Equilibrium Potentials I

Electrochemistry of the Nerve Cell

Neurons are both electrical and chemical devices. Understanding the basic electrochemistry of nerve cells serves as the foundation for understanding all of a neuron's important electrical properties.

Key ideas for this unit:

• Equilibrium versus steady state
• Diffusion and concentration gradient
• Osmotic balance
• Electroneutrality
• Equilibrium potential (the Nernst potential)

Equilibrium versus Steady State

• If the heating or cooling system shut down, would your room stay the same temperature?
• If it were cold outside, it would get very cold; if it were hot outside, it would get very hot.
• How does it stay at almost the same temperature when the system is working?
• A thermostat senses the room temperature, and it adjusts the amount of heating or cooling needed to keep the temperature steady.
• Thus, this is an example of a steady state that requires energy for its maintenance.
• In contrast, if the room reaches the temperature of the outside, no energy is needed to keep it at that temperature; it has reached thermal equilibrium.
• Are living organisms at equilibrium?
• Very far from it; we only reach equilibrium after death. We need constant infusions of energy to maintain a steady body temperature.
• The nerve cell invests a great deal of energy in setting up a steady state electrical potential across its membrane, known as the resting potential.
• The resting potential, however, is built upon several equilibria, which we will need to understand first; once we do, we can understand the electrochemical basis for the resting potential.

A Simulation Program for Understanding Electrochemistry

Figure 1: The Nernst Simulator

• To follow the details of how equilibria, and then the resting potential, are established across the membrane of the nerve cell, it is helpful to visualize the movements of charged entities, ions, across the membrane.
• To see this happening in real time, we have created a simulation, which we call Nernst.
• Download the Nernst program, and run it on your computer:

• Once you have the program running, you will see a screen that looks like Figure 1:
  • On the left-hand side of the screen, you see the settings that you can control.
  • In the middle of the screen, you see the actual simulation of the movements of the different ions.
  • On the right-hand side of the screen, you see some of the measurements that can be made, based on the simulation results.
  • We will use this program to explore and understand the electrochemistry of the nerve cell.

The Three Conditions for Membrane Equilibrium

Three conditions must be satisfied to ensure equilibrium across a membrane:

1. If the membrane is permeable to water,

the concentration of solute particles on one side of the membrane must equal the concentration of solute particles on the other side of the membrane (osmotic balance).

2. If the membrane has charged particles on either side of it,

the concentration of charges on each side of the membrane must sum to zero (bulk electroneutrality).

3. If the membrane is permeable to charged particles,

the chemical gradient must equal the electrical gradient (Nernst or equilibrium potential).

We will now explore each of these conditions by setting up different conditions on either side of a membrane, and see how the results unfold, as well as how we can capture these ideas mathematically so that we can predict the outcome in advance of doing the experiment.

Diffusion and Concentration Gradients

Let's first examine an apparently very simple situation: a membrane that is permeable to water, which is placed between two compartments filled with water.

Figure 2: Water and semi-permeable membrane

• What happens?
  • At the macroscopic level, nothing appears to happen.
  • But at the microscopic level, a great deal is happening: water molecules approach and bounce off the membrane. Some water molecules actually fit through the holes in the membrane, and diffuse across it.
  • As long as the number of water molecules crossing from side 1 to side 2 is, on average, about the same as the number crossing from side 2 to side 1, nothing appears to change macroscopically.

What happens if we have an unequal number of particles dissolved in water on side 1 of a membrane versus side 2?

Figure 3: Glucose and semi-permeable membrane

• Let's begin by assuming that the membrane is equally permeable to both water and glucose molecules. What happens?
  • On the side that has more glucose dissolved in water, the probability of a glucose molecule hitting up against the membrane is higher than on the side that has less dissolved glucose.
  • As a consequence, the probability of a molecule of glucose on the right side diffusing to the left side of the membrane is higher than the probability that this will happen in the opposite direction.
• What happens over time? We can readily calculate this:
  • Assume that the concentration of glucose on side 1 is $C_1$, and that the concentration of glucose on side 2 is $C_2$.
  • Then, the probabilities of diffusion in either direction will become the same only when the concentration on both sides is equal.
  • This will be true when the concentration on each side is the average of the concentrations on the different sides, or $$\frac{C_1+C_2}{2}. \quad\text{(Equation 1)}$$
  • For the specific example in the figure, the concentration at equilibrium is $(100\ \text{mM}+200\ \text{mM})/2 = 150\ \text{mM}$.

