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238 Chapter 7 Electrochemistry
238 Chapter 7 Electrochemistry There is a powerful agent, obedient, rapid, easy, which conforms to every use, and reigns supreme on board my vessel. Everything is done by means of it. It lights, warms, and is the soul of my mechanical apparatus. This agent is electricity.Ž „ Jules Verne, 1870, in Twenty Thousand Leagues under the Sea, translated by Gerard Harbison Concepts The first clear human records of bioelectricity are inscribed on 5 th dynasty tombs in Egypt, dating from about 2400 BCE. These carvings show depictions of the electric Malaptererus electricus , native to the Nile. Our first encounter with bioelectric-ity, therefore, was likely painful; these fish can deliver jolts of up to 350 V. Somewhat later, mysterious earthen jars found in middle-eastern archaeological sites may have been the first chemical batteries, possibly used to do electroplating. However, systematic study of the interaction of electric currents with biological organisms had to wait until the 18 th century and the famous experiments of Luigi Galvani, who showed that electric shocks applied to frog legs caused the muscles to contract, and inspired Mary Shelleys novel Frankenstein, and later an entire genre of horror movies. Galvanis compatriot, Alessandro Volta, was one of the first to appreciate that, while biological organisms generate and are influenced by electricity, electric phenomena are not exclusive to life, and can be generated by an apparatus as simple as two disks of different metal in contact with each other and with an ionic solution. These simple electrochemical cells possess a characteristic potential. Under standard conditions, the electrochemical one of which is oxidized and one reduced, and it is proportional to the standard free energy of the redox (reduction�oxidation) reaction. There is in fact an identity between the electri-cal work done by an electrochemical cell and the free energy. The Nernst equation is the second link between thermodynamics and electricity, and relates the concentrations of chemical compounds at equilibrium and their electrochemical potential under nonstandard conditions. This link underpins the generation of bioelectrical potentials; cells pump ions across cell membranes, leading to electrical potentials, which are in turn sustained by equilibria with ion concentration gradients across these membranes. Biological organisms do not merely do chemistry; they also do electrochemistry. Applications The large voltages electric eels and catfish produce are generated by stacking literally thousands of cell membranes, each of which individually creates a potential of the order of 100 mV. The spectacular (one might even say shocking) voltages these electric fishes create are a bizarre evolutionary adaptation of a much more general phenomenon: the Basic Electricity | 239membrane potential. All cells carry membrane potentials. These potentials are the foun-dation of nerve transmission, and also drive the transport of most chemical compounds across biological membranes. Frogs, seaweed, and other organisms that live in contact with water have semiper-meable skins. Water and some ions and small molecules pass through the skins; macro-molecules generally do not. The frog or the seaweed can selectively concentrate certain mole

cules inside and selectively exclude or
cules inside and selectively exclude or excrete other molecules. How do they do it? If a molecule can easily pass through the skin, how can the inside concentration be main-tained at a value that is different from the outside concentration, and still be consistent with thermodynamics? The answer is to ensure that the free-energy change for transport-ing the molecule inside is negative. For example, the presence of a protein inside seaweed that strongly binds the iodide ion ensures that the iodide concentration in the seaweed is always higher than in the seawater. If the concentration of free iodide is the same inside and outside, the bound iodide would account for the concentrating effect of the seaweed. This effect, known as passive transport , does not depend on whether the seaweed is alive or dead. Similarly, metabolism is not involved in the transport of ligands like O 2 through the walls of the alveoli to bind to macromolecules like hemoglobin in the blood. Cells, however, also perform active transport . A number of different kinds of ion pumps, of which the most important is the sodium�potassium ATPase, actively transport ions across cell membranes, using the chemical energy provided by ATP hydrolysis to over-come unfavorable electrical and chemical potential gradients. The negative potential across the cell membrane allows positive ions such as Ca2+ and Mg2+ to accumulate by simple equilibration; it can also drive the transport of other, sometimes uncharged molecules using specialized membrane proteins such as symporters and antiporters. The electrical potential itself is exquisitely sensitive to the movement of small numbers of ions, and can under the right circumstance change rapidly, allowing the phenomenon of nerve conduction. It can also be brought into and out of equilibrium with much more robust ion concentration gradients, allowing long-term maintenance of stable cell potentials. Even more fundamentally, the processes of respiration and photosynthesis drive hydro-gen ions across the cell membranes of bacteria and the inner membranes of mitochondria and chloroplasts. This gradient of potential and of hydrogen ion concentration is, in turn, used to drive the energetically unfavorable synthesis of ATP from ADP and an inorganic phosphate. Thus the transduction of energy in all but a few organisms uses electrical transmembrane potentials as intermediates. Finally, humans have learned to use the Nernst equation to build sensors to convert chemical concentrations of oxidizable or reducible molecules, such as glucose, into electrical potentials, and thus build sensitive chemical sensors that allow rapid and reliable measure-ments of concentration of single chemical compounds in chemically complex mixtures such as blood; this is the basis of the glucose meters used in management of diabetes. Development of electrochemical sensors is an important contemporary area of applied chemical research. Basic Electricity The base electrical SI unit is the ampere or amp (A), which is technically defined as the electrical current which, passing through two infinitely long parallel wires a meter apart, gives a force between the wires of 2*10-7 N per meter of length. The amp is really not a fundamental unit, however; rather, it corresponds to a current or f

low of electrical charge Q , whic
low of electrical charge Q , which can be expressed as a differential: I=dQdt . (7.1) The SI unit of electrical charge is the coulomb (C): 1 A = 1 C s -1 . 240 Chapter 7 | Electrochemistry The smallest known quantity of charge on a free particle is the charge on the electron, which is often given the special abbreviation e. This charge has been measured to high accuracy; the best current estimate is 1.60217656*10-19 C. Even though the electron is a negatively charged particle, e is defined as a positive quantity; the charge on an electron, strictly speaking, is therefore -e . Ions have charges Z which are integer multiples of e ; the charge on a Mg 2+ ion, for example, is Ze=2e. What about a mole of ions, instead of 1 ion? If a single ion has a charge of e (in other words, Z = 1), the charge of a mole of ions, Qm=NAe=96485.3 C mol-1. This quantity of charge is given a special designation, the Faraday constant, or F . Just as moving a mass in the Earths gravity requires work, so does moving charge in an electrical potential. The SI unit of electrical potential is the volt (V). 1 joule of work is required to bring a charge of one coulomb from an infinitely distant point (which is assumed to have a potential of zero) to a point with an electrical potential of 1 V. In other words, wE=-QV. Electrical potential is a state variable, as is charge, and therefore electrical work w E , unlike mechanical work, is a state variable. We can also define electric power P as electrical work done per second: P=dwEdt . (7.2) Combining this with wE=QV and Eq. 7.1 , we obtain P=VI. The SI unit of power is the watt (W). Capacitance and Electrical Neutrality Just as a mass carries with it its own gravitational field, so a charge creates an electrical potential around it. For example, walking across a carpet in your socks rubs some electrons onto you from the carpet, or onto the carpet from you, so that you become charged up to a potential of perhaps thousands of volts, enough to cause sparks when you touch an object at a different potential and to give you a shock. It is truly shocking how few electrons it takes to give you a potential of thousands of volts. The quantity that determines the potential on an object per unit of charge it carriesis the capacitance ( C ): C = Q � V . The SI unit of capacitance is the farad (F); as you might expect, 1 F = 1 C�1 V. Calculating capacitances of real objects is beyond the scope of this book, but a simple formula gives the capacitance (actually, the self-capacitance) of a conducting sphere: C=4pe0r, where r is the radius of the sphere, and e0 is the electric constant: e0=8.854187817620*10-12 F m-1. A little math indicates that the capacitance of a liter of a solution of an electrolyte in water, in a round flask that occupies a sphere of radius 0.062 m, is 6.9*10-12 F. Therefore, unless they are specially designed to carry large quantities of charge, the capacitances of real objects are tiny! To charge that solution to 1 V requires 6.9*10-12 C of electrons or ions. But 6.9*10-12 C

corresponds to (6.9*10-12
corresponds to (6.9*10-12 C)�F mol of electrons or ions, or 7.2*10-17 mol, somewhat over 43,000,000 particles. A few tens of millions of electrons or ions, in other words, is enough to bring a 1 L solution to a potential of 1 V. So, if we have a 1 L salt solution that contains 1 mol of chloride ions and it contains 1+(7.2*10-17) mol of sodium ions, that solution will have a potential of approximately +1 V. This is a paradox which often confuses even quite advanced students. In electro-chemistry, we always assume there are equal quantities of positive and negative ions (or more correctly, equal positive and negative ionic charges) in a solution. But if the quantities are exactly equal, the solution should be neutral! The resolution of the paradox is that the excess of one species of ion, positive or negative, needed to create a potential of the order of a few volts, is unmeasurably small, and can always be neglected. In all electrochemi-cal calculations, the total Q of the positive ions is equal and opposite to the total Q of the negative ions. The Electrochemical Cell | 241 Ground and the Reference Potential While, in theory, the zero of potential is defined as the potential at an infinite dis-tance, in practice, this zero of potential cannot even be approximated. As we move away from acharge, we encounter other charges. If we leave the Earth for what we ordinar-ily consider the vacuum of space, we are immersed in a wind of positively charged protonsstreamingfrom the Sun. The galaxy itself has a large electromagnetic field, meaning that the zero of potential, if it exists even approximately, is millions of light-years away. For this reason, all practical potentials must be measured not with respect to an absolute zero of potential but to a reference potential. The reference potential in electronics is often the so-called ground potential; basically, we run a wire to a metal post sunk in the ground. In electrochemistry, as we will see, we often instead use a reference electrode to generate a reference potential. But regardless, there is no practical absolute zero of potential. Potential is in this respect like most other thermodynamic properties, such as energy, enthalpy, and so on, which can only practically be defined relative to some standard value. It is unlike entropy, for which a true zero value may be defined. The Electrochemical Cell Each of us is surrounded by dozens of electrochemical cells; we call them batteries, a term coined by Benjamin Franklin. They power our cellphones, smoke detectors, thermostats, our laptop or tablet computers, and our garage door openers. It is hard to believe the first documented electrochemical cell was invented only slightly more than two centuries ago, by the Italian scientist Alessandro Volta. Voltas pile was made up of pairs of silver and zinc disks, spaced by pieces of brine- or lye-soaked cloth. At the surface of the zinc disks, zinc metal dissolved in the brine, leaving two electrons: ZnSZn2++2e- The electrons flowed to the silver surface, where they reacted with water to give hydrogen gas: 2 H2O+2e-S2OH-+H2 Stacks of these disks gave estimated voltages between one end of the pile and the other of up to 50 V. Volt

