Fundamentals of Electrochemistry - W.H. Freeman

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1504T_ch14_270-297 02/15/06 7:12 Page 291 A 2 is the dominant form of ascorbic acid, and no protons are involved in the net redox reaction. Therefore, the potential becomes independent of pH. A biological example of E°¿ is the reduction of Fe(III) in the protein transferrin, which was introduced in Figure 7-4. This protein has two Fe(III)-binding sites, one in each half of the molecule designated C and N for the carboxyl and amino terminals of the peptide chain. Transferrin carries Fe(III) through the blood to cells that require iron. Membranes of these cells have a receptor that binds Fe(III)-transferrin and takes it into a compartment called an endosome into which H is pumped to lower the pH to 5.8. Iron is released from transferrin in the endosome and continues into the cell as Fe(II) attached to an intracellular metaltransport protein. The entire cycle of transferrin uptake, metal removal, and transferrin release back to the bloodstream takes 1–2 min. The time required for Fe(III) to dissociate from transferrin at pH 5.8 is 6 min, which is too long to account for release in the endosome. The reduction potential of Fe(III)-transferrin at pH 5.8 is E°¿ 0.52 V, which is too low for physiologic reductants to reach. The mystery of how Fe(III) is released from transferrin in the endosome was solved by measuring E°¿ for the Fe(III)-transferrin-receptor complex at pH 5.8. To simplify the chemistry, transferrin was cleaved and only the C-terminal half of the protein (designated Trf C ) was used. Figure 14-12 shows measurements of log5[Fe(III)Trf C ]/[Fe(II)Trf C ]6 for free protein and for the protein-receptor complex. In Equation 14-26, E E°¿ when the log term is zero (that is, when [Fe(III)Trf C ] [Fe(II)Trf C ]). Figure 14-12 shows that E°¿ for Fe(III)Trf C is near 0.50 V, but E°¿ for the Fe(III)Trf C -receptor complex is 0.29 V. The reducing agents NADH and NADPH in Table 14-2 are strong enough to reduce Fe(III)Trf C bound to its receptor at pH 5.8, but not strong enough to reduce free Fe(III)-transferrin. log {[oxidized form]/[reduced form]} 2.5 2.0 1.5 1.0 0.5 Free Fe-Trf c Fe-Trf c -receptor complex 0.0 E' ° E' ° –0.5 –0.5 –0.4 –0.3 –0.2 E (V vs. S.H.E.) –0.1 Figure 14-12 Spectroscopic measurement of log {[Fe(III)Trf C ]/[Fe(II)Trf C ]} versus potential at pH 5.8. [S. Dhungana, C. H. Taboy, O. Zak, M. Larvie, A. L. Crumbliss, and P. Aisen,“Redox Properties of Human Transferrin Bound to Its Receptor,” Biochemistry, 2004, 43, 205.] Terms to Understand ampere anode cathode coulomb current E°¿ electric potential electrode Faraday constant formal potential galvanic cell half-reaction joule Latimer diagram Nernst equation ohm Ohm’s law oxidant oxidation oxidizing agent potentiometer power reaction quotient redox reaction reducing agent reductant reduction resistance salt bridge standard hydrogen electrode standard reduction potential volt watt Summary Work done when a charge of q coulombs passes through a potential difference of E volts is work E q. The maximum work that can be done on the surroundings by a spontaneous chemical reaction is related to the free-energy change for the reaction: work G. If the chemical change produces a potential difference, E, the relation between free energy and the potential difference is G nFE. Ohm’s law (I E/R) describes the relation between current, voltage, and resistance in an electric circuit. It can be combined with the definitions of work and power (P work per second) to give P E I I 2 R. A galvanic cell uses a spontaneous redox reaction to produce electricity. The electrode at which oxidation occurs is the anode, and the electrode at which reduction occurs is the cathode. Two half-cells are usually separated by a salt bridge that allows ions to migrate from one side to the other to maintain charge neutrality but prevents reactants in the two half-cells from mixing. The standard reduction potential of a half-reaction is measured by pairing that half-reaction with a standard hydrogen electrode. The term “standard” means that activities of reactants and products are unity. If several half-reactions are added to give another half-reaction, the standard potential of the net half-reaction can be found by equating the free energy of the net half-reaction to the sum of free energies of the component half-reactions. The voltage for a complete reaction is the difference between the potentials of the two half-reactions: E E E , where E is the potential of the half-cell connected to the positive terminal of the potentiometer and E is the potential of the half-cell connected to the negative terminal. The potential of each half-reaction is given by the Nernst equation: E E° (0.059 