Problem 1: Diffusion and Concentration Gradients

You now have enough information to work on Problem Set 1, Problem 1. Please do so now.

Osmotic Balance

The problem shown in Figure 3 becomes more interesting if we assume that water can freely move from one side of the membrane to the other, but glucose cannot.

• How would this work?
  • Recall that the glucose molecule is significantly larger than a water molecule. So if the holes in the membrane were the right size, they would let water molecules through, but prevent glucose molecules from moving from one side to the other.
  • Thus, the size of a membrane channel is one way to create selective permeability for a molecule.
• Recall the first rule we stated above: the concentration of solute particles on either side of the membrane has to be the same.
  • Since the glucose molecules can no longer cross the membrane, the water molecules must move to satisfy this rule.
  • To see which way they will go, ask: where are the water molecules more concentrated?
  • They are more concentrated on the side that has fewer glucose particles dissolved in water (which displace the water molecules).
  • So, for the problem shown in Figure 3, this means that water molecules will tend to diffuse from the left side to the right side.
  • As they do so, the volume of the left side will shrink, and the volume of the right side will increase as more water molecules move to the right.
  • This is the basis for osmotic pressure.
• When will the probabilities of water moving to the left or the right sides again be equal?
  • As the left side shrinks, the concentration of glucose rises; as the right side swells, the concentration of glucose falls.
  • When the volume has changed so that the concentration of glucose is the same on both sides, the concentration of the water on both sides will also become equal, and the system will stop changing.
  • We can calculate when this happens if we let $x$ represent the change in volume, $C_1$ and $V_1$ the initial concentration and volume on the left side, and $C_2$ and $V_2$ the initial concentration and volume on the right side.
    • When the concentrations are equal, the system will stop changing, which we can write symbolically as $$\frac{C_1 V_1}{V_1 - x} = \frac{C_2 V_2}{V_2 + x}. \quad\text{(Equation 2)}$$ Here the quantities $C_1 V_1$ and $C_2 V_2$ are the number of moles of glucose on either side of the membrane because concentration is measured in $\text{millimolar} = \text{millimole/liter}$.
    • We have assumed that the left side has the higher concentration of water in it, so that it will lose volume (which is the reason that $x$ has a negative sign in the denominator on the left), and that the right side will gain an equal amount of volume (which is the reason that $x$ has a positive sign in the denominator on the right). If the assumption is wrong, the equation will still work, but $x$ will have a negative value.
    • We can solve for $x$ to determine the volume change, and we obtain $$x = \frac{C_2 V_2 V_1 - C_1 V_1 V_2}{C_2 V_2 + C_1 V_1}. \quad\text{(Equation 3)}$$
    • For the particular case we are considering, assuming that each initial volume is 1 liter, we find that $$\begin{align} x &= \frac{(200\ \text{millimole/liter}) (1\ \text{liter}) (1\ \text{liter}) - (100\ \text{millimole/liter}) (1\ \text{liter}) (1\ \text{liter})}{(200\ \text{millimole/liter}) (1\ \text{liter}) + (100\ \text{millimole/liter}) (1\ \text{liter})} \\ &= \frac{100\ \text{millimole liter}}{300\ \text{millimole}} \\ &= \frac{1}{3}\ \text{liter}. \end{align}$$
    • At the start of this experiment, the number of glucose molecules on the left side was $100\ \text{millimole/liter} \times 1\ \text{liter} = 100\ \text{millimole}$. Since the membrane is impermeable to glucose, the number of glucose molecules is still $100\ \text{millimole}$ at the end of the experiment. The left side was initially more dilute, so the volume shrinks to $1 - \frac{1}{3}\ \text{liter} = \frac{2}{3}\ \text{liter}$. Thus the final concentration on that side is $\frac{100\ \text{millimole}}{2/3\ \text{liter}} = 150\ \text{millimole/liter}.$
    • The number of glucose molecules on the right side is $200\ \text{millimole/liter} \times 1\ \text{liter} = 200\ \text{millimole}$ (again, this does not change over the course of the experiment). The right side was initially more concentrated, so the volume swells to $1 + \frac{1}{3}\ \text{liter} = \frac{4}{3}\ \text{liter}$. Thus the final concentration on that side is $\frac{200\ \text{millimole}}{4/3\ \text{liter}} = 150\ \text{millimole/liter}.$
  • As promised, the final concentrations are identical on both sides.

