How can a chemical gradient oppose an electrical gradient?

Question

I'm a physicist taking a class on cell biology and we are currently discussing the concept of an electrochemical gradient, the difference in potential (voltage) and chemical concentrations across a membrane. My instructor stated that a neurons resting potential is caused by these two opposing "forces" reaching an equilibrium across a membrane that is permeable to potassium ions, but essentially impermeable to sodium ions.

The electrical gradient is quite straightforward. More positive charge on one side of a membrane would create a net coulombic force towards the other side. Chemical gradient, on the other hand, is not governed by any technical "force." Instead, diffusion down chemical gradients is simply a consequence of entropy, and can be described using Fick's Law.

It is unclear to me how diffusion can in any way oppose electrical force. Allow me to demonstrate this with two thought experiments using a container full of water that is divided in two halves using a membrane that is permeable only to potassium ions. In the first experiment, 10 potassium ions are dropped onto one side of the container. After some time, the potassium ions would have passively diffused across the membrane and, on average, each side of the membrane would have 5 potassium ions at any given moment. In the second experiment, rather than being empty, the other half of the container has 10 (or more!) sodium ions. As the potassium ions begin to diffuse toward the membrane, they would feel the force from the ten (or more) positive sodium ions and, on average, stay on their side of the container. This proves that the chemical gradient cannot oppose an electrical gradient.

Answer 1

Suppose you have a table with two sides labelled A and B. You start off with 6000 dice on side A. The rule is if you you roll a 6 that dice gets moved into side B. But a dice in side B needs a 6 to stay there or it gets moved back into side A.

So you roll 6000 dice. You'd expect about 1000 of them to come up 6. So now you have a 1000 on side B and 5000 on side A. You roll again - about 833 will come up 6 on side A, and 166 will come up 6 on side B - so 833 get moved from A to B, and 1000-166 = 834 get moved from B to A. So now you have 5001 on one side and 999 on the other side. Keep doing this and about the same will happen and at any moment there will be about 5000 on one side and 1000 on the other.

Note the chance of an individual dice moving A to B is only 1 in 6, while the chance of a dice moving B to A is 5 in 6 but the rate of movement of dice from one side to the other is equal because there are more dice on the A side.

This is what is happening with your potassium ions. They have random motion with some fast and some slow at any given moment - distributed according to the Maxwell-Boltzmann distribution with high energies less likely than low energies. so some on the "potassium" side have enough kinetic energy to cross into the "sodium" side against the coulomb force but the probability is low. Potassium ions on the Sodium side on the other hand have a high probability of going back the other way because of the net coulomb force. But that probability is not absolutely 1 - they have to be going in the right direction and are continually randomly hitting other atoms which will cause them to move quickly and slowly in all directions.

So at any given moment there is a non-zero chance of finding a potassium ion on the sodium side - the exact chance will depend on how many of the sodium ions there are. So for a certain sodium concentration there will be an equilibrium condition where by some potassium ions can be found on the sodium side - meaning there will be a net coulomb field across the membrane. That equilibrium represents the concentration at which the rate of crossings in each direction is equal.

That coulomb field of course represents an electrostatic force - and since in equillibrium we speak of forces being equal and opposite we can consider this probabilistic distribution of K ions as also representing a "force" even though it actually is nothing of the kind.