a, in truth, didnt really understand ho
a, in truth, didnt really understand how his battery worked. He made some per-ceptive and accurate observations, such as that zinc gave a larger voltage than tin. But he thought the electrical potential was being generated by the contact between the different metals, rather than the solution. He thought the device could operate indefinitely, and did not appreciate that the corrosion of the zinc, which he saw as a problem to be remedied, was actually what was generating the electricity. Reversibility in the Electrochemical Cell It should be clear that Voltas cell was inherently irreversible. Zinc dissolved in the aque-ous electrolyte, but (at least at the beginning) the reverse process could not occur since there were no zinc ions in the solution. Likewise, there was no supply of hydrogen gas at the cathode, and so hydrogen could be generated but it could not react backwards to give hydrogen ions. Figure 7.2 shows a first attempt at a reversible electrochemical cell. We immerse a zinc electrode „ the anode „ in a solution of zinc ions (zinc sulfate will do nicely). We can similarly immerse a copper electrode „ the cathode „ in a solution of copper ions, and connect the two electrodes with a wire, so electrons can flow between them. Finally, FIGURE 7.1 Alessandro Voltas original drawing of his electrochemical pile „ the first documented electrical battery. The disks labeled Z and A are zinc and silver; the dark disks between them are brine-soakedcloth. 242 Chapter 7 | Electrochemistrysalt bridgeZn metalCu metalZn2+Cu2+V1.1 FIGURE 7.2 A Zn�Cu electro-chemical cell. Zinc dissolves at the cathode, leaving two elec-trons, which flow to the anode, where they react with Cu 2+ to plate out as copper metal. A salt bridge allows ions to flow between the two halves of the apparatus and maintains electri-cal neutrality. If the resistance is sufficient, the cell will develop a potential of approximately 1.1 V, negative on the zinc side. The exact value depends on the concentrations. we need a mechanism for ions to flow between the two solutions; as we calculated in the previous section, a very small number of electrons flowing from one metal to the other will create a substantial potential difference, which will shut the battery down. This is usually done with a salt bridge, a piece of porous material or a U-tube allowing ions to move between the two solutions, restoring electrical neutrality. If we make the zinc and copper solutions equal in concentration and place a voltmeter between the zinc and copper electrodes, the experiment shows us we will develop a potential of &1.1 V between the two electrodes, with the zinc electrode negative. This electrochemical cell is still not reversible; zinc will dissolve from the anode and copper will plate out on the cathode. However, if we apply an external positive potential to the cathode, relative to the anode, we can slow down, stop, or even reverse the electrochemi-cal process. At an external potential V ex of approximately 1.1 V on the cathode, relative to the anode, zinc stops dissolving and copper stops plating out. At a potential of greater than 1.1 V, zinc will, in fact, plate out on the anode and copper will dissolve at the anode. At the point the reaction

stops, we have achieved reversibility.
stops, we have achieved reversibility. The external potential needed to achieve this condition depends on the metals used to build the cell, and to a lesser extent on the concentration of the ions in the solution. We call this value of V ex the electrochemical potential , and give it the symbol E(cursive E) . If the experiment is done in the standard state (1 bar pressure, pure zinc and copper metals, solutions of Zn2+ and Cu2+ activity equal to exactly 1 m ), at reversibility, we measure a standard electrochemical potential E . Electrical Work, Electrochemical Potential, and Free Energy The work done by the cell by moving a charge Q against the external potential V ex is the electrical work, wE=QVex. If both Q and V ex have the same sign, the work is positive, because its being done by the surroundings on the system. Previously, when we discussed the First Law, we wrote U=q+w, with dw=-pex dV. This was the situation when there was only pressure-volume work. Now, however, when we also have electrical work, we have to add it to the pV contribution so that the First Law includes all contributions to the work: U=q+wpV+wE . (7.3) The Electrochemical Cell | 243 We can now use H = U + (pV) and G = H - TS, to obtain G=q+wpV+wE+(pV)-TS . (7.4) At constant pressure, (pV)=pV , and under reversible conditions, wpV =-pV . At constant temperature, under reversible conditions, S=q�T, or q=TS. Making these substitutions: G=TS-pV+wE+pV-TS=wE . (7.5) In other words, the electrical work at constant pressure and temperature, under reversible conditions, is the free energy change of the reaction. Some books refer to this as the useful work, as if pV work isnt useful! It is probably more accurate to say that in the most general case, what we call w E above is the non- pV work. We can now insert our definition of electrical work to obtain G=QVex . (7.6) Since were generally interested in chemical processes, the charge per mole is just 1 Faraday, multiplied by the number of electrons, n e , involved in oxidizing or reducing one mole of reactant, and by -1, since the electronic charge is negative. n e is always a positive integer. If were in an electrochemical cell under reversible conditions, Vex=E, and so rGm=-veFE . (7.7) Its usual to drop the subscripts denoting constant pressure and constant temperature, because we almost always do electrochemistry under these conditions. If G is negative, E is positive, and we can thence deduce that a positive electrochemical potential corresponds to a reaction that goes spontaneously in the forward direction. Finally, under standard conditions: rG=-veFE . (7.8) We can obtain expressions for S and H in a straightforward fashion: a0rGm0Tbp=-rSm rSm=veF a0E0Tb

p (7.9) or rHm=
p (7.9) or rHm=rGm+TrSm=veF cE+Ta0E0Tbpd . (7.10) By measuring the electrochemical potential as a function of temperature, therefore, we can obtain the entropy, enthalpy, and free energy of reaction. Standard Electrochemical Potentials Because the electrochemical potential depends only on the free energy of reaction ( Eq. 7.8 ), it is a state function, and for that reason it is additive. So, for example, if we build a standard cell with zinc as the anode and copper as the cathode „ which is written in conventional electrochemical notation as: ZnZn2+(1m)  Cu2+(1m)Cu E=1.1037 V The single vertical line denotes a phase boundary (in this case, the electrode surface); a double line denotes a salt bridge. We can also build one with a copper anode and a silver cathode: CuCu2+(1m)  Ag+(1m)Ag E=0.4577 V 244 Chapter 7 | Electrochemistry The potential of a zinc�silver cell can be obtained by adding these two together: ZnZn2+(1m)  Ag+(1m)Ag E=1.5614 V Note that it is impossible to measure the potential for just a single reduction or oxidation. A cell must always include a species being oxidized and another being reduced. To discuss the electrochemical potentials of single species „ a half-cell „ therefore, we must define a standard half-cell. Chemists have chosen the reduction of H + to H 2 gas under standard conditions to have a standard electrode potential of zero: H+ (aq)+e-S½ H2 (g) E=0 V To be a standard cell, the hydrogen ions and the hydrogen gas must each have an activ-ity of 1; this means the molality of H + is somewhere close to 1 m and the gas pressure is approximately 1 bar. It is obvious we cannot build an electrode out of hydrogen gas, so instead we use a platinum electrode coated with finely divided platinum (platinum black). The oxidation of platinum metal has an electrochemical potential far more negative (unfavorable) than that of hydrogen gas; in addition, the platinum surface catalyzes the electrochemical reaction between H + and H 2 . The standard hydrogen electrode (SHE) is depicted in figure 7.3 . In practice, because the activity coefficients for hydrogen ions at molal concentrations are quite different from 1, it is preferable to measure cell potentials versus a SHE with a much lower H + concentration (ideal-dilute solution), and correct for concentration using the Nernst equation (see below). The SHE is therefore something of a theoretical construct, although recent research indicates an approximately 0.85 m solution of HCl in water has a  1 for H+. In addition to measuring other half-cells coupled to a SHE, all standard poten-tials are tabulated as reductions, regardless of whether the potential is favorable or unfavorable. So, for example, if we measure the half cell potential for ZnZn2+, it is tabulated as Zn2+(aq, a = 1)+2 e-SZn (s) E=-0.7618 V even though, coupled to a SHE, the zinc half-cell reaction would spontaneously proceed in the other direction, as shown by the sign of the voltage. In order to obtain the electroche