Now, let's extend our analysis a step further. Consider a cell with a membrane permeable to water, containing a large number of impermeable particles (e.g., proteins), immersed in a bath containing the same number of impermeable particles, but of a different kind (e.g., an impermeable sugar such as sucrose), as shown in Figure 4. Assume that the bath's volume is large relative to the volume of the cell.

Figure 4: Cell in an isotonic bath

• Based on the rule that the concentration of solute particles on either side must be identical for osmotic balance, this system will be in equilibrium.
• What will happen if the concentration of the external sucrose is reduced by half, to 0.125 molar?
  • The side on which water is more concentrated is now the outside of the cell; so water will tend to diffuse into the cell, swelling its volume so that it ultimately increases in volume to twice its original volume, reducing the internal concentration of the impermeable particles within it to be the same as those outside ($0.125\ \text{millimolar}$). When the concentration of particles outside the cell is lower than inside the cell, this is known as a hypotonic solution. If the swelling is too extreme, the cell will burst; so this is a serious problem for the survival of the cell!
• What will happen if the concentration of the external sucrose is doubled to 0.5 molar?
  • The side on which water is more concentrated is now the inside of the cell; so water will tend to diffuse out of the cell, reducing its volume so that it ultimately $1/2$ its initial size, increasing the internal concentration of the impermeable ions within it to be the same as those outside ($0.5\ \text{millimolar}$). When the concentration of particles outside the cell is higher than inside the cell, this is known as a hypertonic solution. This can also be lethal to cells; so regulating osmotic balance is very important.
• Note again that this was assuming that the volume of the bath was far larger than the volume of the cell, so that we could ignore the changes in the volume of the bath, which would be negligible compared to those in the cell. If the bath volume was comparable in size to the cell, then the analysis we did above would again be relevant.

Electroneutrality

So far, we have dealt with uncharged particles. In water, however, many molecules that are bound together ionically can form weak bonds with water molecules.

• A negatively-charged ion (such as a chloride ion) can form weak bonds with the positively charged hydrogen atoms within the water molecule.
• A positively-charged ion (such as a sodium ion) can form weak bonds with the negatively charged oxygen atom within the water molecule.
• These associated water molecules, which allow the ions to dissolve into water, are called waters of hydration.
• Separating electronic charges requires large amounts of energy, so if positively charged ions diffuse from one location to another, they will usually be accompanied by negatively charged ions; their charges thus cancel out, leading to electroneutrality.

Given this information, we can make some predictions about what will happen if we create a concentration gradient of charged particles across a membrane. Consider the situation presented in Figure 5.

Figure 5: Unequal concentrations of charged particles on each side of a membrane

• If the membrane is permeable to both sodium and chloride ions, what will happen?
  • We can apply the same logic that we used in the previous section to this problem. Clearly, each negatively charged chloride ion will move with a positively charged sodium ion, and given that the concentration of the ions is much higher on the right side, they will tend to diffuse to the less concentrated left side, until the concentration has become equal on both sides.
  • We can calculate the final concentration, as we did above for the uncharged glucose (Equation 1). For the situation shown in this figure, the final concentration would be $\frac{0.1\ \text{M} + 1.0\ \text{M}}{2} = \frac{1.1\ \text{M}}{2} = 0.55\ \text{M}$.

Balancing Concentration and Electrical Gradients: The Equilibrium or Nernst Potential

How would one create a selectively permeable ion channel?

