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.”

— 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

catfish, 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

potential is an unvarying property of a specific pair of metals or other chemical substances,

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 | 239

membrane 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

molecules 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 flow of electrical

charge  Q , which can be expressed as a differential:

I =dQ

dt . (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 =dwE

dt . (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 carries is 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 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 a charge, 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

protons  streaming  from 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:

Zn S Zn2 +

+ 2e-

The electrons flowed to the silver surface, where they reacted with water to give hydrogen gas:

2 H2O + 2e-S 2OH-

+ H2

Stacks of these disks gave estimated voltages between one end of the pile and the other

of up to 50 V.

Volta, 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-soaked cloth.

242 Chapter 7 | Electrochemistry

salt bridgeZn metal Cu metal

Zn2+ Cu2+

V

1.1 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. 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 - T�S, to obtain

�G = q + wpV + wE + �(pV ) - T�S . (7.4)

At constant pressure, �(pV) = p�V , and under reversible conditions, wpV = -p�V . At

constant temperature, under reversible conditions, �S = q>T, or q = T�S. Making these

substitutions:

�G = T�S - p�V + wE + p�V - T�S = 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 = QV ex . (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:

a 0�rGm

0Tb

p= - �rSm �rSm = veF a 0E

0Tb

p (7.9)

or

�rHm = �rGm + T�rSm = veF cE+T a 0E

0Tb

pd . (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:

Zn � Zn2 +(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:

Cu � Cu2 +(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:

Zn � Zn2 +(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-S ½ 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 Zn � Zn2 + , it is

tabulated as

Zn2 +(aq, a = 1) + 2 e-S Zn (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 electrochemical 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 .

E X A M P L E 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 metal

H+(a = 1)

FIGURE 7.3 The standard hydrogen electrode (SHE).

The Electrochemical Cell | 245

SOLUTION The two half-reactions are:

Zn2 +

+ 2e-S Zn E� = -0.7618

Fe3 +

+ 3e-S Fe E� = -0.037

(1) Reverse the oxidation step and multiply the electrochemical potential by -1:

Zn S Zn2 +

+ 2 e- E� = 0.7618

(2) Multiply the two reactions by integers:

3 Zn S 3 Zn2 +

+ 6e- E� = 0.7618

2 Fe3 +

+ 6e-S 2 Fe E� = -0.037

(3) Add the reactions and the electrochemical potentials:

3 Zn + 2 Fe3 +S 2 Fe + 3 Zn2 + E� = 0.725

Whether we write the equation this way, or as Zn + 2/3 Fe3 +S 2/3 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� -