mical potential for a redox reaction fro
mical potential for a redox reaction from two half-cell potentials, we do the following: 1. Reverse reactants and products for the oxidation half-reaction and multiply the electrochemical potential of the oxidation half-reaction by -1. 2. Multiply the two reactions by integer(s) so as to make the number of elections equal for oxidation and reduction steps. 3. Add the reactions and the electrochemical potentials. When balancing the electrons, we do not multiply the electrochemical potentials by the same integers. The number of electrons per mole is explicitly included in Eq. 7.8 . EXAMPLE 7.1 Obtain a balanced chemical equation, and calculate the standard electrochemical potential for the reduction of Fe3+ to Fe (metallic iron) by Zn metal. H2(1 bar)Pt metalH+(a = 1) FIGURE 7.3 The standard hydrogen electrode (SHE). The Electrochemical Cell | 245 SOLUTION The two half-reactions are: Zn2++2e-SZn E=-0.7618 Fe3++3e-SFe E=-0.037 (1) Reverse the oxidation step and multiply the electrochemical potential by -1: ZnSZn2++2 e- E=0.7618 (2) Multiply the two reactions by integers: 3 ZnS3 Zn2++6e- E=0.7618 2 Fe3++6e-S2 Fe E=-0.037 (3) Add the reactions and the electrochemical potentials: 3 Zn+2 Fe3+S2 Fe+3 Zn2+ E=0.725 Whether we write the equation this way, or as Zn+2/3 Fe3+S2/3 Fe+Zn2+ , the electrochemical potential is the same. Concentration Dependence of E The standard potentials in table 7.1 can be converted to standard free energies using Eq. 7.8 . What if we are interested in the potentials at non-standard concentrations? We start with the relation between standard Gibbs free energy and molar Gibbs free energy: rGm=rG+RT ln Q . (7.9) Substituting Eq. 7.7 for rGm and 7.8 for rG : -veFE=-veFE+RT ln Q . (7.10) Simplifying, we obtain the Nernst equation : E=E-RTveF ln Q . (7.11) The reaction quotient is as defined previously ( Eq. 4.16 ). At 25°C, we can make use of the identity that ln Q = ln 10 log 10 Q = 2.303 log 10 Q , and substitute 298 K for T , to get the useful rule of thumb: E = E-0.0591 Vve log Q . (7.12) The Nernst equation can be used to calculate the potential for any concentrations of reac-tants and products in a cell. At equilibrium, rG=0 and therefore E=0 , so the left hand side of Eq. 7.11 is zero and we can substitute K for Q . Thus, the standard potential gives the equilibrium constant for the reaction in the cell: E=RTveF ln K . (7.13) 246 Chapter 7 | Electrochemistry TABLE 7.1 Standard Reduction Electrode Potentials at 25°C Oxidant/Reductant Electrode reaction E(V) E(V) (pH 7) K+�K K++e-SK …2.931 Ca2+�Ca Ca2++2e-SCa …2.868 Na+�Na Na++e-SNa …2.71 Mg2+�Mg

Mg2++2e-SMg …2.372
Mg2++2e-SMg …2.372 Mn2+�Mn Mn2++2e-SMn …1.185 Zn2+�Zn Zn2++ 2e-SZn …0.7618 Acetate�acetaldehyde CH3COO-+2e-+3H+SCH3CHO+H2O …0.105 …0.586 Fe2+�Fe Fe2++ 2-eSFe …0.447 H+�H2�Pt 2H++2e-SH2 0 …0.414 CO2�formate CO2 (aq)+H++2e-SHCOO- …0.181 …0.409 Ferredoxin Fd[Fe3+]SFd[Fe2+] …0.395 FMN(Mitochondrial complex I) FMN+2H++2e-SFMNH2 …0.380 Gluconolactone�glucose C6H10O6+2H++2e-SC6H12O6 …0.345 L-cystine�L-cysteine L@cystine+2H++2e-S2 L@cysteine …0.355 Fe3+�Fe Fe3++2e-SFe …0.037 Acetoacetate�b-hydroxybutyrate CH3COCH2COO-+2H++2e-SCH3CHOHCH2COO- …0.346 NADP+�NADPH NADP++H++2e-SNADPH …0.339 NAD+�NADH NAD++H++2e-SNADH …0.324 Glutathione(S-S)�GlutathioneSH GSSG+2H++2e-S2 GSH …0.240 FMN�FMNH 2 FMN+2H++2e-SFMNH2 …0.213 FAD�FADH2 FAD+2H++2e-SFADH2 …0.212 Acetaldehyde�ethanol CH3CHO+2e-+2H+SCH3CH2OH +0.221 …0.193 Pyruvate�lactate CH3COCOO-+2H++2e-SCH3CHOHCOO- …0.184 Oxaloacetate�malate -OOCCOCH2COO-+2H++2e-S-OOCCHOHCH2COO- …0.158 FAD�FADH2(Complex II) FAD++2H++2e-SFADH2 …0.05 Fumarate�succinate -OOCCHCHCOO-+2H++2e-S-OOCCH2CH2COO- +0.045 Myoglobin Mb[Fe3+]+e-SMb[Fe2+] +0.046 Ubiquinone UQ+2H++2e-SUQH2 +0.052 Dehydroascorbate� ascorbate DHA+H++2e-SAsc- +0.080 AgCl�Ag�Cl- AgCl+e-SAg+Cl- +0.222 Cytochrome c Cyt@c[Fe3+]+e-SCyt@c[Fe2+] +0.254 Calomel 1�2 Hg2Cl2+e-SHg+Cl- +0.268 O2�H2O2 O2+2H++2e-SH2O2 +0.695 +0.281 (continues)Transmembrane Equilibria | 247 EXAMPLE 7.2 The enzyme glutathione reductase replenishes the cells supply of glutathione (GSH), regenerating two molecules of GSH from a single molecule of oxidized glutathione (GSSG), using NADPH as a source of two reducing equivalents. Using the data in table 7.1 and a typical cellular NADP�NADPH+ ratio of 0.005, calculate the equilibrium cellular concentration of GSSG at pH 7 and 25°C, if the GSH concentration is 4 mM. SOLUTION The two half-reactions are: GSSG+2H++2e-S2 GSH E=-0.240 V NADP++H++2e-SNADPH E=-0.339 V (1) Reverse the second reaction and multiply the electrochemical potential by -1: NADPHSNADP++H++2e- E=+0.339 V (2) The number of electrons is bala

nced. (3) Add the reactions and the el
nced. (3) Add the reactions and the electrochemical potentials: GSSG+H++NADPHS2 GSH+NADP+ E=+0.099 V K=[GSH]2eq[NADPH]eq[GSSG]eq[NADP+]eq=(4*10-3)20.005[GSSG]eq We do not include the H+ concentration, since the activity of H+ is defined as 1 at pH7 in the biochemists standard state. Solving Eq. 7.13 gives us K = 2223, which in turn gives [GSSG]=1.44*10-7 M = 0.144 M. Healthy cells keep GSSG concentrations very low! Transmembrane Equilibria The membranes of cells and organelles, such as chloroplasts and mitochondria, are good electrical insulators and are impervious to ions. They can therefore sustain transmembrane potentials of hundreds of millivolts and large concentration gradients. Cells and organelles in fact dedicate a considerable part of their genomes to proteins that pump ions across cell Oxidant/Reductant Electrode reaction E(V) E(V) (pH 7) Cu2+�Cu Cu2++2e-SCu +0.342 I2�I- I2+2e-S2 I- +0.535 Ag+�Ag Ag++e-SAg +0.800 O2�H2O O2+4H++4e-S2H2O +1.229 +0.815 NO3 -�NO2 - NO3 - + 3H++2e-SHNO2+H2O +0.934 Br2(aq) �Br- Br2+2e-S2Br- +1.087 Cl2(g) �Cl- Cl2+2e-S2Cl- +1.358 Mn3+�Mn2+ Mn3++e-SMn2+ +1.541 F2(g) �F- F2+2e-S2F- +2.866 TABLE 7.1 Standard Reduction Electrode Potentials at 25°C (cont.)248 Chapter 7 | Electrochemistrymembranes, that act as channels (often gated channels) across cell membranes, and that can use these gradients of concentration and potential to transport other materials. As before, the work done in transporting a charge Q across a cell membrane with potential V is wE=QV. As before, we can equate w E with  G , and adapt Eq. 7.7 : rGm=ZFV . (7.14) Since we are now discussing the movement of ions, rather than electrons, we have replaced -ve with Z , the charge on the ion; and E with V , the transmembrane potential. Moving a positive charge to a more positive potential requires work and a positive free energy change. This molar free energy difference can be written as the electrical part of an electrochemical potential difference, m: mE=ZFV . (7.15) If there is also a concentration gradient across the membrane, the electrochemical potential also has a chemical component (equivalent to the regular chemical potential discussed in chapter 4 ) : mC=m+RT ln Q . (7.16) Since both the inside and outside of cells or organelles are aqueous media, the standard chemical potentials are the same. We can therefore write the total electrochemical potential as a sum of electrical and chemical terms: m=mE+mC=ZFV+RT ln Q . (7.17) In this case, Q=ain/aout, the ratio of ion activities inside and

