Objectives
- 1.
Recognize that concentration gradients and electrical potential gradients store chemical and electrical potential energy, respectively.
- 2.
Recognize that electrochemical potential energy drives all transport processes.
- 3.
Use the concept of electrochemical potential energy to analyze transport processes.
Electrochemical potential energy drives all transport processes
In Chapter 2 , by examining the permeability of biological membranes to various solutes, we concluded that, with the exception of simple, small, and typically lipid-soluble molecules (e.g., O 2 , CO 2 , ethanol), most biologically important solutes (e.g., sugars, amino acids, inorganic ions) cannot readily traverse cellular membranes. Therefore special transport mechanisms are required to move these impermeant solutes from one side of a membrane to the other. An important class of special transport mechanisms—the ion channels—has already been discussed in Chapter 5, Chapter 6, Chapter 7, Chapter 8 . There we observed that differences in ion concentrations and electrical potential across a membrane can drive the movement of ions through channels. In this chapter we introduce the concepts of chemical and electrical potential energy, which are stored in concentration gradients and electrical potential gradients, respectively. We also demonstrate that electrochemical potential energy drives all solute transport processes.
The relationship between force and potential energy is revealed by examining gravity
Experience tells us that the gravitational force acts on an object at any height so that, when the object is released, it is pulled toward the ground. Because force is mass ( m ) times acceleration ( a ), the gravitational force ( F G ) must be:
FG=-maG
where a G is the acceleration caused by gravity and the minus sign indicates that the force is directed downward (in the negative y direction).
Lifting an object of mass, m , from the ground to some height, y , requires an investment of energy. The amount of energy invested in lifting an object is directly proportional to the mass of the object and the height to which the object is lifted. One way to conceptualize this is to say that the object has greater potential energy when it is at a greater height from the surface of the Earth. These considerations are neatly summarized in the definition of gravitational potential energy (PE G ):
PEG=maGy
Implicit in this equation is the fact that ground level is the reference point against which gravitational potential energy is measured; that is, at ground level ( y = 0), PE G = 0. From this definition, it is clear that a change in height, Δ y , causes a corresponding change in gravitational potential energy, ΔPE G = ma G Δ y. In other words, a gradient of gravitational potential energy exists in the y direction. The gravitational potential energy gradient is ΔPE G /Δ y, or, when written as a derivative, d PE G / dy :
dPEGdy=maG=-FG
We interpret this equation as saying that a gradient in gravitational potential energy gives rise to the gravitational force, which moves an object down the potential energy gradient, from some height toward ground level. This example provides an important and general insight: a gradient of potential energy gives rise to a force that will tend to move material down the potential energy gradient.
A gradient in chemical potential energy gives rise to a chemical force that drives the movement of molecules
Fick’s First Law of Diffusion ( Chapter 2 ) describes how a concentration gradient drives the diffusion of molecules down the concentration gradient. This diffusive movement of molecules may be viewed as being driven by a “chemical force.” Because a potential energy gradient gives rise to a force, if a concentration gradient can give rise to a chemical force, the concentration gradient must also embody a gradient of chemical potential energy . Chemical potential energy is commonly represented by the symbol μ. For one mole of any solute, S, at concentration [S], the chemical potential energy has the form ( Box 9.1 )
In diffusion, a gradient in the concentration of a solute ([S]) drives a net flux, J , of the solute, as described by Fick’s First Law ( Chapter 2 ):
J= -Dd[S]dx
Alternatively, we can view the diffusive movement as the result of a chemical force pushing on the solute molecules. The velocity of the molecule, v , should be proportional to the magnitude of the chemical force, F c :
v=uFc
where u is the mobility coefficient of the molecule. We can determine how the flux is related to the molecular velocity by examining Fig. 9.1 , which shows molecules moving at velocity, v , along a cylinder of solution. The flux, J , is simply the number of moles of molecules passing through the cross-sectional area, A , per unit time. Because the molecules are moving at velocity, v , in a time period, Δ t , they would move a distance v Δ t , which is the distance between the two shaded areas. Molecules initially located between these two areas will pass through the area, A , during the time interval, Δ t . The volume contained between the two shaded areas is A × ( v Δ t ), and the number of moles of solute in this volume is [S] × A × ( v Δt). The flux is this number of moles of solute passing through area, A , in time interval, Δ t :
J=(moles of S)/AΔt=[S]×A×(vΔt)/AΔt=[S]v=[S]uFc
We can verify that the concentration, [S], times the velocity, v , is a flux by checking the units. Because [S] has units of moles per volume (e.g., moles/cm 3 ), and v is expressed as distance per unit time (e.g., cm/sec), [S] × v must have units of moles per unit area per unit time, such as (moles/cm 2 )/sec. These units imply the movement of a certain amount of material through a certain area in a certain amount of time, which is indeed the definition of flux.