• The waters of hydration around an ion create an entity of a certain size; the water molecules bind to the ion with a strength related to the ionic charge.
• If a membrane contains a channel that has a certain size, and that channel is lined with appropriate kinds of charges that can take the place of the waters of hydration as an ion passes through it, then one ion will be more likely to permeate that channel than others.
• In general, channels are not perfectly selective; but they are sufficiently selective that one can reasonably speak of a "chloride channel" or a "sodium channel", i.e., a channel that primarily allows chloride ions or sodium ions through (respectively), and does not let other ions through to any appreciable extent.

What would happen to the system shown in Figure 5 if the membrane were only permeable to the chloride ion?

• If only chloride ions are permeable, they will begin to diffuse from the side on which they are more concentrated (the right side) to the side on which they are less concentrated (the left side).
• However, since the sodium ions are not permeable, they can no longer follow the chloride ions across the membrane.
• As chloride ions diffuse to the left side, excess negative charges build up near the membrane on the left (because each chloride ion carries a negative charge to that side), and positive charges build up near the membrane on the right (because, as each chloride ion leaves this side, a positive charge is left behind). These excess charges tend to collect near the membrane because the negative charges on the left are attracted to the positive charges on the right.
• The separated charges across the membrane begin to create an electrical field gradient that applies a force on the charged particles, and this gradient is oriented so that it will increasingly oppose the additional flow of negative charges to the left side of the membrane.
• Thus, we have created a situation in which the concentration gradient sets up an opposing electrical gradient (the net effect of these two gradients is referred to as the electrochemical gradient). This occurs spontaneously, without an input of energy. When the opposing gradients exactly balance one another (or equivalently, when the electrochemical gradient is zero), the system is in equilibrium.
• When equilibrium occurs, the concentrations of chloride ions on either side of the membrane will not change appreciably, but there will be more negative charges on the left and more positive charges on the right. This separation of charges creates a difference in the potential energy an ion has when placed on the left versus on the right. This potential difference is called the membrane potential, denoted $V_m$. The potential difference across a membrane can be measured with a volt meter by placing its electrodes on either side of the membrane.
• When the membrane is permeable to only one ion (such as chloride ions, as in this scenario), we call the membrane potential at equilibrium the equilibrium potential or Nernst potential for that ion. We will not provide the derivation here, but the equilibrium potential, $E_{\text{ion}}$, can be shown to equal the following: $$E_{\text{ion}} = \frac{RT}{z\mathcal{F}} \ln \frac{\left[\text{ion}\right]_\text{out}}{\left[\text{ion}\right]_\text{in}}, \quad\text{(Equation 4)}$$ where $E_{\text{ion}}$ is the potential across the membrane created by the separation of charges (measured in volts), $R$ is the universal gas constant (in joules per kelvin per mole), $T$ is the absolute temperature (in kelvin), $\mathcal{F}$ is Faraday's constant (which represents the number of coulombs of charge per mole of electrons), $z$ is the charge on the permeable ion, $\left[\text{ion}\right]_\text{out}$ is the concentration of the permeable ion outside the cell (in Figure 5, this is the right side), and $\left[\text{ion}\right]_\text{in}$ is the concentration of the permeable ion on the inside of the cell (the left side of Figure 5).
• Given the equation, and the data for the initial condition of the system, we can actually calculate the equilibrium or Nernst potential for our model cell.
  • At about $19^\circ\ \text{C}$ (a few degrees below room temperature), the value of $\frac{RT}{\mathcal{F}} \ln[x]$ is equal to $58\ \text{mV} \times \log_{10}[x]$. Note that for chloride, the value of $z$ is $-1$, because the chloride ion has a net charge of $-1$. So, applying Equation 4 using the values shown in Figure 5, we obtain $$E_{\text{Cl}^-} = -1 \times 58\ \text{mV} \times \log_{10} \frac{1.0\ \text{M}}{0.1\ \text{M}}.$$ Given the rules for logarithms, we eliminate the minus sign in front by flipping the numerator and the denominator, so that we obtain $$= 58\ \text{mV} \times \log_{10} \frac{0.1\ \text{M}}{1.0\ \text{M}},$$ which simplifies further to $$= 58\ \text{mV} \times -1,$$ or $E_{\text{Cl}^-} = -58\ \text{mV}$.
• In summary, what this means is that when our model cell is set up with the same ionic concentrations as in Figure 5 and with a membrane that is selectively permeable to chloride ions, the following will happen:
  • The chloride ions will diffuse down their concentration gradient, reducing the gradient as they do so.
  • An electrical gradient that opposes the movement of the chloride ions will form and increase in strength due to the imbalance in changes created by the movement.
  • The two gradients will eventually balance one another, and net movement of the chloride ions will cease; this is equilibrium.
  • At equilibrium, there is still a charge imbalance, and it is quantified by the equilibrium potential. Under the conditions given here, the equilibrium potential is $-58\ \text{mV}$.
• We have just solved the Nernst equation for a specific ion, and so it would be worth recording the equation for doing this for future reference (note that this equation applies only for chloride ions that have a charge of $-1$, and at about $19^\circ\ \text{C}$, i.e., a few degrees below room temperature): $$E_{\text{Cl}^-} = 58\ \text{mV} \times \log_{10} \frac{\left[\text{Cl}^-\right]_\text{in}}{\left[\text{Cl}^-\right]_\text{out}}. \quad\text{(Equation 5)}$$ Note that at mammalian body temperature, which is about $37^\circ\ \text{C}$, the constant $58\ \text{mV}$ would be replaced by $62\ \text{mV}$.
• What about osmotic balance? By the first rule, water should move from the side that has more water molecules (the left side) to the side that has less (the right side), shrinking the size of the cell. Using Equation 3, and assuming for simplicity that each volume is 1 liter initially, it is easy to show that that the volume change is 9/11 liter (please do the calculation to make sure you see that this is true! Please include your work and the results of your calculation in your notebook.), and using this result in Equation 2, it is easy to show that the total concentration of particles on each side of the membrane will now be equal.
• What happens to the Nernst potential for chloride under those circumstances? Examine Equation 5. The concentration of the chloride ion will now be the same inside and out. What will the resulting voltage be? Please do this calculation. Please include your work and the results of your calculation in your notebook. If you don't find it easy, please review your understanding of logarithms.