outside the cell. Again, since these en
outside the cell. Again, since these environments are often very similar, the activity coefficients are often nearly identi-cal, and we can replace activities by concentrations. Finally, at equilibrium, the difference in chemical potentials is zero, and we have: ZFV=-RT ln K or V=-(RT/ZF) ln K (7.18) Compare this with Eq. 7.13 . EXAMPLE 7.3 The plasma membrane of mammalian cells typically has a potential V=Vin-Vout  -70 mV. It also contains pores that allow potassium to selectively equilibrate across the membrane. If extracellular potassium is typically 5 mM, calculate the concentration of potassium inside the cells at 37°C, assuming activities can be replaced by concentrations. SOLUTION For K+, Z=+1, and therefore RT�ZF=8.31447 * 310K�(1 * 96485.3) = 0.0257 V. Since V = -0.07 V, by Eq. 7.18 , ln Q = 2.62, or Q=[K+ in]�[K+ out]=13.7. This means [K+ in]=13.7*0.005=0.069 M or 69 mM. Donnan Effect and Donnan Potential In chapter 6 , we described e quilibrium dialysis as an easy way to measure binding of a small molecule ligand by a macromolecule. The method becomes more complicated if the macromolecule and ligand are charged. The requirement that the solutions on each side Transmembrane Equilibria | 249of the dialysis membrane must be electrically neutral means that there can be an apparent increase in binding of a ligand with the opposite charge to that of the macromolecule and a decrease in binding of a ligand with the same charge. These effects depend on the net charge on the macromolecule and are not caused by binding at specific sites. The effect of the net charge of the macromolecule on the apparent binding of a ligand can be minimized by using high concentrations of a salt not involved in binding. Thus, the charged macro-molecule and ligands are not the main contributors to the total concentration of ions in the solutions. When equilibrium (except for the macromolecule) is reached for charged species, a voltage is developed across the membrane. The asymmetric distribution of ions caused by the charged macromolecule is called the Donnan effect, and the transmembrane potential is called the Donnan potential. It is observed experimentally when a charged macromolecule is dialyzed in the presence of an electrolyte; it also contributes slightly to the membrane potential of most cells. The experimental apparatus is shown in figure 7.4 . A dialysis mem-brane (a semipermeable membrane with a pore size that allows small molecules, typically under 3000 Da, to pass through, but impermeable to large molecules) encloses a solution of a macromolecule in an electrolyte solution; outside the membrane is the same solution without the macromolecule. For simplicity, we consider the dialysis of a macromolecule with a net charge of ZM and concentration cM against NaCl solution. Outside the membrane, there must be charge balance, and so the electrolyte (say, NaCl) must have equal numbers of positive ( Na+ ) and negative ( Cl- ) ions. Electrical neutrality requires that cNa, out=cCl, out=c , (7.19) where c is the concentration of NaCl outside the membrane at

equilibrium. Inside the membrane, we a
equilibrium. Inside the membrane, we also have electrical neutrality, but the macromolecule charge contributes to it. Summing up over all charges inside: cNa, in-cCl, in+ZMcM=0 . (7.20) At equilibrium, because the membrane is permeable to Na + and Cl- , mNa, in=mNa, out (7.21a) mCl, in=mCl, out . (7.21b) From Eq. 7.17 , we get: mNa=mNa, in-mNa, out=FV+RT ln (aNa, in�aNa, out)=0 (7.22a) mCl=mCl, in-mCl, out=-FV+RT ln (aCl, in�aCl, out)=0 . (7.22b) If we make the reasonable assumption that the activity coefficients are approximately equal inside and outside the membrane, then: FV=-RT ln (cNa, in�cNa, out)=RT ln (cCl, in�cCl, out) . (7.23) and therefore -ln (cNa, in�cNa, out)=ln (cNa, out�cNa, in)=ln (cCl, in�cCl, out) . (7.24) mVelectrolyte electrolyte+macromolecule FIGURE 7.4 Equilibrium dialysis apparatus for measuring the Donnan effect. 250 Chapter 7 | Electrochemistry Exponentiating and cross multiplying, and substituting Eq. 7.19 : cNa, out cCl, out=cNa, in cCl, in=c2 . From Eq. 7.20 we have cCl, in = cNa, in + ZMcM, so c2 - cNa, in(cNa, in + ZMcM) = 0 . (7.25) This is a quadratic equation, whose positive root is cNa, in=-ZMcM+2(ZMcM)2+4c22 . The ratio r of cNa, in to cNa, out is r = cNa, inc = -ZMcM2c + BaZMcM2cb2+1 . (7.26a) From Eq. 5.31 , w e can also calculate the ratio of Cl- inside to outside: cCl-, incCl-, out=ccNa, in=1r . (7.26b) The equations above show that for a positively charged macromolecule the concentration of positive ions inside will be less than that outside (r61) and the concentration of negative ions inside will be greater than outside. For a negatively charged macromolecule, r71, and the concentration of positive ions is greater inside than outside. For example, for a macromolecule with a net positive charge ZM=10 at a concentration of 1 mM (cM=10-3), dialyzed against 0.100 M NaCl ( c = 0.100), Eq. 7.26a gives r = 0.95. This means that the ratio of Na+ inside to outside is 0.95, and the ratio of Cl- inside to outside is 1/r=1.053. Increasing the positive charge or concentration of a macromolecule will make r decrease; increasing the concentration of NaCl will make r approach 1. The asymmetry in Na+ or Cl- concentration on the two sides of the membrane is surpris-ing because the membrane is permeable to Na+ and Cl- and we might think that the relations mNa, in=mNa, out and mCl, in=mCl, out would predict equal concentrations of Na+ and Cl- on the two sides of the membrane. There is, however, a potential difference V, the Donnan potential, across the membrane , as illustrated in figure 5.13 . From Eq. 7.22a we obtain V=-RTF ln aNa, inaNa, out-RTF ln cNa, inc

Na, out-RTF ln r . (7.27
Na, out-RTF ln r . (7.27) From the example above with r = 0.95, for Na+ at 298 K, we can calculate a transmembrane voltage of +1.3 mV. The plus sign means that the electrical potential is higher inside than outside the membrane because the positively charged macromolecule is inside. We of course calculate the same value for the voltage if we use the ratio of Cl- concentration inside and outside. Plasma Membrane Potentials and the Na - K ATPase In the example above, potassium concentrates passively inside the cell, responding to a gradient of electrical potential. This raises the question, however, of how the membrane potential itself is set up. The answer is that we have identified systems that undertake active transport of ions, a process that is dependent on active cellular metabolism. Active transport Transmembrane Equilibria | 251is defined as the transport of a substance from a lower to a higher chemical potential. Because the total free-energy change of the process must be negative, active transport is tied to a chemical reaction that has a negative free-energy change. In a biological system, this means that metabolism is occurring and driving the pump. Therefore, an experimental test of whether active transport is involved is to poison the metabolic activity to see whether the transport also stops. A classic example is the membrane Na+ -K + ATPase, sometimes called the sodium…potassium pump , which uses the free energy of hydrolysis of ATP to pump Na+ ions out of the cell and K+ ions into the cell. The net reaction for the active transport is thought to be: 3Na+in+2K+out+ATPS2 K+in+3Na+out+ADP+Pi Representative concentration differences across a typical animal cell membrane are illus-trated in figure 7.5 . There is also a voltage difference of about -0.07 V across the cell membrane; the inside is negative relative to the outside as shown. This membrane potential arises from the action of the Na+ - K+ ATPase pump, and also from the permeability of themembrane to potassium and the impermeability to sodium. Thus, the magnitude of the potential is determined largely by the ratio [K+ in]�[K+ out]. The Na+ -K + ATPase is inhibited by cardiotonic steroids, such as digoxin, obtained from the foxglove plant, or ouabain, a dart poison obtained from African trees. Cardiotonic steroids at low levels make the heartbeat more powerful; at high levels, they shut down the ion pump completely, paralyzing the heart. The crystal structures of the Na+ -K + ATPase in various stages of its reaction cycle, and in its ouabain-inhibited form, have recently been determined. A mechanism for the reaction is given in figure 7.6 . Binding of ATP causes a conformational change in the protein that narrows the ion channel, thus favoring the exchange of smaller Na+ ions for larger K+ ions inside the cell. The Na+ ions catalyze covalent attachment of the ATP, whose hydrolysis, followed by detachment of the ADP, causes the channel to widen and open on the outside, favoring exchange of the bound Na + for extracellular K+ . Detachment K+K+ATP(f)PiK+K+

