The Nernst potential


An essential characteristic of the functioning of membranes is the fact that they maintain an electric potential difference across them. This is important to many biochemical processes, including the electrical signals in neurons. Somehow, it seems as if membranes can act like "batteries", maintaining an electric potential differences (voltage) across themselves. There is a deep and interesting  physical mechanism that is responsible for creating this potential difference.

The potential arises from a balance between an unbalanced diffusion of ions across the membrane arising from a concentration difference — a kind of entropic force, and the electric force that arises from the resulting separation of charge. We'll make sense of this by telling the story of how a charge separation arises from diffusion and entropic effects, starting with a membrane that is impenetrable to ions but dissolved salt (NaCl) present on both sides, but with different concentrations on each side.

An unbalanced diffusion of ions 

There are lots of ions (Na+ and Cl-) on both sides of the membrane, and each side has a roughly equal number of positive and negative ions.

But if the concentrations are not the same, there will be more positive ions and more negative ions on one side of a membrane than the other. Positive ions are shown as blue and negative ions are shown as red. There is a higher concentrationon the right and a lower concentration on the left.

On both sides the ions are moving randomly in all directions. Now let's suppose that there are passive ion channels in the membrane that only permit positive (blue) ions to pass through. Any positive ion entering them from either side can passes through, but negative ions cannot. By Fick's law of diffusion, more positive ions will move from the right to the left than from the left to the right.

This movement from high to low concentrations is not caused by any physical force, but is a probabilistic phenomenon that results from random motion. Each ion moves randomly, so a positive ion close to the membrane on either side has an equal probability of crossing the membrane, but since one side of the membrane starts with more ions, there are more opportunities for ions on that side to cross over. We can also think about this in terms of entropy:  having an equal concentration of ions on each side of the membrane maximizes the number of possible ways to arrange them, so the tendency is for the number to move towards equality. But they can't go very far in that direction! 

The excess blue charges that begin to appear on the left side of the membrane as a result of the concentration difference will leave a deficit of positive charges on the right and attract the excess negative charges on the right up to the membrane. But the negative ions can't pass through. 

This is like a capacitor supporting a separation of charge! As in a capacitor, this separation results in an electric field and a potential difference. The electric field builds up as the charge separation grows and acts to oppose the flow of charges due to the concentration difference. When the magnitude of this electric field is just right, then the charges reach equilibrium. This is shown in the figure at the right where the balanced charges have been suppressed and only the excess ions (unbalanced charges) are shown. If you count the blue and red charges on either side of the membrane, you'll see an excess of blue on the left and red on the right. The electric field that leads to the potential difference and stops the flow is shown as black arrows.

A balance occurs when the entropic effects (i.e. the diffusion resulting from the concentration difference) are exactly countered by the electric forces that arise as a result of the diffusion. Other examples of this sort of effect are: Debye screening (where random thermal motion of ions in solution is balanced by electrostatic attraction) and the isothermal atmosphere (where random thermal motion of air molecules is balanced by gravity).

The number of ions that need to move across the membrane for the system to reach equilibrium is very small, so the change in the concentrations is negligible.  This makes the math (below) much easier, unlike in the case of Debye screening (where we skipped the details of the math).

Using the Boltzmann factor to figure the probable balance

If we can figure out how much applied potential difference it would take to keep the charges in equilibrium, that will tell us the Nernst potential across this membrane.  (Similarly, when we put something on a scale to measure its weight, we're not directly measuring the gravitational pull of the Earth on the object!  Instead, we're measuring how much upward force we have to exert on the object to balance out the Earth's pull and keep it in equilibrium.  But it still works.)

Without any applied potential difference, the equilibrium state would be equal concentrations on both sides of the membrane.  What potential difference would it take to make the concentration $c_1$on one side and $c_2$ on the other side? Ions are racing up to the membrane, but those on one side have to climb a potential energy hill. Only the faster ones will be able to make it! How to tell how many succeed?

Recall the Boltzmann distribution!  If there is an energy difference between the two sides of the membrane, then ions have a different probability to end up on each side of the membrane (when the system reaches equilibrium), and the ratio of these probabilities is the same as the ratio of the concentrations.

The Boltzmann factor tells us that the ratio

$$\frac{c_1}{c_2} = e^{-\Delta U/k_BT}$$

where $ΔU$ is the energy difference, and that's equal to $qΔV$, where $q$ is the charge of each ion, and $ΔV$ is the potential difference across the membrane.  We can use this to solve for the Nernst potential, $ΔV$:

$$-\frac{q\Delta V}{k_BT} = \ln{\frac{c_1}{c_2}}$$


$$\Delta V = \frac{k_BT}{q} \ln{\frac{c_2}{c_1}}$$

This gives us the voltage across a membrane if we know the concentrations on each side.  Here, $q$ is the charge of a single ion. This is sometimes written as $ze$, where $z$ is the valence of the ion (for example, for Na+, $z = 1$), and $e$ is the elementary charge (magnitude of the charge on an electron).

Let's check that this equation makes sense: If $c_1 = c_2$, then $c_1/c_2 = 1$, so $\ln{(c_1/c_2)} = 0$, so there's no potential difference.  That's what we expect. If there's no concentration difference, then there's no voltage. As the concentration difference gets larger, this fraction gets larger, so the potential difference also gets larger.

What's the sign of the potential difference?

Here, we were mainly paying attention to the magnitude of the voltage difference, and not the sign.  This was on purpose:  though it's possible to figure out the sign of the potential difference from the equation, it's better (and less error-prone) to think it through qualitatively and determine which sign makes sense.

In this example, where the positive ions are the ones that can move through the membrane, positive current flows from high concentration to low concentration, so the concentration gradient acts as if the high-concentration side of the membrane is at a higher potential. This effective potential difference is the "Nernst potential". As the current flows, positive charge builds up on the low-concentration side (and therefore the high-concentration side becomes more negatively charged), which means there is an electric potential difference, and the electric potential is higher on the low-concentration side. A net current continues to flow until the electric potential balances out the Nernst potential, and the system reaches equilibrium.

In the situation where the membrane is permeable to negative ions but not to positive ions, negative ions would flow from higher to lower concentration, so, since the direction of the current is the direction that positive charge flows, this means that the current flows from the lower-concentration side to the higher-concentration side! Therefore, in that case, the concentration gradient acts as if the lower-concentration side of the membrane is at a higher potential.  Again, this is the Nernst potential. Therefore, a higher electric potential builds up on the high-concentration side.

It's worthwhile to think this through each time!

Workout: The Nernst potential


Ben Dreyfus 3/1/12 & 3/23/16, Joe Redish 5/4/19

Article 629
Last Modified: May 14, 2019