RT

veF 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� =RT

veF 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-S K –2.931

Ca2 + >Ca Ca2 +

+ 2e-S Ca –2.868

Na+ >Na Na+

+ e-S Na –2.71

Mg2 + >Mg Mg2 +

+ 2e-S Mg –2.372

Mn2 + >Mn Mn2 +

+ 2e-S Mn –1.185

Zn2 + >Zn Zn2 +

+ 2e-S Zn –0.7618

Acetate> acetaldehyde CH3COO-

+ 2e-

+ 3H+S CH3CHO + H2O –0.105 –0.586

Fe2 + >Fe Fe2 +

+ 2 -e S Fe –0.447

H+ >H2>Pt 2H+

+ 2e-S H2 0 –0.414

CO2>formate CO2 (aq) + H+

+ 2e-S HCOO- –0.181 –0.409

Ferredoxin Fd[Fe3 +] S Fd[Fe2 +] –0.395

FMN(Mitochondrial

complex I) FMN + 2H+

+ 2e-S FMNH2 –0.380

Gluconolactone>glucose C6H10O6 + 2H+

+ 2e-S C6H12O6 –0.345

L-cystine>L-cysteine L@cystine + 2H+

+ 2e-S 2 L@cysteine –0.355

Fe3 + >Fe Fe3 +

+ 2e-S Fe –0.037

Acetoacetate>b-hydroxybutyrate

CH3COCH2COO-

+ 2H+

+ 2e-S CH3CHOHCH2COO- –0.346

NADP+>NADPH NADP+

+ H+

+ 2e-S NADPH –0.339

NAD+>NADH NAD+

+ H+

+ 2e-S NADH –0.324

Glutathione(S-S)>GlutathioneSH

GSSG + 2H+

+ 2e-S 2 GSH –0.240

FMN>FMNH 2 FMN + 2H+

+ 2e-S FMNH2 –0.213

FAD>FADH2 FAD + 2H+

+ 2e-S FADH2 –0.212

Acetaldehyde> ethanol CH3CHO + 2e-

+ 2H+S CH3CH2OH +0.221 –0.193

Pyruvate> lactate CH3COCOO-

+ 2H+

+ 2e-S CH3CHOHCOO- –0.184

Oxaloacetate>malate -OOCCOCH2COO-

+ 2H+

+ 2e-S

-OOCCHOHCH2COO- –0.158

FAD>FADH2(Complex II) FAD+

+ 2H+

+ 2e-S FADH2 –0.05

Fumarate> succinate -OOCCH “ CHCOO-

+ 2H+

+ 2e-S

-OOCCH2CH2COO- +0.045

Myoglobin Mb[Fe3 +] + e-S Mb[Fe2 +] +0.046

Ubiquinone UQ + 2H+

+ 2e-S UQH2 +0.052

Dehydroascorbate> ascorbate

DHA + H+

+ 2e-S Asc- +0.080

AgCl>Ag>Cl- AgCl + e-S Ag + Cl- +0.222

Cytochrome c Cyt@c[Fe3 +] + e-S Cyt@c[Fe2 +] +0.254

Calomel 1>2 Hg2Cl2 + e-S Hg + Cl- +0.268

O2>H2O2 O2 + 2H+

+ 2e-S H2O2 +0.695 +0.281

(continues)

Transmembrane Equilibria | 247

E X A M P L E 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-S 2 GSH E�� = -0.240 V

NADP+

+ H+

+ 2e-S NADPH E�� = -0.339 V

(1) Reverse the second reaction and multiply the electrochemical potential by -1:

NADPH S NADP+

+ H+

+ 2e- E�� = +0.339 V

(2) The number of electrons is balanced.

(3) Add the reactions and the electrochemical potentials:

GSSG + H+

+ NADPH S 2 GSH + NADP+ E�� = +0.099 V

K =[GSH]2

eq[NADPH]eq

[GSSG]eq[NADP+]eq=

(4 * 10- 3)2

0.005[GSSG]eq

We do not include the H+ concentration, since the activity of H+ is defined as 1 at

pH 7 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 μM. 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-S Cu +0.342

I2>I- I2 + 2e-S 2 I- +0.535

Ag+ >Ag Ag+

+ e-S Ag +0.800

O2>H2O O2 + 4H+

+ 4e-S 2H2O +1.229 +0.815

NO3 - >NO2 - NO3 - + 3H+

+ 2e-S HNO2 + H2O +0.934

Br2(aq) >Br- Br2 + 2e-S 2Br- +1.087

Cl2(g) >Cl- Cl2 + 2e-S 2Cl- +1.358

Mn3 + >Mn2 + Mn3 +

+ e-S Mn2 + +1.541

F2(g) >F- F2 + 2e-S 2F- +2.866

TABLE 7.1 Standard Reduction Electrode Potentials at 25°C (cont.)

248 Chapter 7 | Electrochemistry

membranes, 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 Δ 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 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 .

E X A M P L E 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 | 249

of 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 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)

mV

electrolyte 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

+ 4c2

2 .

The ratio r of cNa, in to cNa, out is

r = cNa, in

c =

-ZMcM

2c + B aZMcM

2cb2

+1 . (7.26a)

From Eq. 5.31 , w e can also calculate the ratio of Cl- inside to outside:

cCl- , in

cCl- , 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 (r 6 1) and the concentration of negative

ions inside will be greater than outside.

For a negatively charged macromolecule, r 7 1, 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 = -

RT

F ln

aNa, in

aNa, out� -

RT

F ln

cNa, in

cNa, out� -

RT

F 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 | 251

is 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 + ATP S 2 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

the membrane 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)

Pi

K+

K+

(e)

ATP

Na+

Na+

Na+

(a)

Pi

K+

K+

(d)

Pi

Na+

Na+

Na+

ADP

(b)

Na+

Na+

(g)

Pi

Na+

Na+

ADP

(c)

Na+

Extracellular fluid

Cytosol

ATPNa+

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 = –70 mV

Outside[Na+] = 140 mM [K+] = 5 mM

Cell membraneCell membrane

FIGURE 7.5 Typical ion concentrations inside and outside the plasma membrane.

252 Chapter 7 | Electrochemistry

of 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 ain

aout . (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 10

140= -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 100

5= 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 μM, and [P i ] = 25 mM. This gives Q = 0.001:

�m = -31300 + 8.31447 * 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+ in S Na+ out 3 13.56 40.7

K+ out S K+ in 2 0.97 1.9

ATP + H2O S ADP + Pi -49.1

ATP + H2O + 3Na+

in+ 2K+

out S ADP + 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 | 253

high intracellular potassium, and low intracellular sodium is usually a sign of imminent

cell death!

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 opening

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 mV

0 ms 2.5 ms

5 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 (ΔpH) 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 ΔpH 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:

C

COO–

COO–C

H

H

HO

H+ NAD+

C

COO–

COO–C

O

HH+ NADH + H+

malatedehydrogenase

malate oxaloacetate

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-S oxaloacetate E�� = -0.158 V

NAD+

+ H+

+ 2e-S NADH 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-S NADH E�� = -0.324 V

UQ + 2H+ in + 2e-S UQH2 E�� = +0.052 V

complex I

succinate

FADFADH2

UQH2UQ

fumarate

4 H+

NAD+ NADH + H+

FMNFMNH2

UQH2UQ

UQH2 UQ

2 cyt-c (Fe3+)

2 cyt-c (Fe2+)

2 H+

½ O2

+ 2H+H2O

2 cyt-c (Fe2+)

2 cyt-c (Fe3+)

2 H+

Mitochondrial intermembrane space

Mitochondrial matrix

complex 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.