16/n) log Q (at 25°C), where each reaction is written as a reduction and Q is the reaction quotient. The reaction quotient has the same form as the equilibrium constant, but it is evaluated with concentrations existing at the time of interest. Electrons flow through the circuit from the electrode with the more negative potential to the electrode with the more positive potential. Complex equilibria can be studied by making them part of an electrochemical cell. If we measure the voltage and know the concentrations (activities) of all but one of the reactants and products, the Nernst equation allows us to compute the concentration of the unknown species. The electrochemical cell serves as a probe for that species. Biochemists use the formal potential of a half-reaction at pH 7 (E°¿) instead of the standard potential (E°), which applies at pH 0. E°¿ is found by writing the Nernst equation for the half-reaction and grouping together all terms except the logarithm containing the formal concentrations of reactant and product. The combination of terms, evaluated at pH 7, is E°¿. Summary 291
1504T_ch14_270-297 1/23/06 11:35 Page 292 Exercises 14-A. A mercury cell formerly used to power heart pacemakers has the following reaction: Zn(s) HgO(s) S ZnO(s) Hg(l) If the power required to operate the pacemaker is 0.010 0 W, how many kilograms of HgO (FM 216.59) will be consumed in 365 days? How many pounds of HgO is this? (1 pound 453.6 g) 14-B. Calculate E° and K for each of the following reactions. (a) I 2 (s) 5Br 2 (aq) 6H 2 O T 2IO 3 10Br 12H (b) Cr 2 Fe(s) T Fe 2 Cr(s) (c) Mg(s) Cl 2 (g) T Mg 2 2Cl (d) 5MnO 2 (s) 4H T 3Mn 2 2MnO 4 2H 2 O (e) Ag 2S 2 O 2 3 T Ag(S 2 O 3 ) 3 2 (f) CuI(s) T Cu I 14-C. Calculate the voltage of each of the following cells. With the reasoning in Figure 14-8, state the direction of electron flow. (a) Fe(s) 0 FeBr 2 (0.010 M) NaBr(0.050 M) 0 Br 2 (l) 0 Pt(s) (b) Cu(s) 0 Cu(NO 3 ) 2 (0.020 M) Fe(NO 3 ) 2 (0.050 M) 0 Fe(s) (c) Hg(l) 0 Hg 2 Cl 2 (s) 0 KCl(0.060 M) KCl(0.040 M) 0 Cl 2 (g, 0.50 bar) 0 Pt(s) 14-D. The left half-reaction of the cell drawn here can be written in either of two ways: AgI(s) e T Ag(s) I (1) or Ag e T Ag(s) (2) The right half-cell reaction is H e T 1 2 H 2(g) (3) I − (aq) Ag wire coated with Agl(s) V – + Salt bridge H + (aq) Ag(s) Agl(s) Nal (0.10 M) HCI (0.10 M) H 2 (g, 0.20 bar) Pt(s) (a) Using Reactions 2 and 3, calculate and write the Nernst equation for the cell. (b) Use the value of K sp for AgI to compute E°[Ag ] and find the cell voltage. By the reasoning in Figure 14-8, in which direction do electrons flow? (c) Suppose, instead, that you wish to describe the cell with Reactions 1 and 3. We know that the cell voltage (E, not E°) must be the same, no matter which description we use. Write the Nernst equa- Pt E° 1.35 V Pt H 2 (g) in tion for Reactions 1 and 3 and use it to find E° in Reaction 1. Compare your answer with the value in Appendix H. 14-E. Calculate the voltage of the cell Cu(s) 0 Cu 2 (0.030 M) K Ag(CN) 2 (0.010 M), HCN(0.10 F), buffer to pH 8.21 0 Ag(s) by considering the following reactions: Ag(CN) 2 By the reasoning in Figure 14-8, in which direction do electrons flow? 14-F. (a) Write a balanced equation for the reaction PuO2 S Pu 4 and calculate E° for the reaction. 0.966 ? 1.006 PuO2 2 PuO ¡ 2 ¡ Pu 4 ¡ Pu 3 1.021 c (b) Predict whether an equimolar mixture of PuO 2 2 and PuO 2 will oxidize H 2 O to O 2 at a pH of 2.00 and P O2 0.20 bar. Will O 2 be liberated at pH 7.00? 14-G. Calculate the voltage of the following cell, in which KHP is potassium hydrogen phthalate, the monopotassium salt of phthalic acid. Hg(l) 0 Hg 2 Cl 2 (s) 0 KCl(0.10 M) 7 KHP(0.050 M) 0 H 2 (g, 1.00 bar) 0 Pt(s) 14-H. The following cell has a voltage of 0.083 V: Hg(l) 0 Hg(NO 3 ) 2 (0.001 0 M), KI(0.500 M) 7 S.H.E. From this voltage, calculate the equilibrium constant for the reaction In 0.5 M KI, virtually all the mercury is present as HgI 2 4 . 14-I. The formation constant for Cu(EDTA) 2 is 6.3 10 18 , and E° is 0.339 V for the reaction Cu 2 2e T Cu(s). From this information, find E° for the reaction 14-J. On the basis of the following reaction, state which compound, H 2 (g) or glucose, is the more powerful reducing agent at pH 0.00. CO 2 H HCOH HOCH 2H 2e HCOH HCOH CH 2 OH Gluconic acid pK a 3.56 e T Ag(s) 2CN E° 0.310 V HCN T H CN pK a 9.21 Hg 2 4I T HgI 2 4 CuY 2 2e T Cu(s) Y 4 E¿ 0.45 V CHO HCOH HOCH HCOH H 2 O HCOH CH 2 OH Glucose (no acidic protons) 14-K. Living cells convert energy derived from sunlight or combustion of food into energy-rich ATP (adenosine triphosphate) 292 CHAPTER 14 Fundamentals of Electrochemistry
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Fundamentals of Electrochemistry - W.H. Freeman

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