(e)ATPNa+Na+Na+(a)PiK+K+(
(e)ATPNa+Na+Na+(a)PiK+K+(d)PiNa+Na+Na+ADP(b)Na+Na+(g)PiNa+Na+ADP(c)Na+Extracellular fluidCytosolATPNa+ FIGURE 7.6 The Na+ - K+ ATPase pump cycle. In state (a), the pore contains three sodium ions and is cova-lently linked to an ATP. In step (b), the ATP is hydrolyzed to bound ADP and P i . Dissociation of the bound ADP in step (c) leads to a conformational change where the pore widens and opens up on the extracellular face of the membrane, allowing Na+ ions to escape and be replaced by K+ ions (d). These induce dephos-phorylation (e) and cause the pore to close. An ATP can then bind (f) causing the pore to open on the cyto-solic side, and narrow, forcing out the K+ ions and allowing Na+ to bind (g). These then catalyze a covalent link between ATP and the protein, which in turn causes the pore to close again (back to a). Inside[Na+] = 10 mM [K+] = 70 mM V = Ð70 mV Outside[Na+] = 140 mM [K+] = 5 mMCell membraneCell membrane FIGURE 7.5 Typical ion concentrations inside and outside the plasma membrane. 252 Chapter 7 | Electrochemistryof the covalently bound phosphate then closes the channel on the outside and readies it for binding of another ATP. It is important to make sure that the detailed mechanism is consistent with the laws of thermodynamics. If it is not, we know that we have left something out. The free energy for the net reaction must be negative. This means that the positive free energy of actively transporting ions against concentration and voltage gradients must be more than balanced by the negative free energy of ATP hydrolysis. Lets divide the process into three parts: the transport of Na+ ions out, the transport of K+ ions in, and the hydrolysis of ATP. The free energy per mole difference between Na+ inside and outside is given by Eq. 7.19 : m=ZFV+RT ln ainaout . (7.19) We replace ratios of activities by ratios of concentrations and use a temperature of 37°C: m=1 * 96485.3 * -0.07 + 8.31447 * 310.15 * ln 10140=-13559 J�mol But the pump works in the opposite direction (it propels the ion from inside to outside) and so m=13.6 kJ�mol. The free energy per mole for the transport of K+ ions is m=1*96485.3*-0.07+8.31447*310.15*ln 1005=971 J�mol . The electrical potential difference of -70 mV favors transport of positive ions in; thus, the favorable effect of the electric field, attracting positive ions toward the inside of the cell, nearly cancels the effect of the concentration gradient for potassium ions, which tends to cause them to diffuse out of the cell. In contrast, moving positively charged sodium ions out of the negatively charged cell interior is unfavorable. The free energy for the hydrolysis of ATP to ADP and inorganic phosphate (P i ) is m=m+RT ln [ADP][Pi][ATP] . m is -31.3 kJ mol -1 for ATP hydrolysis at 310 K. Typical concentrations in healthy muscle cells are [ATP] = 1 mM, [ADP] = 40 M, and [P i ] = 25 mM. This gives Q = 0.001: m=-31300+8.31447

*310.15*ln(0.001)=-49133 J�
*310.15*ln(0.001)=-49133 J�mol We can now account for all three free-energy changes: n m (kJ mol -1 ) rGm (kJ mol -1) Na+ inSNa+ out 3 13.56 40.7 K+ outSK+ in 2 0.97 1.9 ATP+H2OSADP+Pi -49.1 ATP + H2O + 3Na+in+ 2K+outSADP + Pi + 3Na+out + 2K+in -6.5 Clearly, the free energy of ATP hydrolysis is sufficient to account for the active trans-port of the ions. However, the free-energy calculation does not prove the mechanism. It only shows that the mechanism is a possible one that does not violate thermodynamic principles. This is a very important test, however, because if the calculation did not give a net negative free energy, the mechanism would have been immediately disproved. In fact, any sign this reaction is departing from the canonical conditions of low ADP, Transmembrane Equilibria | 253high intracellular potassium, and low intracellular sodium is usually a sign of imminent celldeath! The transmembrane electrical and chemical potential differences are linked but com-plementary in function. The electrical potential acts on all ions, and can therefore facilitate the accumulation of positively charged ions such as Ca2+ and Mg2+ in the cytoplasm. On the other hand, it is exquisitely fragile; movement of a few thousand ions can collapse or even reverse the membrane potential (see problem 7.1)! The chemical potential difference is far more robust, since it involves a macroscopic concentration difference and therefore trillions or quadrillions of ions per cell. In some cases, it is directly coupled to the electri-cal potential difference by, for example, potassium channels that allow equilibration of K+ across the membrane. This allows the robust chemical potential difference to buffer the electrical potential. In other cases, the chemical potential for a particular ion is decoupled from the electrical potential, as is the case most of the time for Na+ ions. Excitable membranes, such as those in the sensory nervous system of animals, take advantage of the possibilities available by alternately coupling and decoupling the electri-cal potential with the chemical potential of an ion. These membranes contain separate Na+ and K+ channels, both of which are voltage-gated . In response to the membrane potential reaching a certain value, the Na+ channels open, allowing inflow of the far-from- equilibrium Na+ ions and causing the membrane potential to collapse and even reverse. This causes the gate to close, and potassium channel gates to open, restoring the normal cell potential. The resulting wave of (relatively) positive electrical potential can travel down the nerve cell axon, carrying a neuronal signal. These channels can also respond to activators such as Ca2+, and can be blocked by inhibitors such as tetrodotoxin (the puffer fish poison). Using the techniques of modern electrophysiology, the currents associated with single channels can be monitored, as illustrated in figure 7.7 . By recording the current flowing across a small region of membrane-containing neuronal channels, the pattern of o

pening and closing of individual channe
pening and closing of individual channels is seen as pulses of current that occur intermittently and with a range of duration. When one channel is open, an ionic current (upward deflection) flows; when two channels are open simultaneously, twice as much current flows. When the channel gates close, the current falls to zero. The effect of applying a depolarizing transmembrane potential on the magnitude of current flowing in an open channel is also seen in figure 7.7 . +20 mV+50 mV+100 mV0 ms2.5 ms5 pA FIGURE 7.7 Simulation of single channel currents in a small area of cell membrane that includes two potassium channels. A single channel opening is seen as a current step upward from zero (the baseline value). As the applied (depolarizing) potential increases, the channels are more likely to open, and also the flow of ions increases, because the potassium electrochemical potential is further from equilibrium. Occasionally, two channels open at the same time, particularly on the bottom trace. 254 Chapter 7 | Electrochemistry Photosynthetic energy conversion involves H + gradients (pH) across the thylakoid membrane (see figure 6.21 ) . Light activation of PS1 and PS2 pumps H + ions from the stromal to the lumenal phase. A transmembrane electric field V (positive inside) also results from the electron transport induced by PS1 and PS2. Both the pH and the V provide the source of electrochemical potential to power the ATPase, a tiny molecular machine that covalently links ADP with phosphate to produce high chemical potential ATP. The ATP in turn provides a source of chemical potential for a variety of biochemi-cal reactions, including those involved in converting CO 2 to sucrose, cellulose, and other plant materials. Biological Redox Reactions and Membranes Oxidation…reduction reactions are essential for energy storage and conversion in biologi-cal organisms. For example, the pyruvate produced as a product of glycolysis ( figure 4.13 ) undergoes oxidative decarboxylation to form acetyl coenzyme A, which then enters the citric acid cycle. In the citric acid cycle, the acetate is coupled to oxaloacetate to form the six-carbon molecule, citrate. Citrate then undergoes a series of at least eight reac-tions that involve progressive oxidation of two of the carbon atoms to CO 2 and return of the remaining four-carbon portion as oxaloacetate for reentry into the cycle. Coupled to the oxidation of one mol of acetate to 2 mol CO 2 is the reduction of 3 mol of NAD+ to NADH and the production of 1 mol of reduced flavin adenine dinucleotide (FADH 2 ). This reducing power generated in the citric acid cycle is subsequently used for biosyn-thetic reactions and for the formation of ATP by oxidative phosphorylation (respiration) in mitochondria. The step that completes the citric acid cycle is the oxidation of malate to oxa-loacetate coupled with the reduction of NAD+ to NADH by the enzyme malate dehydrogenase: CCOOÐCOOÐCHHHOH+ NAD+CCOOÐCOOÐCOHH+ NADH + H+malatedehydrogenasemalateoxaloacetate To calculate the rG associated with this reaction, we can use E values for the corresponding redox half-reactions listed in table 7.1

: Malate+2H++2e-Soxaloaceta
: Malate+2H++2e-Soxaloacetate E=-0.158 V NAD++H++2e-SNADH E=-0.324 V E=-0.324-(-0.158)=-0.166 V rG=-2*(96485.3 C mol-1)*-0.166 V=32.0 kJ�mol This step in the cycle is not spontaneous under standard conditions. However, it will become spontaneous under conditions where the demand for NADH is high; that is, when [NADH]/[NAD+] is low. Under these conditions, the malate will undergo conversion to oxaloacetate, which serves to turn on the citric acid cycle. This is an example of metabolic regulation whereby supply and demand are kept in reasonable balance. Biological Redox Reactions and Membranes | 255 Oxidative Phosphorylation A principal source of cellular ATP is the process of oxidative phosphorylation (respira-tory electron transport) carried out in mitochondria. * This is a series of electron transfer reactions catalyzed by four membrane-associated enzyme complexes whereby NADH is oxidized, ultimately by O 2 . These complexes are depicted in fig 7.8 . The complexes I, III, and IV form a super-complex in the mitochondrial inner mem-brane called a respirasome . However, it is useful to analyze the thermodynamics of the respirasome in terms of each of the four enzyme complexes individually. NADH-Q Reductase (Complex I) NADH, which is a soluble small molecule reductant produced in the citric acid cycle and elsewhere, is imported into the mitochondria. It undergoes oxidation at the enzyme complex I, formally called NADH-Q reductase. Complex I is a large (100 kDa) , complex protein, with up to 46 subunits, that contains a flavin mononucleotide, (FMN) cofactor, and nine iron-sulfur centers. Initially, NADH reduces the FMN, which then feeds the pair of electrons singly down a chain of iron-sulfur (FeS) centers. Finally, the electrons are passed on to reduce ubiquinone (UQ). The reduced ubiquinone is then released to travel to complex III; being hydrophobic, it probably diffuses predominantly within the membrane itself. All of the redox activity takes place in the peripheral part of the complex, which protrudes into the mitochondrial matrix (the center of the mitochondrion); it is however coupled to a membrane-bound region that translocates four protons from the matrix to the outside of the mitochondrial inner membrane (the intermembrane space ). The reduction half-reactions taking place in the complex are: NAD++H+ in+2e-SNADH E=-0.324 V UQ+2H+ in+2e-SUQH2 E=+0.052 V complex IsuccinateFADFADH2UQH2UQfumarate4 H+NAD+NADH + H+FMNFMNH2UQH2UQUQH2UQ2 cyt-c (Fe3+)2 cyt-c (Fe2+)2 H+½ O2+ 2H+H2O2 cyt-c (Fe2+)2 cyt-c (Fe3+)2 H+Mitochondrial intermembrane spaceMitochondrial matrixcomplex IVcomplex IIIcomplex II FIGURE 7.8 The mitochondrial respiratory chain, which contains two pathways that merge at complex III. The first pathway begins at NADH, proceeds through reduced ubiquinone ( UQH2 ) and reduced cytochrome- c (cyt- c ) and ends in the reduction of oxygen to water. The second pathway begins at succinate, but also ends in the reduction of water using UQH2 and cyt-