We now have enough information to take on a somewhat more challenging problem. There are generally impermeable particles on both sides of a cell membrane, usually proteins. Some of these have net charges, and some are uncharged. Consider the situation shown in Figure 6, in which the membrane is permeable only to water and to chloride ions; 100 mM of impermeable proteins, represented as 100 mM P, are also present inside the cell.

Figure 6: Balancing osmotic pressure, bulk electroneutrality, and concentration and electrical gradients

• Here is how to analyze the problem:
  • We need to ensure bulk electroneutrality within the cell, that is, there have to be as many positive as negative charges inside. What is the internal concentration of chloride ions based on this rule (i.e., the concentration of chloride ions on the left side of Figure 6)?
  • Since we've filled in the concentrations of all components within the cell, what is the total concentration of solute particles inside the cell (left side of Figure 6)?
  • We have to maintain bulk electroneutrality outside of the cell as well; additionally, to guarantee osmotic balance, we have to ensure that the concentration of solute particles outside the cell equals the concentration of solute particles inside the cell. Given this information, what are the concentrations of the sodium and chloride ions outside the cell (right side of Figure 6)?
  • Now that you have the concentrations of chloride ions both outside and inside the cell, what is the equilibrium or Nernst potential across the cell membrane, i.e., please calculate $E_{\text{Cl}^-}$ using Equation 5.
• The only way to be sure that you understand the material so far is to actually try to do the calculations that have just been outlined. Please do so now.
• Once you've tried the calculations, you can check the results by clicking here. Please include your work and the results of your calculation in your notebook.
• Those of you who think carefully about what is happening may have the following very important question:
  • We need to maintain equal total concentrations of solute particles on both sides of the membrane to ensure osmotic balance.
  • But, to set up the equilibrium or Nernst potential, some chloride ions need to move from the region where they are more concentrated to the region in which they are less concentrated.
  • How would this be possible without altering the total concentrations?
  • The answer is that in fact the concentrations are slightly altered; but the number of ions that need to move from one side to the other before equilibrium is established is very, very small relative to the much larger number in the bulk solution.
  • As a consequence, they have a minuscule impact on the total concentrations on either side.
  • This is an important question, however, because sometimes if the volume of the cell region is small (e.g., a small spine on a cortical dendritic tree), then the movement of a relatively small number of ions can make a difference in the total concentrations.
  • For most of the problems you will need to think about this semester, you will not have to worry about this inaccuracy.

