Fundamentals of Electrochemistry - W.H. Freeman

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1504T_ch14_270-297 1/23/06 11:35 Page 283 To find the concentration of Ag , we use the solubility product for AgCl. Because the cell contains 0.033 4 M Cl and solid AgCl, we can say Putting this value into the Nernst equation gives which differs from the value calculated in Equation 14-18 because of the accuracy of and the neglect of activity coefficients. Equations 14-17 and 14-19 give the same voltage because they describe the same cell. Advice for Finding Relevant Half-Reactions When faced with a cell drawing or a line diagram, first write reduction reactions for each half-cell. To do this, look in the cell for an element in two oxidation states. For the cell we see Pb in two oxidation states, as Pb(s) and PbF 2 (s), and Cu in two oxidation states, as and Cu(s). Thus, the half-reactions are Cu 2 [Ag ] K sp(for AgCl) [Cl ] Right half-cell: Left half-cell: 1.8 1010 0.033 4 Cu 2 2e T Cu(s) 5.4 10 9 M 1 E 0.799 0.059 16 log 5.4 10 0.309 9 9 V Pb(s) 0 PbF 2 (s) 0 F (aq) Cu 2 (aq) 0 Cu(s) PbF 2 (s) 2e T Pb(s) 2F K sp (14-20) K sp [Ag ] [CI ] The cell voltage cannot depend on how we write the reaction! How to figure out the half-cell reactions You might have chosen to write the Pb half-reaction as Left half-cell: Pb 2 2e T Pb(s) (14-21) because you know that, if PbF is present, there must be some Pb 2 2 (s) in the solution. Reactions 14-20 and 14-21 are equally valid and should predict the same cell voltage. Your choice of reactions depends on whether the F or Pb 2 concentration is easier to figure out. We described the left half-cell in terms of a redox reaction involving Pb because Pb is the element that appears in two oxidation states. We would not write a reaction such as F 2 (g) 2e T 2F , because F 2 (g) is not shown in the line diagram of the cell. The Nernst Equation Is Used in Measuring Standard Reduction Potentials The standard reduction potential would be observed if the half-cell of interest (with unit activities) were connected to a standard hydrogen electrode, as it is in Figure 14-7. It is nearly impossible to construct such a cell, because we have no way to adjust concentrations and ionic strength to give unit activities. In reality, activities less than unity are used in each half-cell, and the Nernst equation is employed to extract the value of E° from the cell voltage. 12 In the hydrogen electrode, standard buffers with known pH (Table 15-3) are used to obtain known activities of H . Don’t invent species not shown in the cell. Use what is shown in the line diagram to select the half-reactions. Problem 14-20 gives an example of the use of the Nernst equation to find E°. 14-5 E and the Equilibrium Constant A galvanic cell produces electricity because the cell reaction is not at equilibrium. The potentiometer allows negligible current (Box 14-4), so concentrations in the cell remain unchanged. If we replaced the potentiometer with a wire, there would be much more current and concentrations would change until the cell reached equilibrium. At that point, nothing would drive the reaction, and E would be 0. When a battery (which is a galvanic cell) runs down to 0 V, the chemicals inside have reached equilibrium and the battery is “dead.” Now let’s relate E for a whole cell to the reaction quotient, Q, for the net cell reaction. For the two half-reactions Right electrode: Left electrode: aA ne T cC dD ne T bB E° E° At equilibrium, E (not E°) 0. 14-5 E° and the Equilibrium Constant 283
1504T_ch14_270-297 01/28/06 14:05 Page 284 Box 14-4 Concentrations in the Operating Cell Why doesn’t operation of a cell change the concentrations in the cell? Cell voltage is measured under conditions of negligible current flow. The resistance of a high-quality pH meter is 10 13 . If you use this meter to measure a potential of 1 V, the current is I E R 1 V 10 13 1013 A If the cell in Figure 14-6 produces 50 mV, the current through the circuit is 0.050 V/10 13 5 10 15 A. This value corresponds to a flow of 5 10 15 C/s 9.649 10 4 C/mol 5 1020 mol e /s The rate at which Cd 2 is produced is 2.5 10 20 mol/s, which has a negligible effect on the cadmium concentration in the cell. The meter measures the voltage of the cell without affecting concentrations in the cell. If the salt bridge were left in a real cell for very long, concentrations and ionic strength would change because of diffusion between each compartment and the salt bridge. We assume that cells are set up for such a short time that mixing does not happen. the Nernst equation looks like this: E E E E° 0.059 16 n A c C log aE° A a A 0.059 16 n log A b B A d D b log a log b log ab To go from Equation 14-23 to 14-24: 0.059 16 log K E° n log K nE° 0.059 16 10 logK 10nE°/0.059 16 K 10nE°/0.059 16 0.059 16 E (E° E° ) log Ac C A d D E° n A a AA b B 14243 123 E° Q 0.059 16 log Q n (14-22) Equation 14-22 is true at any time. In the special case when the cell is at equilibrium, E 0 and Q K, the equilibrium constant. Therefore, Equation 14-22 is transformed into these most important forms at equilibrium: 0.059 16 Finding E° from K: E° log K (at 25°C) (14-23) n Finding K from E°: K 10nE°/0.059 16 (at 25°C) (14-24) Equation 14-24 allows us to deduce the equilibrium constant from E°. Alternatively, we can find E° from K with Equation 14-23. Example Using E to Find the Equilibrium Constant Find the equilibrium constant for the reaction Cu(s) 2Fe 3 T 2Fe 2 Cu 2 We associate E° with the half-reaction that must be reversed to get the desired net reaction. Solution The reaction is divided into two half-reactions found in Appendix H: 2Fe 3 2e T 2Fe 2 E° 0.771 V Cu 2 2e T Cu(s) E° 0.339 V Cu(s) 2Fe 3 T 2Fe 2 Cu 2 Then we find E° for the net reaction E° E° E° 0.771 0.339 0.432 V and compute the equilibrium constant with Equation 14-24: K 10 (2)(0.432)/(0.059 16) 4 10 14 Significant figures for logs and exponents were discussed in Section 3-2. A modest value of E° produces a large equilibrium constant. The value of K is correctly expressed with one significant figure, because E° has three digits. Two are used for the exponent (14), and only one is left for the multiplier (4). 284 CHAPTER 14 Fundamentals of Electrochemistry
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Fundamentals of Electrochemistry - W.H. Freeman

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