![Concentration and Ecell
• Consider the following redox reaction:
Zn(s) + 2H+ (aq) Zn2+(aq) + H2(g) E°cell = 0.76 V
DG°= -nFE°cell < 0 (spontaneous)
• What if [H+] = 2 M?
Expect driving force for product formation to increase.
Therefore DG decreases, and Ecell increases
How does Ecell dependend on concentration?](https://image.slidesharecdn.com/chapter-19-moor-221026161218-4d3432c8/85/Chapter-19-moor-ppt-2-320.jpg)
![Concentration and Ecell (cont.)
• With the Nernst Eq., we can determine the effect
of concentration on cell potentials.
Ecell = E°cell - (0.0591/n)log(Q)
• Example. Calculate the cell potential for the
following:
Fe(s) + Cu2+(aq) Fe2+(aq) + Cu(s)
Where [Cu2+] = 0.3 M and [Fe2+] = 0.1 M](https://image.slidesharecdn.com/chapter-19-moor-221026161218-4d3432c8/85/Chapter-19-moor-ppt-4-320.jpg)
![Concentration and Ecell (cont.)
• If [Cu2+] = 0.3 M, what [Fe2+] is needed so that
Ecell = 0.76 V?
Fe(s) + Cu2+(aq) Fe2+(aq) + Cu(s) E°cell = +0.78 V
Ecell = 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](https://image.slidesharecdn.com/chapter-19-moor-221026161218-4d3432c8/85/Chapter-19-moor-ppt-7-320.jpg)
![Concentration and Ecell (cont.)
Fe(s) + Cu2+(aq) Fe2+(aq) + Cu(s)
4.7 = Q
Q
Fe2
Cu2
4.7
Q
Fe2
0.3
4.7
[Fe2+] = 1.4 M](https://image.slidesharecdn.com/chapter-19-moor-221026161218-4d3432c8/85/Chapter-19-moor-ppt-8-320.jpg)
![Measurement of pH (cont.)
Hg2Cl2(s) + H2(g) 2Hg(l) + 2H+(aq) + 2Cl-(aq)
• What if we let [H+] vary?
Q
H
2
Cl
2
PH2
H
2
Cl
2
Ecell = E°cell - (0.0591/2)log(Q)
Ecell = E°cell - (0.0591/2)(2log[H+] + 2log[Cl-])
Ecell = E°cell - (0.0591)(log[H+] + log[Cl-])
saturate
constant](https://image.slidesharecdn.com/chapter-19-moor-221026161218-4d3432c8/85/Chapter-19-moor-ppt-15-320.jpg)
![Measurement of pH (cont.)
Ecell = E°cell - (0.0591)log[H+] + constant
• Ecell is directly proportional to log [H+]
electrode](https://image.slidesharecdn.com/chapter-19-moor-221026161218-4d3432c8/85/Chapter-19-moor-ppt-16-320.jpg)
Chapter-19-moor.ppt
- 1. The Nernst Equation • Outline: – Why would concentration matter in electrochem.? – The Nernst equation. – Applications
- 2. Concentration and Ecell • Consider the following redox reaction: Zn(s) + 2H+ (aq) Zn2+(aq) + H2(g) E°cell = 0.76 V DG°= -nFE°cell < 0 (spontaneous) • What if [H+] = 2 M? Expect driving force for product formation to increase. Therefore DG decreases, and Ecell increases How does Ecell dependend on concentration?
- 3. Concentration and Ecell (cont.) • Recall, in general: DG = DG° + RTln(Q) • However: DG = -nFEcell -nFEcell = -nFE°cell + RTln(Q) Ecell = E°cell - (RT/nF)ln(Q) Ecell = E°cell - (0.0591/n)log(Q) The Nernst Equation
- 4. Concentration and Ecell (cont.) • With the Nernst Eq., we can determine the effect of concentration on cell potentials. Ecell = E°cell - (0.0591/n)log(Q) • Example. Calculate the cell potential for the following: Fe(s) + Cu2+(aq) Fe2+(aq) + Cu(s) Where [Cu2+] = 0.3 M and [Fe2+] = 0.1 M
- 5. Concentration and Ecell (cont.) Fe(s) + Cu2+(aq) Fe2+(aq) + Cu(s) • First, need to identify the 1/2 cells Cu2+(aq) + 2e- Cu(s) E°1/2 = 0.34 V Fe2+(aq) + 2e- Fe(s) E°1/2 = -0.44 V Fe(s) Fe 2+(aq) + 2e- E°1/2 = +0.44 V Fe(s) + Cu2+(aq) Fe2+(aq) + Cu(s) E°cell = +0.78 V
- 6. Concentration and Ecell (cont.) • Now, calculate Ecell Fe(s) + Cu2+(aq) Fe2+(aq) + Cu(s) E°cell = +0.78 V Ecell = E°cell - (0.0591/n)log(Q) Q Fe2 Cu2 (0.1) (0.3) 0.33 Ecell = 0.78 V - (0.0591 /2)log(0.33) Ecell = 0.78 V - (-0.014 V) = 0.794 V
- 7. Concentration and Ecell (cont.) • If [Cu2+] = 0.3 M, what [Fe2+] is needed so that Ecell = 0.76 V? Fe(s) + Cu2+(aq) Fe2+(aq) + Cu(s) E°cell = +0.78 V Ecell = 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
- 8. Concentration and Ecell (cont.) Fe(s) + Cu2+(aq) Fe2+(aq) + Cu(s) 4.7 = Q Q Fe2 Cu2 4.7 Q Fe2 0.3 4.7 [Fe2+] = 1.4 M
- 9. 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?
- 10. 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.
- 11. Concentration Cells (cont.) Ag Ag+ + e- E1/2 = ? V Anode: Ag+ + e- Ag E1/2 = 0.80 V Cathode: Q Ag anode Ag cathode 0.1 1 0.1 Ecell = E°cell - (0.0591/n)log(Q) 0 V Ecell = - (0.0591)log(0.1) = 0.0591 V 1
- 12. Concentration Cells (cont.) Another Example: What is Ecell?
- 13. Concentration Cells (cont.) Ecell = E°cell - (0.0591/n)log(Q) 0 Fe2+ + 2e- Fe 2 e- transferred…n = 2 2 Q Fe2 anode Fe2 cathode 0.01 .1 0.1 Ecell = -(0.0296)log(.1) = 0.0296 V anode cathode e-
- 14. 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: Hg2Cl2(s) + 2e- 2Hg(l) + 2Cl-(aq) E°1/2 = 0.27 V Hg2Cl2(s) + H2(g) 2Hg(l) + 2H+(aq) + 2Cl-(aq) H2(g) 2H+(aq) + 2e- E°1/2 = 0.0 V E°cell = 0.27 V
- 15. Measurement of pH (cont.) Hg2Cl2(s) + H2(g) 2Hg(l) + 2H+(aq) + 2Cl-(aq) • What if we let [H+] vary? Q H 2 Cl 2 PH2 H 2 Cl 2 Ecell = E°cell - (0.0591/2)log(Q) Ecell = E°cell - (0.0591/2)(2log[H+] + 2log[Cl-]) Ecell = E°cell - (0.0591)(log[H+] + log[Cl-]) saturate constant
- 16. Measurement of pH (cont.) Ecell = E°cell - (0.0591)log[H+] + constant • Ecell is directly proportional to log [H+] electrode
- 17. Summary DG = DG° + RTln(Q) DG = -nFEcell Ecell = E°cell - (0.0591/n)log(Q) • None of these ideas is separate. They are all connected, and are all derived directly from thermodynamics.