At equilibrium, the fluxes of chloride ions into and out of the cell are in balance (the net flux is zero). We can readily demonstrate, both experimentally and theoretically, that this balance is stable, i.e., if it is slightly perturbed, the membrane will return to the equilibrium value.

• Imagine for a moment that the cell in Figure 6 is in equilibrium, and you could experimentally move a small number of chloride ions from the inside of the cell to the outside. This would create more negative charges on the outer surface of the membrane, and fewer on the inside surface of the membrane. Such a change corresponds to an increase in the membrane potential, and we say that the cell depolarizes.
  • What would this change do to the concentration gradients of the chloride ions? At equilibrium, the concentration of chloride ions outside the cell was twice the concentration inside the cell. When we move a small number of chloride ions out of the cell, the concentration gradient becomes even greater than it was before.
  • What would this change do to the electrical gradient? At equilibrium, the membrane potential was $-17.5\ \text{mV}$, and so the negatively-charged chloride ions were repelled from the inside of the cell. When we make our small change, the membrane depolarizes. This means the electric field becomes weaker, and the repulsion from the inside of the cell felt by the chloride ions is weakened. (In fact, if we depolarize the cell enough so that the membrane potential becomes positive, the chloride ions will actually become attracted to the inside of the cell.)
  • Both of these effects favor the movement of chloride ions from the outside of the cell to the inside. Thus, the small change we made will be undone automatically. The distribution of ions will return to what it was before, at equilibrium.
• Imagine instead that the cell in Figure 6 is in equilibrium, and you move a small number of chloride ions from the outside of the cell to the inside. This would create more negative charges on the inside surface of the membrane, and fewer on the outside surface of the membrane. Such a change corresponds to a decrease in the membrane potential, and we say that the cell hyperpolarizes.
  • What would this change do to the concentration gradients of the chloride ions? At equilibrium, the concentration of chloride ions outside the cell was twice the concentration inside the cell. When we move a small number of chloride ions into the cell, the concentration gradient is decreased.
  • What would this change do to the electrical gradient? At equilibrium, the membrane potential was $-17.5\ \text{mV}$, and so the negatively-charged chloride ions were repelled from the inside of the cell. When we make our small change, the membrane hyperpolarizes. This means the electric field becomes stronger (more negative), and the repulsion from the inside of the cell felt by the chloride ions is stronger.
  • Both of these effects favor the movement of chloride ions from the inside of the cell to the outside. Thus, the small change we made will be undone automatically. The distribution of ions will return to what it was before, at equilibrium.
• If a system that is in equilibrium is perturbed and it returns back to its previous value, one refers to that equilibrium as stable. We can see from this thought experiment that the Nernst or equilibrium potential is a stable equilibrium: small perturbations to the value of the membrane potential will be rapidly undone by the underlying fluxes.
• Note that once the concentration difference is set up in Figure 6, the equilibrium requires no further energy for its maintenance. The flux of chloride into the cell balances the flux of chloride out of the cell, so the overall concentration will not change, even if the potential of the membrane is briefly perturbed.

• Once you've mastered the material in this unit, you are ready to go on to understand the electrochemical basis of the resting potential.

Setting Up an Equilibrium Potential

The process by which a single ion (potassium) sets up an equilibrium (Nernst) potential across the membrane

Understanding Hyperpolarization and Depolarization

Charge separation across the membrane, hyperpolarization and depolarization

Membrane Potential and the Effects of Changing Concentrations vs. Injecting Current

Charge separation across the membrane, hyperpolarization and depolarization