1. Electrochemistry Electrochemical Cells –Galvanic cells –Voltaic cells –Nernst Equation –Free Energy

2. Galvanic Cells Device in which chemical energy is changed into electrical energy Oxidation at the anode; reduction at the cathode.

7. Complete Galvanic Cell

8. Fe(s) ⃒ Fe 2+ (aq) ‖ MnO 4 - (aq), Mn 2+ (aq) ⃒ Pt(s) Cell Notation Anode Cathode Phase change Physical barrier-salt bridge or porous disc

9. Forming a Galvanic Cell Also need to balance the half-reactions to match electrons lost and gained. Cell potential is an intensive property. Do not multiply it by any coefficient used to balance the electrons.

10. Standard Reduction Potentials Convenient to assign a potential to each half-reaction Use 2H + + 2e -  H 2 E  = 0 as a standard. Can then find standard reduction potentials (E  ) for any half-reaction.

11. Cell Potential “Pull” of electrons from the anode to the cathode Also known as electromotive force Measured in volts (V) 1 V = 1 J/C

12. Standard cell potential

13. Forming a Galvanic Cell Potentials are all listed as reduction potentials To have a spontaneous reaction, one half- reaction must be reversed to be an oxidation. Reaction will be spontaneous in the direction that produces a positive cell potential. When the reaction is reversed, the sign of E  is reversed as well.

14. Comparing Potentials 2H + + 2e -  H 2 0.00 V Ag + + 1e -  Ag 0.80 V Ni 2+ + 2e -  Ni -0.23 V Fe 3+ + 1e -  Fe 2+ 0.77 V VO 2 + + 2H + + 1e -  VO 2+ + H 2 O 1.00 V Cr 3+ + 1e -  Cr 2+ -0.50 V Can Fe 2+ reduce VO 2 + ? Can Fe 2+ reduce Cr 3+ ?

15. A Question For You Write the balanced equation for the combination of the half-reactions below. Cu + + 1e -  Cu  = 0.52V Ni 2+ + 2e -  Ni  = - 0.23VNi  Ni 2+ + 2e - E  = 0.23V Cu + + e -  Cu E  =0.52V

16. Answer Write the balanced equation for the combination of the half-reactions below. Cu + + 1e -  Cu  = 0.52V Ni 2+ + 2e -  Ni  = - 0.23VNi  Ni 2+ + 2e - E  = 0.23V 2Cu + + 2e -  2Cu E  = 0.52V 2Cu + + Ni  2Cu + Ni 2+ E  = 0.75V

17. What is the voltage of this cell? Fe 2+ + 2e -  Fe E  = -0.44V MnO 4 -  Mn 2+ + 4H 2 O8H + +5e - + E  = 1.51 V 16H + + 2MnO 4 - + 5Fe  5Fe 2+ 2Mn 2+ + 8H 2 O Fe(s) ⃒ Fe 2+ (aq) ‖ MnO 4 - (aq), Mn 2+ (aq) ⃒ Pt(s) Cell Notation Anode Cathode

18. Answer Fe 2+ + 2e -  Fe E  = -0.44V MnO 4 -  Mn 2+ + 4H 2 O8H + +5e - + E  = 1.51 V 16H + + 2MnO 4 - + 5Fe  5Fe 2+ 2Mn 2+ + 8H 2 O E  =1.07 V Fe(s) ⃒ Fe 2+ (aq) ‖ MnO 4 - (aq), Mn 2+ (aq) ⃒ Pt(s) Cell Notation Anode Cathode

19. Complete Cell Description Write a complete balanced reaction Identify direction of electron flow Designate anode and cathode Include each electrode and all relevant ions

20. Cell Potential Units Amount of work that can be done depends on the cell potential 1 V = 1 J/C (n=moles of e - and F=96,485 C/mol e - )

21. Cell Potential &  G w max =  G  G = -nFE  G  = -nFE 

22. A Question For You Calculate  G  for the reaction involving the half-reactions below. Cu 2+ + 1e -  Cu +  = 0.16V Au 3+ + 3e -  Au E  = 1.50V Cu +  Cu 2+ + 1e -  = - 0.16V Cu +  Cu 2+ + e - E  = - 0.16V

23. Solution-Part 1 Calculate  G  for the reaction involving the half-reactions below. Cu 2+ + 1e -  Cu +  = 0.16V Au 3+ + 3e -  Au E  = 1.50V Cu +  Cu 2+ + 1e -  = - 0.16V3Cu +  3Cu 2+ + 3e - E  = - 0.16V 3Cu + + Au 3+  Au + 3Cu 2+ E  = 1.34V

24. Solution-Part 2 E  = 1.34 V n = 3 mol e - F = 96,485 C/mol e -  G  = -nFE  = - (3mol)(96,485C/mol)(1.34V)  G  = - 388 kJ

25. Nernst Equation Dependence of E  on concentration is similar to dependence of  G  on concentration.  G =  G  + RT ln(Q)  G = -nFE and  G  = -nFE  -nFE = - nFE  + RT ln(Q) E = E  - [(RT)/(nF)] ln(Q) E = E  - [0.0591 / n] log(Q) (at 25  C)

26. Nernst Equation Example 2Al + 3Mn 2+  2Al 3+ + 3Mn E  = 0.48V Calculate E at 25  C when [Mn 2+ ]=0.50M and [Al 3+ ]=1.50M. Q = [Al 3+ ] 2 / [Mn 2+ ] 3 = (1.50) 2 /(0.50) 3 Q = 18 E= 0.48V – (0.0591/6) log (18) = 0.467V

27. Electromotive force (emf) or cell potential (E) – driving force that moves electrons from the anode to the cathode Units: E is measured in volts (V) Current is measured in amps (A) 1 A = 1 C/s 1 V = 1 J/C Coulomb (C) – charge transferred when 1 A flows for 1 s Faraday constant (F) = 96,500 C/mol e -

28. Derivation of Nernst Equation

29. At 25 o C: Nernst equation

30. At 25 o C:

31. Relationship between E and  G

32. Nonstandard Concentrations Table gives standard reduction potentials What happens to cell potential when concentrations aren’t 1M? Cu + 2Ce 4+  Cu 2+ + 2Ce 3+ E  = 1.36V [Ce 4+ ] = 3.0M [Cu 2+ ] = 2.0M

33. Challenge Problem Predict whether E is larger or smaller than E  for the following cases. 2Al + 3Mn 2+  2Al 3+ + 3Mn E  = 0.48V [Al 3+ ] = 0.25M[Mn 2+ ] = 1.0M [Al 3+ ] =2.50M[Mn 2+ ] = 1.0M [Al 3+ ] = 1.0M[Mn 2+ ] = 0.3M [Al 3+ ] = 1.0M[Mn 2+ ] = 3.0M

34. Concentration Cells We can build a galvanic cell where both compartments contain the same components but at different concentrations. Ag + Ag +  Ag + + Ag 1.0M 0.1M

35. Equilibrium Is the system at equilibrium under standard states? What happens to E as the reaction progresses? At equilibrium, Q=K and E = 0. Can use the Nernst equation to calculate K.

36. Another Question For You Calculate K at 25  C for the reaction below. 2Al + 3Mn 2+  2Al 3+ + 3Mn E  = 0.48V E= E  - [0.0591 / n] log(Q) E  = [0.0591 / n] log(K) log (K) = nE  / 0.0591 log (K) = (6)(0.48)/0.0591 = 48.7 K = 5.38  10 48