1. Electrochemistry

2. #13 Electrochemistry and the Nernst Equation Goals: To determine reduction potentials of metals To measure the effect of concentration on reduction potential

4. Redox Chemistry/Electrochemistry  Thermodynamics of redox reactions  Chemical / electrical work interchange  Involves transfer of electrons or electron density Oxidation: Loss of electrons Reduction: Gain of electrons

6. Zn (s) + Cu 2+ (aq)  Cu (s) + Zn 2+ (aq) 2e -

7. Electrochemistry/Electrochemical Cells Redox reaction produces or uses electrical energy Voltaic (galvanic) cell: spontaneous reaction generates electrical energy (battery) Electrolytic cell: absorbs energy from an electrical source to drive nonspontaneous reaction (recharge)

8. Cell Components Electrodes: conduct electricity between cell and surroundings Anode:oxidation site“AO”, “an ox” Cathode:reduction site“CR”, “red cat” Electrolyte: ion mixture involved in reaction or carrying charge Salt bridge: completes circuit (provides charge balance)

9. Experimental Set-up Electrochemical cell  Separated ½ reactions (ox and red)  “Driving force” for electron transfer is measurable  What is +1.10?

10. Calculating cell potential

11. Calculating the cell potential for a zinc and copper cell Zn +2 + 2e-  Zn (s) - 0.76 Cu +2 + 2e-  Cu (s) 0.34 Zn (s)  Zn +2 + 2e- + 0.76 Cu +2 + 2e-  Cu (s) 0.34 1.10

12. Electrochemical Cell Set-up e – M ox (s) Cathode Ions are reduced M y+ (aq) + y e -  M (s) M y+ Oxidizing agent M red (s) e – e – e – voltmeter Salt Bridge (KNO 3 ) M x+ Reducing agent Anode Electrode is oxidized M (s)  M x+ (aq) + x e -

13. Example Set-up e – Zn(s) Cathode Ions are reduced Cu 2+ (aq) + 2e -  Cu (s) Cu 2 + Oxidizing agent Cu(s) e – e – e – +1.10V Salt Bridge (KNO 3 ) Zn 2+ Reducing agent Anode Electrode is oxidized Zn (s)  Zn 2+ (aq) + 2e -

14. Zn (s) + Cu 2+ (aq)  Cu (s) + Zn 2+ (aq) Zn gives up e - to Cu spontaneously Zn “pushes harder” on e - e - on Zn – greater potential energy – greater electrical potential +1.10 V is a measure of this

15. Electrochemical potential – measured voltage Voltage difference in energy of the e - on the metals or relative difference in metals’ abilities to give e - different metals → different e - energy → different “push” on e - Electromotive force (EMF; cell potential), E cell Driving force on electrons Measured voltage= potential difference Higher E cell =larger “drive”

16. Energy, E 0, and Spontaneity Cell potential Free Energy Spontaneity Positive E 0 cell  G 0 < 0Spontaneous Negative E 0 cell  G 0 > 0Not Zero E 0 cell  G 0 = 0Equilibrium  G 0 :free energy of change amount of available (electrical) work

17. Standard Reduction Potentials, E 0 E 0 cell cell potential under standard conditions (reference tables) elements in standard states: s, l, g solutions:1 M gases:1 atm Relative to standard hydrogen electrode, “SHE” 2H + (aq) + 2 e -  H 2(g) E 0 cell = 0.00 V Overall E 0 cell :combine E 0 ’s for half-reactions

18. Example E 0 values Reduction reaction E 0 Mg 2+ + 2e -  Mg−2.30 V Zn 2+ + 2e -  Zn−0.76 V Ni 2+ + 2e -  Ni−0.23 V 2H 2+ + 2e -  H 2 0.00 V Cu 2+ + 2e -   Cu+0.34 V Ag + + e -   Ag +0.80 V Au 3+ + 3e -   Au +1.50 V  More positive E 0 = greater reduction potential  The push on e - relative to H 2 /2H +

19. E 0 values More positive: Stronger oxidizing agent Easier to reduce More readily accepts e - More negative: Stronger reducing agent More easily oxidized More readily gives e - In a spontaneous reaction Stronger R.A. + O.A.  Weaker R.A. + O.A.

20. Calculating E 0 cell Reaction:Zn (s) + Cu 2+ (aq)  Cu (s) + Zn 2+ (aq) red. Zn (s)  Zn 2+ (aq) + 2 e - E 0 = +0.76 V ox. Cu 2+ (aq) + 2 e -  Cu (s) E 0 = +0.34 V Zn (s) + Cu 2+ (aq)  Cu (s) + Zn 2+ (aq) E 0 = +1.10 V Assumes 1 M Cu 2+ and Zn 2+ solutions under standard conditions

21. Part 1 Examples Copper and Lead: Pb  Pb 2+  Cu 2+  Cu oxidation:Pb metal/solution reduction:Cu metal/solution Reduction potentials (table) Pb 2+ (aq) + 2 e -  Pb (s) E 0 = –0.13 V Cu 2+ (aq) + 2 e -  Cu (s) E 0 = +0.34 V For spontaneous reaction, E 0 > 0so calculated E 0 is Pb (s) + Cu 2+ (aq)  Cu (s) + Pb 2+ (aq) E 0 = +0.47 V