2. Lecture 14: The Nernst Equation Reading: Zumdahl 11.4 Outline: –Why would concentration matter in electrochem.? –The Nernst equation. –Applications

3. Concentration and E cell Consider the following redox reaction: Zn(s) + 2H + (aq) Zn 2+ (aq) + H 2 (g)E° cell = 0.76 V  G°= -nFE° cell < 0 (spontaneous) What if [H + ] = 2 M? Expect driving force for product formation to increase. Therefore  G decreases, and E cell increases How does E cell dependend on concentration?

4. Concentration and E cell (cont.) Recall, in general:  G =  G° + RTln(Q) However:  G = -nFE cell -nFE cell = -nFE° cell + RTln(Q) E cell = E° cell - (RT/nF)ln(Q) E cell = E° cell - (0.0591/n)log(Q) The Nernst Equation

5. Concentration and E cell (cont.) With the Nernst Eq., we can determine the effect of concentration on cell potentials. E cell = E° cell - (0.0591/n)log(Q) Example. Calculate the cell potential for the following: Fe(s) + Cu 2+ (aq) Fe 2+ (aq) + Cu(s) Where [Cu 2+ ] = 0.3 M and [Fe 2+ ] = 0.1 M

6. Concentration and E cell (cont.) Fe(s) + Cu 2+ (aq) Fe 2+ (aq) + Cu(s) First, need to identify the 1/2 cells Cu 2+ (aq) + 2e- Cu(s) E° 1/2 = 0.34 V Fe 2+ (aq) + 2e- Fe(s) E° 1/2 = -0.44 V Fe(s) Fe 2+ (aq) + 2e-E° 1/2 = +0.44 V Fe(s) + Cu 2+ (aq) Fe 2+ (aq) + Cu(s)E° cell = +0.78 V

7. Concentration and E cell (cont.) Now, calculate E cell Fe(s) + Cu 2+ (aq) Fe 2+ (aq) + Cu(s)E° cell = +0.78 V E cell = E° cell - (0.0591/n)log(Q) E cell = 0.78 V - (0.0591 /2)log(0.33) E cell = 0.78 V - (-0.014 V) = 0.794 V

8. Concentration and E cell (cont.) If [Cu 2+ ] = 0.3 M, what [Fe 2+ ] is needed so that E cell = 0.76 V? Fe(s) + Cu 2+ (aq) Fe 2+ (aq) + Cu(s)E° cell = +0.78 V E cell = E° cell - (0.0591/n)log(Q) 0.76 V = 0.78 V - (0.0591/2)log(Q) 0.02 V = (0.0591/2)log(Q) 0.676 = log(Q) 4.7 = Q

9. Concentration and E cell (cont.) Fe(s) + Cu 2+ (aq) Fe 2+ (aq) + Cu(s) 4.7 = Q [Fe 2+ ] = 1.4 M

10. Concentration Cells Consider the cell presented on the left. The 1/2 cell reactions are the same, it is just the concentrations that differ. Will there be electron flow?

11. Concentration Cells (cont.) Ag + + e - Ag E° 1/2 = 0.80 V What if both sides had 1 M concentrations of Ag + ? E° 1/2 would be the same; therefore, E° cell = 0.

12. Concentration Cells (cont.) AgAg + + e - E 1/2 = ? V Anode: Ag + + e - Ag E 1/2 = 0.80 VCathode: E cell = E° cell - (0.0591/n)log(Q) 0 V E cell = - (0.0591)log(0.1) = 0.0591 V 1

13. Concentration Cells (cont.) Another Example: What is E cell ?

14. Concentration Cells (cont.) E cell = E° cell - (0.0591/n)log(Q) 0 Fe 2+ + 2e - Fe 2 e - transferred…n = 2 2 E cell = -(0.0296)log(.1) = 0.0296 V anodecathode e-e-

15. Measurement of pH pH meters use electrochemical reactions. Ion selective probes: respond to the presence of a specific ion. pH probes are sensitive to H +. Specific reactions: Hg 2 Cl 2 (s) + 2e - 2Hg(l) + 2Cl - (aq) E° 1/2 = 0.27 V Hg 2 Cl 2 (s) + H 2 (g) 2Hg(l) + 2H + (aq) + 2Cl - (aq) H 2 (g)2H + (aq) + 2e - E° 1/2 = 0.0 V E° cell = 0.27 V

16. Measurement of pH (cont.) Hg 2 Cl 2 (s) + H 2 (g) 2Hg(l) + 2H + (aq) + 2Cl - (aq) What if we let [H + ] vary? E cell = E° cell - (0.0591/2)log(Q) E cell = E° cell - (0.0591/2)(2log[H + ] + 2log[Cl - ]) E cell = E° cell - (0.0591)(log[H + ] + log[Cl - ]) saturate constant

17. Measurement of pH (cont.) E cell = E° cell - (0.0591)log[H + ] + constant E cell is directly proportional to log [H + ] electrode

18. Summary  G =  G° + RTln(Q)  G = -nFE cell E cell = E° cell - (0.0591/n)log(Q) None of these ideas is separate. They are all connected, and are all derived directly from thermodynamics.