Living organisms and thermodynamics (Pentcho Valev)

Bryant Fujimoto fujimoto at u.washington.edu
Sat Oct 18 13:22:30 EST 1997

```<RUMYM at BGEARN.ACAD.BG> writes:

>In reply to Lukasz Salwinski and Bryant Fujimoto

>I still claim that the problem how the potential is distributed in a
>Donnan system is extremely difficult, despite the deceiving simplicity
>of the system. I have discussed it many times, with rather competent
>people, but the number of unsurmountable contradictions is indeed great.
>Still there is a line of argument which, in my opinion, could lead to the
>desired solution. Please tell me what you accept and what you don't.

What I don't accept is your definition of the electric potential.

>1. The electrical double layer at a Donnan membrane is capacitor-like.

>2. For an infinite-sized capacitor, the field outside the capacitor is
>zero, whereas the potential on the two sides is constant and different.
>(This is a standard result in electrostatics so please do not say that
>you do not accept it). Except for points very close to the membrane,
>thermodynamics predicts the same potential far from the membrane. This
>coincidence is suspicious - any real membrane is of course finite.

The prediction has nothing to do with the size of the membrane.  See
below.

>3. For a finite-sized capacitor, along its central axis, the field is
>very weak outside it, near the plates, and decreases in magnitude with
>distance. Accordingly, the potential also decreases in magnitude with
>distance, but very slowly. (This is relatively easy to calculate
>for a finite-sized capacitor in a vacuum). If the situation in a Donnan
>system is analogous, we can understand why bulk measurements seem to
>support the thermodynamic prediction - the slow decrease of the potential
>simply cannot be detected. If, however, the system is long enough and
>if this analogy works, the potential at both ends of the system must be
>zero.

Gouy-Chapman theory states that the field due to the unequal distribution
of charge at the double layer decays approximately exponentially, with
a decay length approximately equal to the Debye length.  In 0.1 M NaCl
in water at 25C, the Debye length is about 10 angstroms.  So the field due
to the double layer decays away very quickly.  (The Debye length is
proportional to the inverse of the square root of the concentration, so
a 100 fold decrease in concentration increases the Debye length by a
factor of 10.)

The problem is that the electric potential in this case is not defined the
way you seem to think it is.  You want to consider only the contributions
from the nonuniform distribution of ions.  However, _as_defined_, the
electric potential also includes the effect of the concentration differences
between the two halves of your system.  Once you move away from the
membrane (and the Debye length tells you how far you have to move) the
concentrations are constant on each side, and so the potential difference
between the two halves of the system is constant.  You may not like this
particular definition, but if you are going to discuss electrochemistry,
you are simply going to have to learn to work with it.

>4. If the analogy with the finite-sized capacitor is valid, another very
>interesting implication follows. The potential is greater in magnitude
>along the central axis, but quickly goes to zero in a perpendicular
>direction, e.g. for points lying on the capacitor plane but ouside the
>capacitor.
>The biological implication is even more interesting: If, at equilibrium,
>a double layer is formed at a semipermeable spot on the membrane, the
>potential quickly goes to zero for adjacent impermeable spots, i.e.
>there is no transmembrane potential difference at the impermeable spots
>and the concentration difference there can be used, e.g. for ATP synthesis.

Why does the transmembrane potential difference have to be zero in order
for the concentration difference to be usable for ATP synthesis?  I see
no particular reason why it should be.  Even though the field outside
a capacitor decreases when you move away from its edge, that doesn't
mean the potential difference between the two halves of the capacitor
decreases.

Bryant Fujimoto
fujimoto at u.washington.edu

```