2 log 10 [H + ] out 2 [H + ] in 2 E cell = constant + (0.0592 v) * pH Notice that E cell is uniformly sensitive to pH over 14 orders of magnitude!!

13. Adding Half-Cell Potentials 12.23 a. Calculate the half-cell potential E o for the half-reaction Cu 2+ + e - = Cu + E 3 o = ?? using these tabulated potentials: Cu 2+ + 2 e - = Cu(s)E 1 o = +0.340 Cu + + e - = Cu(s)E 2 o = +0.522 You can’t simply add the potentials because the numbers of electrons differ. However, you can always add  G o ’s: Cu 2+ + 2 e - = Cu(s)  G 1 o = -n F E 1 o Cu + + e - = Cu(s)  G 2 o = -n F E 2 o Cu 2+ + e - = Cu +  G 3 o =  G 1 o -  G 2 o

14. Adding Half-Cell Potentials (Problem 12.23 a) 12.23 a. Calculate the half-cell potential E o for the half-reaction Cu 2+ + e - = Cu + E 3 o = ?? using these tabulated potentials: Cu 2+ + 2 e - = Cu(s)E 1 o = +0.340 Cu + + e - = Cu(s)E 2 o = +0.522 n 3 F E 3 o = n 1 F E 1 o - n 2 F E 2 o  G 3 o =  G 1 o -  G 2 o E 3 o = n 1 E 1 o - n 2 E 2 o n 3 Ans: + 0.158 v

15. Reduction Potential Diagrams and Disproportionation 12.23 b. Will Cu + (aq) disproportionate in aqueous solution? 2 Cu + (aq) = Cu(s) + Cu 2+ (aq) The oxidation-reduction reactions of Cu 2+ /Cu + /Cu(s) are described on a reduction potential diagram of the form: Cu 2+ Cu + Cu(s) 0.158 v 0.522 v 0.342 v Disproportionation occurs only if the potential to the right of Cu + is greater than that to the left of Cu + Cu 2+ + e - = Cu + Cu + + e - = Cu(s) same species is both oxidized and reduced

16. Reduction Potential Diagrams and Disproportionation (ii) Cu 2+ Cu + Cu(s) 0.158 v 0.522 v 0.342 v Disproportionation occurs only if the potential to the right of Cu + is greater than that to the left of Cu + Cu 2+ + e - = Cu + Cu + + e - = Cu(s) Why? Disproportionation requires that you run this backward and this forward which occurs only if this E o is more positive than this E o Ans: yes