c as intermediates. * Mitochon
c as intermediates. * Mitochondria are found in eukaryotic cells. They are membrane-surrounded structures in which oxidative metabolism occurs. 256 Chapter 7 | Electrochemistry Reversing the first half-reaction, and adding, we see the overall reaction is strongly favorable: UQ+H+ in+NADHSNAD++UQH2 E=+0.376 V The standard free-energy difference rG=-veFE=-2*96485.3 C mol-1*0.376 V=-72.6 kJ mol-1 . Q for this reaction is surprisingly hard to measure in viable mitochondria, but it is probably close to 1. This highly favorable reaction is now thought to be coupled to the translocation of no fewer than four protons from the mitochondrial matrix to the intermembrane space (which for small molecules is in equilibrium with the cytoplasm). Using typical values of -160 mV for the membrane potential (Vin-Vout ) and a pH of 0.5 (Matrix pH 7.8; cytoplasmic pH 7.3), we can obtain the free-energy change for the translocation of a single proton from the mitochondrial membrane interior to the exterior from Eq. 7.19 , assuming a temperature of 25°C. The chemical potential difference is m=ZFV + RT ln ainaout=ZFV-2.303 RTpH =1*96485.3*-0.16 +8.31447*298.15*2.303*-0.5=-18292 J mol-1. Translocating a mol of protons from inside to out therefore takes -m , or 18.3 kJ�mol. Adding the two coupled processes: n m (kJ mol -1 ) rGm (kJ mol -1) H+inSH+out 4 +18.3 -73.2 UQ+H+in+NADHSNAD++UQH2 1 -72.6 -72.6 UQ+5H+in+NADHSNAD++UQH2+4H+out -0.6 The rGm value is near zero, and so the reaction is essentially at equilibrium. This can be confirmed experimentally; it has been shown that artificially increasing the membrane potential in mitochondria can cause the complex I reaction to run in reverse, reducing NAD+ using UQH2 . That reaction would ordinarily be overwhelmingly unfavorable. The gradient of proton electrochemical potential is used to drive ATP synthesis (seebelow). Succinate Dehydrogenase (Complex II) The electrochemical potential for the fumarate reduction half-reaction to succinate -OOCCH=CHCOO-+2H++2e-S-OOCCH2CH2COO- E= +0.040V is positive, and if reversed, could not feasibly drive the reduction of NAD+ . The mitochon-drion therefore has a separate membrane-bound succinate dehydrogenase complex that uses reducing equivalents from succinate to reduce ubiquinone, via a flavin-adenine dinucleo-tide (FAD) intermediate (and a chain of FeS centers similar to those seen in complex I). The full reaction is UQ+-OOCCH2CH2COO-SOOCCH=CHCOO-+UQH2 E=+0.012 V . The reaction is obviously very close to equilibrium at Q ~ 1 , and therefore does not drive any proton translocation. In fact, the membrane-bound part of complex II seems to be little more than a stalk or anchor for the enzyme, all of whose reactive elements lie in the mitochondrial matrix. Biological Redox Reactions and Membranes | 257 Coenzyme Q … Cytochrome c Oxid

oreductase (Complex III) Both complex
oreductase (Complex III) Both complex I and complex II produce UQH2 , a hydrophobic quinol that probably par-titions into the mitochondrial inner membrane, and thus diffuses to the next element in the respiratory chain, complex III. This enzyme, via a complex mechanism involving half-reduced ubisemiquinone intermediates, uses one molecule of UQH2 to reduce two molecules of cytochrome- c , a relatively small, soluble, basic protein that resides in the mitochondrial intermembrane space. Cytochrome- c contains a heme iron that has an Fe3+ (oxidized) and an Fe2+ (reduced) state. We can once again determine the standard bio-chemists electrochemical potential for the full reaction from the two-half reactions: UQ+2H+in+2e-SUQH2 E=+0.052 V cyt@c(Fe3+)+e-Scyt@c(Fe2+) E=+0.254 V Reversing the first half-reaction, multiplying the second by two, and adding: 2 cyt@c(Fe3+)+UQH2Scyt@c(Fe2+)+UQ+2H+out E=+0.202 V Notice the two protons are released on the intermembrane side. Notice also that while we reverse the sign of the electrochemical potential, we do not multiply its value by 2! The standard free-energy difference rG=-veFE=-2*96485.3 C mol-1*0.202 V=-39.0 kJ mol-1 . Q is once again probably close to 1. This reaction is coupled with the transport of two protons across the cell membrane per UQH2 oxidized. Again assuming each mol of protons translocated requires 18.3 kJ, the overall process is favored by -39.0+2*18.3=2.4 kJ mol-1 under standard conditions. The reaction, like that of complex I, runs close to equilibrium. Cytochrome c Oxidase (Complex IV) The fourth mitochondrial membrane complex, cytochrome c oxidase, receives electrons one at a time from reduced cytochrome- c [cyt@c(Fe2+)] , which diffuses approximately 10nm from the active site of complex III, and transmits them to molecular oxygen, which requires four electrons overall to become reduced to two water molecules. The cytochrome c oxidase complex accomplishes this one-electron to four-electron transfer through the mediation of two heme centers. Each heme is associated with a copper ion near the heme iron. The net reaction (per two electrons transferred) is 2 cyt @c(Fe2+)+1�2O2+2H+inS2 cyt@c(Fe3+)+H2O . The protons come from the matrix side of the membrane via a proton channel. From the two half-reactions cyt@c(Fe3+)+e-Scyt@c(Fe2+) E=+0.254 V 1�2O2+2H++2e-SH2O E=+0.815 V we obtain a value for E=0.815-0.254=0.541V . Since the oxygen pressure is certainly less than the 0.2 bar of atmospheric oxygen, and typical mitochondrial matrix pHs are 7.8, not 7.0, E is probably considerably lower than E for the second half reaction. The stan-dard free energy difference rG=-veFE=-2*96485.3 C mol-1*0.561 V=-109.3 kJ mol-1. This reaction is coupled with the transport of two mol protons across the cell membrane per mol water generated. Since each proton transloc

ated requires approximately 18.3 kJ mol
ated requires approximately 18.3 kJ mol -1 , the reaction is favored by -54.3 kJ mol -1 under standard conditions, although it is probably somewhat less favored in vivo . 258 Chapter 7 | Electrochemistry Mitochondrial Oxidation of NAD Adding together the overall reactions of complexes I, III, and IV, we obtain the deceptively simple result NADH+1�2 O2+11 H+inSNAD+ + H2O+10 H+out . Four of the protons come from complex I, two from complex III, two from complex IV, and two because UQ is reduced on the matrix side and oxidized on the intermembrane side. The standard electrochemical potential for the chemical part of the reaction is NADH+H+in+1�2 O2SNAD++H2O E=+1.139 V. This corresponds to a rG value of 219.8 kJ mol -1 . Approximately 183.0 kJ mol -1 (10 * 18.3) is stored as proton electrochemical potential energy. As we have written it (without the comparatively small corrections for Q ), the efficiency is 83%. By the standards of most energy transducing systems, this is a marvel of efficiency! ATP Synthase While many of the protein complexes involved in oxidative phosphorylation are exqui-sitely complex, there is surely none so beautiful as the F 1 F 0 ATP synthase, a tiny molecu-lar machine that couples electrochemical potential, mechanical energy, and chemical synthesis. The enzyme is depicted in figure 7.9 . F 1 (short for fraction 1) is the mushroom-shaped, soluble component, whose subunits are labeled in Greek letters, and has the stoichiometry a3b3gde. F 0 is the membrane-bound component, labeled in Roman letters, and has the stoichiometry abc n , where n can be anywhere from 10 to 14. F 1 binds to F 0 on the inner face of the mitochondrial inner membrane. The b subunit is catalytically active in ATP synthesis, and research by the Boyer group showed it has three states: an open state (O) in which ATP can dissociate and ADP and P i can bind, a loose state (L) in which the ADP and P i are enclosed by the protein but loosely bound, and a tight (T) state in which they have reacted to form a very tightly bound ATP. Each F 1 unit has one b subunit in each of the three states. Boyer proposed that rotation of the F 1 unit switched the b units between the three states cyclically in the order OSLST. This was beautifully demonstrated by Yoshida, who attached the F 1 unit top down on a glass slide, linked the g or e subunits to a fluorescently labeled actin filament, and watched the tiny filament rotate by fluorescence microscopy when ATP was added. Even more elegantly, by attaching a magnetic bead to the particle and applying a rotating magnetic field, it was possible to drive ATP synthesis by mechanically rotating the F 1 unit. The direction of rotation required for synthesis is the opposite of the direction in which the unit spins when ATP is applied. This shows that ATP synthesis and mechanical rotation of the head relative to the stalk are coupled. Where does the rotation come from? Again, single-molecule microscopy showed that the membrane embedded ring of 10…15 c subunits rotates with the F 1 unit when F 1 is anchored to F 0 , as it is