* 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 + NADH S NAD+

+ 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 ain

aout= ZFV - 2.303 RT�pH

= 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+

in S H+

out 4 +18.3 -73.2

UQ + H+

in + NADH S NAD+

+ UQH2 1 -72.6 -72.6

UQ + 5H+

in + NADH S NAD+

+ 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

(see below).

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-

S OOCCH= 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 Oxidoreductase (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-S UQH2 E�� = +0.052 V

cyt@c(Fe3 +) + e-S cyt@c(Fe2 +) E�� = +0.254 V

Reversing the first half-reaction, multiplying the second by two, and adding:

2 cyt@c(Fe3 +) + UQH2 S cyt@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

10 nm 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+

in S 2 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-S cyt@c(Fe2 +) E�� = +0.254 V

1>2O2 + 2H+

+ 2e-S H2O 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 translocated 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+

in S NAD+ + 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 O2 S NAD+

+ 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 O S L S T. 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. 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

c subunit 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 space

Mitochondrial matrix

α αβ β

γε

δ

b

a ccc c c

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+

out S H+

in 31>3 -18.3 -61.0

ADP + Pi S ATP + H2O 1 +49.1 +49.1

ADP + Pi + 31>3H+

out S ATP + 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 Δμ 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)

a 0�rGm

0Tb

p= - �rSm

or�rSm = ve Fa 0E

0Tb

p(7.9)

�rHm = �rGm + T�rSm = ve F cE + Ta 0E

0Tb

pd (7.10)

The Nernst equation:

E = E�-

RT

veF ln Q (7.11)

E� =RT

veF ln K (7.13)

Transmembrane Electrochemical Potential �m = �mE + �mC = ZFV + RT ln Q (7.17)

V = - (RT/ZF) ln K (7.18)

Mathematics Needed for Chapter 7 Just algebra and logarithms!

260 Chapter 7 | Electrochemistry

References

Suggested Reading

1. Ruma Banerjee et al., 2008. Redox Biochemistry , New York,

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 material.

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 D � 5 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-S Mn2 +(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 pH 1

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

0.2 bar, 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-S MDHA-

+ 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-S DHA + H+

+ 2e-

This half-reaction has a standard reduction potential E�� of

+0.08 V. Finally, MDHA – undergoes a disproportionation

reaction:

2 MDHA-

+ H+S DHA + Asc-

Determine the biochemical standard free-energy change

�G�� for this reaction.

4. The standard electrochemical potential for the reduction

of nitrate ion (NO3-

+ 3H+

+ 2e-S HNO2 + 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-S Fe(CN)6

4 -

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)6

3-

] = 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

by  molecular oxygen in a one-electron process to give

Fe3 + -Mb and superoxide, O 2 - . The reaction can be written

Fe2 +@Mb + O2 S Fe3 +@Mb + O2-

E�� = -0.971 V .

The biochemists’ (pH 7) reduction potential of Fe3 + -Mb is

Fe3 +@Mb + e-S Fe2 +@Mb

E�� = +0.046 V .

O 2 can be electrochemically reduced to hydrogen superox-

ide, a weak acid (pKa � 4.9):

O2 + H+

+ e-S HO2

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+

+ Pi S

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 + ADP S

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 pH 7,

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

0.1 M 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[Fe3+] + lactate

a. What is E�� for this reaction at pH 7 and 25°C?

b. Calculate the equilibrium constant for the reaction at

pH 7 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

Fdred S Fdox + 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

oxidized form.

a. Calculate K� , the equilibrium constant, for the system

1>2 H2 + Fdox S Fdred + 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) S AA-

+ 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?

262 Chapter 7 | Electrochemistry

13. Consider the oxidation of ethanol to acetaldehyde:

CH3CH2OH +1>2 O2(g) S CH3CHO + 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 +S Mg@ATP

a. 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 + A S P700+

+ 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) S NADPH + 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-S MBred(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

of  methylene 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-S 2H2O (E�� = +0.815 V)

cystine + 2H+

+ 2e-S 2 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 2.2 mM.

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) S 1 mol NaCl(0.20M outside)

a. Write an expression for the free-energy change for this

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 + H2O S ADP + 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 40  mV ( negative inside). What was the intracellular

lithium concentration?

22. Ascorbate and copper ions have the following electro-

chemical potentials:

Dehydroascorbate + 2H+

+ 2e-S Ascorbate

E�� = +0.08 V

Cu2 +

+ e-S Cu+ 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 0.02 M 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 +S NADP+

+ H+

+ Hg.

The electrochemical potentials for the two reduction half

reactions are:

NADP+

+ H+

+ 2e-S NADPH E�� = -0.339V

Hg2+

+ 2e-S Hg 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 μM.

24. Seawater contains approximately 5 μg of Zn 2+ per liter.

Calculate what voltage must be put on a zinc anode,

relative to seawater, to completely prevent zinc from

dissolving in seawater.