The flux driven by the chemical force is the same flux as that specified by Fick’s First Law, so the expressions for J in Equations B9.1 and B9.3 can be equated. Making use of the relationship between the ionic mobility, u , and the diffusion coefficient, D , u = D / RT ( Box 4.1 ), and solving for the chemical force, yields:
Fc= -RT1[S]d[S]dx= -RTdln[S]dx
The last step of Equation B9.4 required us to recognize the derivative of the natural logarithm function ( Appendix B ).
We know that the chemical force, F c , arises from a gradient in chemical potential energy d μ s / dx, and tends to push molecules down the gradient: d μ s / dx = − F c . Combining this with Equation B9.4 gives:
dμSdx=RTdln[S]dx
Integrating Equation B9.5 , we find that the chemical potential energy must have the form:
μS=constant+RTIn[S]
where constant is the constant of integration, which must be determined independently. Examining Equation B9.6 shows that when [S] = 1 M, μ S = constant . In other words, the constant is the chemical potential energy when the solute is at a concentration of 1 M. This special value is given the symbol, μ 0 s . Therefore the complete expression for the chemical potential energy of a mole of solute, S, at concentration, [S], is:
μS=μS0+RTIn[S]
Although μ S has the units of energy, it is commonly referred to as the chemical potential .
μS=μS0+RTIn[S]
where μ 0 S is the chemical potential energy of the solute at the reference concentration of 1 M; thus, at [S] = 1 M, μ S = μ 0 S . Equation 9.4 tells us that solute molecules located in a region of high concentration are at a higher chemical potential energy than the same molecules located in a region of low concentration.
An ion can have both electrical and chemical potential energy
All molecules have chemical potential energy. Ions, because they carry electric charge, can also have electrical potential energy . At any electrical potential, V , the electrical potential energy associated with a single ion carrying z charges (e.g., z = +2 for Ca 2+ ; z = –1 for Cl − ) is zeV , where e is the magnitude of a single electric charge ( e = 1.602 × 10 − 19 coulomb). For a mole of such ions, the electrical potential energy should be multiplied by Avogadro’s number ( N A = 6.022 × 10 23 /mol). The magnitude of one mole of elementary charges is given a special name, the Faraday (symbol F ), and has the value 96,485 coulombs/mol. Thus the electrical potential energy for one mole of ions, each carrying z charges, is zFV .
Because an ionic solute, S z , can be at a particular concentration and electrical potential, its potential energy is the sum of both chemical and electrical components. The complete expression for the electrochemical potential energy of one mole of an ionic solute, S z , at concentration, [S z ], and at an electrical potential, V , is:
μSz=(μSz0+RTIn[Sz])+zFV
Again, although μ s z has the units of energy, it is more commonly referred to as the electrochemical potential . Box 9.2 shows examples of expressions for the electrochemical potential of two common solutes.
The general expression for the electrochemical potential of a solute is given by text equation 9.5:
μSz=(μSz0+RTIn[Sz])+zFV
For a neutral solute like glucose (G), which carries no electrical charge ( z = 0), its potential energy should not depend on the membrane potential ( V m ) of the cell. Therefore the electrochemical potentials of glucose inside and outside the cell are given by the simple expressions:
μG,i=μG0+RTIn[G]i
and
μG,o=μG0+RTIn[G]o
respectively.
For an ionic solute such as the sulfate ion, SO 4 2− , which carries a charge ( z = –2), its potential energy does depend on V m . Because V m is the electrical potential inside relative to that outside (defined to be zero), the electrochemical potential for sulfate inside the cell is given by:
μSO42-,i=(μSO42-0+RTln[SO42-]i)+(-2)FVm=μSO42-0+RTln[SO42-]i-2FVm
and that for sulfate outside the cell is given by:
μSO42-,o=(μSO42-0+RTln[SO42-]o)+(-2)F(0)=μSO42-0+RTln[SO42-]o
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