on the mitochondrial inner membrane. Rot
on the mitochondrial inner membrane. Rotation of this ring is driven by an electrochemical gradient of H + ions across the membrane ( the protonmotive force ). Protons enter through a pore on the outer face of the a subunit, and protonate a carboxylic acid side chain on the c subunit immediately adjacent to it. This causes the csubunit to rotate away from the a subunit, moving the next c subunit into position. This deposits a proton from the corresponding carboxylic acid to the inner side of the membrane, and then receives a proton from the outer side, causing it in turn to rotate. For each proton, the c ring undergoes a 360�n rotation. H+Intermembrane spaceMitochondrial matrixbaccccc FIGURE 7.9 The mitochondrial�chloroplast F 1 F 0 ATP synthase. Summary | 259 Since each proton induces a 360/n rotation, and it requires one-third of a full rotation, or 120° rotation, to release an ATP from the T site, the number of protons passing through the membrane per ATP synthesized is 120�(360�n)=n�3 . Depending on the species and the organelle, n�3 can vary between 31�3 and 5. If we take a minimum value of n =10, and our previous mH+ of 18.3 kJ mol -1 , this means that 18.3*3 1�3=61 kJ mol-1 of free energy is available to turn the rotor by 120°, and thus synthesize one mol of ATP. We have previously calculated that at Q = 1000, 49.1 kJ mol -1 of free energy is needed for ATP synthesis. Thus, n m (kJ mol -1 ) mG (kJ mol -1 ) H+outSH+in 31�3 -18.3 -61.0 ADP+PiSATP+H2O 1 +49.1 +49.1 ADP+Pi+31�3H+outSATP+H2O+31�3H+in -11.9 and the free energy from the proton electrochemical potential difference is more than ade-quate to drive the synthesis of ATP forward. However, if  H+ drops below 14.7 kJ�mol, the sign of the net free-energy change will reverse, and ATP hydrolysis will drive protons out of the mitochondrial interior. Summary Galvanic Cells Electrical work and free energy: G=QVex=we (7.5)…(7.6) G=-veFE (7.8)a0rGm0Tbp=-rSmorrSm=veFa0E0Tbp(7.9)rHm=rGm+TrSm=ve FcE+Ta0E0Tbpd (7.10) The Nernst equation: E=E-RTveF ln Q (7.11)E=RTveF ln K (7.13) Transmembrane Electrochemical Potential m=mE+mC=ZFV+RTln Q (7.17)V=-(RT/ZF) ln K (7.18) Mathematics Needed for Chapter 7 Just algebra and logarithms! 260Chapter 7|Electrochemistry References Suggested Reading 1. Ruma Banerjee et al., 2008. Redox Biochemistry , NewYork, Wiley, is an excellent and comprehensive guide to biologi-cal oxidation and reduction. 2. The online Electrochemical Encyclopedia at http:// electrochem.cwru.edu/encycl/ is hosted by the Ernest B.Yeager Center for Electrochemical Sciences (YCES) at Case Western Reserve university, and has lots of valuable reference m

aterial. Moser, C. C., J. M. Keske
aterial. Moser, C. C., J. M. Keske, K. Warnke, R. S. Farid, and P. L. Dutton. 1992. Nature of Biological Electron Transfer. Nature 355:796…802. Morth, J. P., Pedersen, B.P., Toustrup-Jensen, M.S., Sørensen, T. L.M., Petersen, J., Andersen, J. P., Vilsen, B., and Nissen,P. 2007. Crystal structure of the sodium…potassium pump. Nature (London) 450:1043…1049. Brandt, U. 2006. Energy Converting NADH:Quinone Oxidoreductase (Complex I) Annu. Rev. Biochem. 75:69…92. Yankovskaya, V., Horsefield, R., Törnroth, S., Luna-Chavez, C., Miyoshi, H., Léger, C., Byrne, B., Cecchini, G., Iwata, S. 2003. Architecture of Succinate Dehydrogenase and Reactive Oxygen Species Generation. Science 299: 700…704. Dudkinaa, N. V., Kudryashev, M., Stahlberg, H., and Boekemaa, E. J. 2011. Interaction of complexes I, III, and IV within the bovine respirasome by single particle cryoelectron tomography. Proc. Natl. Acad. Sci. USA 37:15196…15200. Nakamoto, R. K., Scanlon, J. A. B., Al-Shawi, M. K. 2008. The rotary mechanism of the ATP synthase. Arch. Biochem. Biophys. 476: 43…50. Problems 1. The cytoplasm of a cell is a (moderately) conducting fluid, surrounded by an insulating membrane, and embedded in a solution that is usually at least somewhat conducting. It can therefore be treated as a capacitor. Assume the cell is spheri-cal with a radius r=10 mm, the cell membrane has an area A=4pr2, and a thickness D5 nm. The capacitance should then be given by C=eA/D, where the permittivity e is given by ere0, with er, the dielectric constant, having a value of approximately 5, and e0=8.8542*10-12 F m-1.Calculate the capacitance of the cell membrane, and cal-culate how many unipositive ions need to cross the cell membrane to give the interior a potential of 100 mV. 2. This is the half-cell reaction for the reduction of solid manganese dioxide: MnO2(s)+4H+(aq)+2e-SMn2+(aq)+2H2O (l) E=+1.23 V Determine the electrochemical potential and the free-energy change for a system where a solution of 0.1 M Mn2+ at pH1 is shaken in air at 0.2 bar pressure of O 2 , resulting in produc-tion of MnO2 ( s ), with the reduction of the oxygen to water. If the pH is held constant at 1 and the oxygen pressure at 0.2bar, what will be the equilibrium concentration of Mn2+in the presence of excess MnO2 (s) ? 3. Ascorbic acid (Asc; Vitamin C) is a monoanion at neutral pH. It is a powerful antioxidant, which scavenges free radicals. It is oxidized in a single-electron oxidation to the monodehydroascorbate radical anion (MDHA - ) according to the reaction Asc-SMDHA-+H++e- . The half-reaction has a standard reduction potential E of +0.330 V. Ascorbate also undergoes a two-electron oxida-tion to dehydroascorbate (DHA): Asc-SDHA+H++2e- This half-reaction has a standard reduction potential E of +0.08 V. Finally, MDHA … undergoes a disproportionation reaction: 2 MDHA-+H+SDHA+Asc- Determine the biochemical standard free-energy change G for this reaction. 4.

The standard electrochemical potential
The standard electrochemical potential for the reduction of nitrate ion (NO3-+3H++2e-SHNO2+H2O) is E=+0.934. The equation reflects the fact that nitric acid (HNO 3 ) is a strong acid, fully dissociated at pH 0, while nitrous acid (HNO 2 pK a = 3.398) is a weak acid. Calculate E for the half-cell reaction at pH 7. 5. Cytochromes are iron…heme proteins in which a porphyrin ring is coordinated through its central nitrogens to an iron atom that can undergo a one-electron oxidation… reduction reaction. Cytochrome f is an example of this class of mol-ecules, and it operates as a redox agent in chloroplast Problems | 261 photosynthesis. The standard reduction potential E of cytochrome f at pH 7 can be determined by coupling it to an agent of known E , such as ferricyanide�ferrocyanide: Fe(CN)63-+e-SFe(CN)64- E=+0.440 V In a typical experiment, carried out spectrophotometrically, a solution at 25°C and pH 7 containing a ratio [Fe(CN)64-]�[Fe(CN)63-]=2 is found to have a ratio [Cyt fred]�[ Cyt fox]=0.10 at equi-librium. a. Calculate E (reduction) for cytochrome f. b. On the basis of the standard reduction potential E for the reduction of O 2 to H 2 O at pH 7 and 25°C, is oxidized cytochrome f a good enough oxidant to cause the forma-tion of O 2 to H 2 O at pH 7? 6. Fe2+ -myoglobin ( Fe2+ -Mb) is spontaneously oxidized bymolecular oxygen in a one-electron process to give Fe3+ -Mb and superoxide, O 2 - . The reaction can be written Fe2+@Mb+O2SFe3+@Mb+O2- E=-0.971 V . The biochemists (pH 7) reduction potential of Fe3+ -Mb is Fe3+@Mb+e-SFe2+@Mb E=+0.046 V . O 2 can be electrochemically reduced to hydrogen superox-ide, a weak acid (pKa4.9): O2+H++e-SHO2 E=-1.215 V a. Calculate the pH 7 reduction potential for oxygen to superoxide. b. Calculate the potential for the one-electron oxidation of myoglobin by oxygen at an oxygen pressure of 0.02 bar and pH 7. 7. The reaction glyceraldehyde@3@phosphate+NAD++PiS 1,3 diphosphoglycerate+NADH+H+ has rG=6.3 kJ mol-1 . If the standard reduction poten-tial E of NAD + is -0.324 V and the reaction 1,3 diphosphoglycerate+ADPS 3@phosphoglycerate+ATP has rG=-18.8 kJ�mol, calculate the standard reduc-tion potential E for the reaction 3@phosphoglycerate+2e-+3H+S glyceraldehyde@3@phosphate+H2O . 8. Lysozyme (m.w. 14.3 kD) is a rather basic protein; at pH7, it has a net positive charge of +18. If we dissolve 5 g of lysozyme in 100 mL of 0.1 M KCl, and dialyze against 0.1M KCl, calculate the Donnan potential and the concen-tration of K+ and Cl - inside the membrane. 9. Consider the following reaction, in which two electrons are transferred from cytochrome- c (reduced):

