Notes and a Worked Example on Membrane Potentials

Consider two compartments, designated by ′ and ′′, each of which contain one liter of water, separated by a rigid membrane which is permeable to sodium ion, chloride ion and water, but not to the P^{2−} ion. Let 0.02 moles of NaCl and 0.005 moles of Na_{2}P be dissolved in the lefthand (′) compartment. We wish to calculated the composition of both compartments wchen equilibrium is established. We will then procede to discuss briefly how the potential might be measured and to consider in the context of the simplest reasonably soundly-based expression for the activity coefficients in dilute solutions of electrolytes (the Debye-Huckle theory) the magnitude of the errors involved in the ideal solution approximation.

mNa^{+} = X ; mCl^{−} = Y ; m_{P_{2−}} = 0.005  mNa^{+} = mCl^{−} = Z

Conditions for Equilibrium

T′ =T′′
μNa^{+}^{′} =μNa^{+}^{′′}
μCl^{−}^{′} =μCl^{−}^{′′}
μH_{2}O^{′} =μH_{2}O^{′′}

μ_{i} =μ_{i}^{∘}(T, P = 0) + P V_{i} + z_{i}F ϕ_{+}RT ln a_{i}
≈μ_{i}^{∘}(T, P = 0) + P V_{i} + z_{i}F_{ϕ} + RT ln m_{i}  (1)

A. Calculation of the Equilibrium Composition

The chemical potential of sodium chloride is the sum of the chemical potential of Na^{+} and Cl^{−}. With the definitions:

μNaCl^{∘} = (μNa^{+}^{∘}) + (μCl^{−}^{∘})

VNaCl =VNa^{+} + VCl^{−}
aNaCl = (aNa^{+}) + (aCl^{−})
μNaCl =μNaCl^{∘}(T, P = 0) + P VNaCl + RT ln[(aNa^{+})(aCl^{−})]  (2)

Note that the electrical terms cancel (zNa^{+} = 1 zCl^{−} = −1) since NaCl is electrically neutral. Equating the chemical potential of NaCl on both sides of the membrane:

μNaCl^{∘} + P′VNaCl + RT ln[(aNa^{+}^{′})(aCl^{−}^{′})] = μNaCl^{∘} + P′′VNaCl + RT ln[(aNa^{+}^{′′})(aCl^{−}^{′′})]

(3)

cancelling the μ^{∘}, dividing by RT and rearranging

ln[(aNa^{+}^{′})(aCl^{−}^{′})] = ln[(aNa^{+}^{′′})(aCl^{−}^{′′})] + (P′′ − P′)VNaCl/RT  (3′)

Temporarily neglecting the osmotic term and approximating the activity by the molar concentration

XY =Z^{2}  (4)
X =Y + 0.01  (4.1 − electrical neutrality)
X + Z =0.03  (4.2 − conservation of Na^{+})

The rest is algebra! Substitution of (4.1) and (4.2) into (4) gives

X(X − 0.01) =(X − 0.03)^{2}

or

0.05X =(0.03)^{2}

or

X = 0.018  Y =0.008  Z = 0.012

B. Calculation of the Osmotic Pressure

The osmotic pressure can be calculated as usual by equating the chemical potential of water on opposite sides of the membrane. With the usual approximations

Π = (c′ − c′′)RT = (X + Y + 0.005 − 2Z) = (0.007)RT = 0.171atm

The osmotic effect on the equilibrium concentrations (ΠV_{NaCl}/RT) is small, in cgs units

Π ≈1.71 × 10^{5}dynes/cm^{2}
VNaCl ≈20 cm^{3}
RT ≈2.5 × 10^{10}ergs
ΠVNaCl/RT =1.4 × 10^{−4}

C. Calculation of the Electrostatic Potential Difference

The electrostatic potential can be obtained by equating the chemical potential of either one of the charged species on both sides of the membrane. For sodium z = 1 so

μNa^{+}^{∘} + P′VNa^{+} + F ϕ′ + RT ln mNa^{+}^{′} = μNa^{+}^{∘} + P′′VNa^{+} + F ϕ′′ + RT ln mNa^{+}^{′′} (5)

ϕ′′ − ϕ′ =(RT/F) ln (\frac{m′}{m′′} ) + (P′ − P′′)VNa^{+}/F
=0.0104 volts + 3.5 × 10^{−6} volts = 0.0104 volts

The final situation is as illustrated in the figure below. Note that the righthand compartment has assumed a positive potential and that the chloride ion has moved against its concentration gradient.

mNa^{+} = 0.018
mCl^{−} = 0.008
m_{P_{2−}} = 0.005
Π = 0.171 atm

mNa^{+} = mCl^{−} = 0.012
ϕ = 0.0104 volt

D. Calculation of the pH Gradient

To get the pH difference across the membrane, equate chemical potentials of H^{+}

μH^{+}^{∘} + P′VH^{+} + F ϕ′ + RT ln mH^{+}^{′} = μH^{+}^{∘} + P′′VH^{+} + F ϕ′′ + RT ln mH^{+}^{′′}

\begin{matrix}ln (\frac{aH^{+}^{′′}}{aH^{+}^{′}} ) = & − 2.303Δ(pH) = −ΠVH^{+}/RT − ϕF/RT \\ = & (−0.0104 × 96500)/(8.314 × 298.15) = −0.405 \end{matrix}

(The osmotic pressure term is negligible).

