1. Lecture 13: E° cell and  G Reading: Zuhdahl: 11.3 Outline –E° cell and work –E° cell and  G

2. E° cell and work We saw in Lecture 12 how to construct a galvanic electrochemical cell that was capable of generating a flow of electrons. This flow of electrons (current) can be used to perform work on the surroundings.

3. E° cell and work (cont.) From the definition of electromotive force (emf): In English: 1 J of work is done when 1 C of charge is transferred between a potential difference of 1 V. Volt = work (J)/charge (C)

4. E° cell and work (cont.) Recall, we have a system-based perspective such that work done by the system is negative: E° cell = -w/q w = work q = charge (not heat!) Rearranging: qE° cell = -w

5. An example: What is the amount of work that is done by a galvanic cell in which 1.5 moles of electrons are passed between a potential of 2 V? E° cell and work (cont.) qE° cell = -w (1.5 mol e - )(2 V) = -w What do we do with this?

6. Define the “Faraday” (F): 1 F = the amount of charge on 1 mol of e - E° cell and work (cont.) 1 F = 96,485 C/mol e - Then: q = (1.5 mol e - ) x (1 F = 96,485 C/mol e - ) = 144,728 C Finally: -w = qE° cell = (144,728 C)(2 J/C) = 298.5 kJ

7. E° cell and  G Since there is a relationship between  G and work, there is also a relationship between  G and E° cell.  G° = -nFE° cell The above relationship states that there is a direct relationship between free energy and cell potential. For a galvanic cell: E° cell > 0 Therefore,  G° < 0 (spontaneous)

8. E° cell and  G (cont.) For the following reaction, determine the overall standard cell potential and determine  G° Cd +2 (aq) + Cu(s) Cd(s) + Cu +2 (aq) Cd +2 + 2e - Cd E° 1/2 = -0.40 V Cu +2 + 2e - Cu E° 1/2 = +0.34 V Cu Cu +2 + 2e - E° 1/2 = -0.34 V

9. E° cell and  G (cont.) Cd +2 (aq) + Cu(s) Cd(s) + Cu +2 (aq) Cd +2 + 2e - Cd E° 1/2 = -0.40 V Cu Cu +2 + 2e - E° 1/2 = -0.34 V E° cell = -0.74 V  G° = -nFE° cell = (-2 mol e - )(96485 C/mol e - )(-0.74 J/C) = 142.8 kJ (NOT spontaneous, NOT galvanic) 2

10. E° cell and  G (cont.) We now have two relationships for  G°:  G° = -nFE° cell = -RTln(K) -nFE° cell = -RTln(K) E° cell = (RT/nF) ln(K) E° cell = (0.0257 V) ln(K) n

11. E° cell and  G (cont.) The above relationship states that by measuring E° cell, we can determine K. E° cell = (0.0257 V) ln(K) = (0.0591) log(K) n The above relationship illustrates that electrochemical cells are a venue in which thermodynamics is readily evident n

12. E° cell and  G (cont.) Developing the “big picture” E° cell = (0.0591 V) log(K) n  G° = -RTln(K)  G° = -nFE° cell It is important to see how all of these ideas interrelate.

13. An Example Balance, determine E° cell and K for the following: S 4 O 6 2- (aq) + Cr 2+ (aq) Cr 3+ (aq) + S 2 O 3 2- (aq) S 4 O 6 2- S 2 O 3 2- Cr 2+ Cr 3+ + e - 22e - + x 2 S 4 O 6 2- + 2Cr 2+ 2Cr 3+ + 2S 2 O 3 2-

14. An Example (cont.) Determining E° cell S 4 O 6 2- S 2 O 3 2- 2Cr 2+ 2Cr 3+ + 2e - 22e - + S 4 O 6 2- + 2Cr 2+ 2Cr 3+ + 2S 2 O 3 2- E° 1/2 = 0.17 V E° 1/2 = 0.50 V E° cell = 0.67 V

15. An Example (cont.) Determining K S 4 O 6 2- + 2Cr 2+ 2Cr 3+ + 2S 2 O 3 2- E° cell = 0.67 V E° cell = (0.0257 V) ln(K) n = (0.059 V) log K n n(E° cell ) (0.059 V) 2 (0.67 V) (0.059 V) = =22.7 = log K K = 10 22.7 = 5 x 10 22