2cyt c[Fe2+]+pyruvate+2H+S
2cyt c[Fe2+]+pyruvate+2H+S 2cyt c[Fe3+]+lactate a. What is E for this reaction at pH 7 and 25°C? b. Calculate the equilibrium constant for the reaction at pH7 and 25°C. c. Calculate the standard Gibbs free-energy change for the reaction at pH 7 and 25°C. d. Calculate the Gibbs free-energy change (at pH 7 and 25°C) if the lactate concentration is five times the pyru-vate concentration and the cytochrome c ( Fe3+ ) is ten times the cytochrome c ( Fe2+ ). 10. The cell Ag(s), AgI(s)KI(10-2 M)KCl(10-3 M)Cl2 (g, 1 bar), Pt(s) has the voltage 1.5702 V at 298 K. a. Write the cell reaction. b. What is rG at 298 K? c. What is G at 298 K? d. Calculate the standard reduction potential for the half-cell on the left. e. Calculate the solubility product of AgI: KAgI=aAg+aI- f. The cell has a potential of 1.5797 V at 288 K. Estimate rS at 298 K for the reaction. 11. Ferredoxins (Fd) are iron- and sulfur-containing proteins that undergo redox reactions in a variety of microorgan-isms. A particular ferredoxin is oxidized in a one-electron reaction, independent of pH, according to the equation FdredSFdox+e- . To determine the standard potential of Fdred�Fdox a known amount was placed in a buffer at pH 7.0 and bubbled with H 2 at 1 bar pressure. (Finely divided platinum cata-lyst was present to ensure reversibility.) At equilibrium, the ferredoxin was found spectrophotometrically to be exactly one-third in the reduced form and two-thirds in the oxidizedform. a. Calculate K , the equilibrium constant, for the system 1�2 H2+FdoxSFdred+H+ . b. Calculate E for the Fdred/Fdox half-reaction at 25°C. 12. The conversion of b-hydroxybutyrate (b@HB-) to acetoac-etate (AA-) is an important biochemical redox reaction that uses molecular oxygen as the ultimate oxidizing agent: b@HB-+1�2 O2(g)SAA-+H2O a. Using the standard reduction potentials given in table 7.1 , calculate rG and the equilibrium constant for this sys-tem at pH 7 and 25°C. b. In a solution at pH 7 and 25°C saturated at 1 bar with respect to dissolved air (which is 20% oxygen), what is the ratio of AA- to b@HB- at equilibrium? 262Chapter 7|Electrochemistry 13. Consider the oxidation of ethanol to acetaldehyde: CH3CH2OH+1�2 O2(g)SCH3CHO+H2O a. Calculate E for this reaction at 25°C. b. Calculate the standard Gibbs free energy (in kJ) for the reaction at 25°C. c. Calculate the equilibrium constant at 25°C for the reaction. d. Calculate E for the reaction at 25°C when aethanol=0.1, pO2=4bar, aacetaldehyde=1, and aH2O=1. e. Calculate rG for the reaction in part (d). 14. Magnesium ion and other divalent ions form complexes with adenosine triphosphate, ATP: ATP+Mg2+SMg@ATP a.

Describe an electrochemical cell tha
Describe an electrochemical cell that would allow you to measure the activity of Mg2+ at any concentration in a 0.100 M ATP solution. b. Describe how you could measure with an electrochemi-cal cell the thermodynamic equilibrium constant for binding of Mg2+ by ATP. 15. Photosystem 1, in higher plants, converts light into chemical energy. Energy, in the form of photons, is absorbed by a chlorophyll complex, P700, which donates an electron to A. The electron is passed down an electron-transport chain, at the end of which NADP+ is reduced. The reduction poten-tials of P700+, A, and NADP+, at pH 7.0 and 25°C are 0.490 V, 0.900 V, and -0.350 V, respectively. a. Calculate, at pH 7.0 and 25°C, E of the reaction P700+ASP700++A- . b. What is G in kJ mol -1 , for the same reaction? c. At pH 7.0 and 25°C find rG in kJ mol -1 for the reaction NADP++H2(g)SNADPH+H+. 16. Certain dyes can exist in oxidized or reduced form in solu-tion. The half-reaction for one such dye, methylene blue (MB), is MBox(blue)+2H++2e-SMBred(colorless) E=+0.400 V. As indicated, the oxidized form is blue, and the reduced form is colorless. From the color of the solution, the relative amounts of the two forms can be estimated. a. Write the equation for the half-cell reduction potential ofmethylene blue in terms of MB ox , [MB red ], [H + ], and E . b. A very small amount of MB ox is added to a solution containing an unknown substance. The pH = 7.0. From the color of the solution, it was estimated that the ratio of concentrations [MB ox ]�[MB red ] = 1.00 * 10 -3 at equi-librium. Assuming that all activity coefficients are equal to 1, determine the half-cell potential of the unknown substance in solution. 17. Consider the following half-cell reactions and their standard reduction potentials at 298 K and pH 7.0 in aqueous solution: O2+4H++4e-S2H2O (E=+0.815 V) cystine+2H++2e-S2 cysteine (E=-0.34 V) a. If you prepare a 0.010 M solution of cysteine at pH 7.0 and let it stand in contact with air at 298 K, what will be the ratio of cysteine�cysteine at equilibrium? The partial pressure of oxygen in the air is 0.20 bar. The activity coefficients may be taken as unity. b. What is G for the reaction when the activities of the reactants and products are the equilibrium values? 18. Typical Mg2+ concentrations in blood plasma are 2.2mM. If the membrane potential of erythrocytes is -90 mV (- inside), calculate the equilibrium concentration of Mg2+ inside the red blood cell at 37°C. 19. In living biological cells, the concentration of sodium ions inside the cell is kept at a lower concentration than the con-centration outside the cell, because sodium ions are actively transported from the cell. Consider the following process at 37°C and 1 bar: 1 mol NaCl (0.05M inside)S1 mol NaCl(0.20M outside) a. Write an expression for the free-energy change for thi

s process in terms of activities. Defin
s process in terms of activities. Define all symbols used. b. Calculate m for the process. You may approximate the activities by concentrations in M. Will the process proceed spontaneously? c. Calculate G for moving 3 mol of NaCl from inside to outside under these conditions. d. Calculate m for the process if the activity of NaCl inside is equal to that outside. e. Calculate m for the process at equilibrium. f. The standard free energy for hydrolysis of ATP to ADP (ATP+H2OSADP+Pi) in solution is rG=-31.3 kJ�mol-1 at 37°C and pH 7. The free energy of this reaction can be used to power the sodium-ion pump. For a ratio of ATP to ADP of 10, what must be the concentration of phosphate to obtain -40 kJ mol -1 for the hydrolysis? Assume that activity coefficients are 1 for the calculation. g. If the ratio of ATP to ADP is 10, what is the concentra-tion of phosphate at equilibrium? Assume ideal solution behavior. What do you conclude from your answer? 20. A cell membrane at 37°C is found to be permeable to Ca2+ but not to anions, and analysis shows the inside concentration to be 0.100 M and the outside concentration to be 0.001 M in Ca2+ . a. What potential difference in volts would have to exist across the membrane for Ca2+ to be in equilibrium at the stated concentrations? Assume that activity coefficients are equal to 1. Give the sign of the potential inside with respect to that outside. b. If the measured inside potential is +100 mV with respect to the outside, what is the minimum (reversible) work required to transfer 1 mol of Ca2+ from outside to inside under these conditions? Problems | 263 21. When heart muscle is treated with external lithium ions, the ions pass through the cell membrane and equili-brate. Muscle cells treated with 150 mM lithium chlo-ride solution achieved a steady-state membrane potential of 40mV ( negative inside). What was the intracellular lithium concentration? 22. Ascorbate and copper ions have the following electro-chemical potentials: Dehydroascorbate+2H++2e-SAscorbate E=+0.08 V Cu2++e-SCu+ E=0.159 V 10 mL of a 0.02 M solution of ascorbate in a buffered solution at pH 7 at 25°C is mixed with 10 mL of 0.02M of Cu 2+ . Write a balanced chemical equation for the reaction, and figure out the equilibrium concentration of dehydro-ascorbate, ascorbate, Cu 2+ , and Cu + . 23. Mercuric reductase carries out the reaction NADPH+Hg2+SNADP++H++Hg. The electrochemical potentials for the two reduction half reactions are: NADP++H++2e-SNADPH E=-0.339V Hg2++2e-SHg E=+0.85 V Calculate the standard free energy of this reaction, and the free energy if the NADP+ : NADPH ratio is 1, the Hg 2+ concentration is 1 mM, and the Hg concentration is 0.3 M. 24. Seawater contains approximately 5 g of Zn 2+ per liter. Calculate what voltage must be put on a zinc anode, relative to seawater, to completely prevent zinc from dissolving