Δ(pH) = 0.405/2.303 = 0.176

The right hand (′′) compartment is at a higher pH as one would expect for a positive potential.

E. Measurement of the Potential Difference

It is particularly instructive to consider how one might measure ϕ′′ − ϕ′. Suppose, for example, one immersed Ag-AgCl electrodes in both compartments and determined the potential difference E between the external metallic conductors which would be necessary to prevent the flow of current. The change in state, for the transfer of one electron from the right compartment to the left is

Ag(s) + Cl^{−}(0.012)′′ →AgCl(s) + e^{−}  (at anode)
AgCl(s) + e^{−} →Ag(s) + Cl^{−}(0.008)′  (at cathode)
Cl^{−}(0.012)′′ →Cl^{−}(0.008)′  (net change at electrodes)

There is a net transfer of Cl^{−}(and charge) from the right to the left. The primes indicate which compartment is being referred to. Across the membrane there must be a net transfer of Cl^{−}from left to right or Na^{+}from right to left. Let t be the number of Cl^{−}transfered across the membrane per electron. The three transfers of ions are

Cl^{−}(0.012)′′ →Cl^{−}(0.008)′ (net change at electrodes)
−t(Cl^{−}(0.012)′′ →Cl^{−}(0.008)′  (across membrane)
(1 − t)(Na^{+}(0.012)′′ →Na^{+}(0.018)′  (across membrane)

(Note that second transfer is writen backwards so a minus sign is used). The net change is

(1 − t)(Na^{+}(0.012)′′ + Cl^{−}(0.012)′′ → Na^{+}(0.018)′ + Cl^{−}(0.008)′)

The free energy for this change is

ΔG = RT ln (\frac{(0.018)(0.008)}{(0.012)^{2}}) = 0

(after all, one doesn’t expect to get electrical work out of a system at equilibrium).

If we placed salt bridges between each electrode and its compartment, the free energy changes at the electrodes would cancel out and only the transfers across the membrane would be left. The net change would be

(1 − t)(Na^{+}(0.012)′′ →Na^{+}(0.018)′)
−t(Cl^{−}(0.012)′′ →Cl^{−}(0.008)′)

or

(Na^{+}(0.012)′′ →Na^{+}(0.018)′)
−t{Na^{+}(0.012)′′+Cl^{−}(0.012)′′ → Na^{+}(0.018)′ + Cl^{−}(0.008)′}

The term in {} is zero as before. This leaves only

ΔG = RT ln (\frac{0.018}{0.012}) = 0.0104volts

F. Activity coefficients

The ionic strength in the left-hand compartment is

I′ = (0.018 + 0.008 + (2^{2})(0.005))/2

For a univalent ion,

\begin{matrix}log γ_{±} = & − 0.509\frac{I^{\frac{1}{2}}}{1 + BI^{\frac{1}{2}}} \\ = & − 0.0662 \\ γ_{±} = & 0.859 \end{matrix}

Where B is assumed to be 1. For the right hand compartment, I = 0.012 and γ_{±} = 0.892.

Equation (4) becomes

(0.859)^{2}XY = (0.892)^{2}Z^{2}

When this is used instead of equation (4), we get X = 0.182. The presumption of ideality introduces approximately a 1% error in these rather dilute solutions.

An electrochemical system which exhibits, in oversimplified form, some aspects of the distribution and mobility of principal ionic species between the interior and exterior of certain cells is sketched below. All concentrations are in moles/liter and the temperature is 300^{∘}K. P^{3−} may be thought of as representing a variety of high molecular weight charged species (proteins, nucleotides, etc.) which can not pass through the membrane. The membrane is supposed to be rigid and passively permeable to H_{2}O, K^{+} and the minority species C^{2+} all of which are distributed at equilibrium. The concentration of C^{2+} need not be considered in the calculation of the osmotic pressure, electrical neutrality or the ionic strength.

mNa^{+} = 0.050 M ; m_{K_{+}} = 0.400 M ; mCl^{−} = 0.090 M ; m_{P_{3−}} = 0.12 M ; m_{C_{2+}} = 10^{−6} M  mNa^{+} = 0.450 M ; m_{K_{+}} = 0.020 M ; mCl^{−} = 0.470 M ; m_{P_{3−}} = 0.000 M ; m_{C_{2+}} =  ? M

a) Calculate the osmotic pressure across the membrane.

b) Calculate the difference in electrostatic potential between the interior of the cell and its surroundings. (Assume activity = molarity).

c) Calculate the equilibrium concentration of C^{2+} outside the membrane.

Suppose one were to introduce silver-silver chloride electrodes into the interior of the cell and the surrounding medium and so set up the galvanic cell

Ag(s) + Cl^{−}(outside) →AgCl(s) + e^{−}  (anode half reaction)
K^{+}(outside) →K^{+}(inside)  (liquid junction)
AgCl(s) + e^{−} →Ag(s) + Cl^{−} (cathode half reaction)

d) What is the free energy change for the process at the liquid junction?

e) What is ΔG for the overall change in state?