The Nernst Equation

The past two weeks I have learned a lot about how we look at neurons and, more specifically, the neuronal membrane as electrical circuits. It turns out that the largest factor influencing the membrane potential is the conductance of the ion channels embedded in the membrane. Perhaps a future post will be dedicated to understanding the contribution of ion channels to membrane potential. In order to understand the resting state of a membrane, however, we do not need to worry about the effects of the channels. Last week we learned the derivation for an important equation used to predict membrane potentials for different ions. This is called the Nernst Equation. The information and equations given here are taken from notes of a lecture given by Dr. John White to our cellular biophysics class.

We begin with Fick's First Law, which describes random thermal motion

\vec{\varphi}_{n(D)} = -D_{n} \nabla c_{n} \mbox{,} \; \nabla c_{n} = \mbox{gradient of } c_{n}=\frac{\partial c_{n}}{\partial x} \mathbf{i} + \frac{\partial c_{n}}{\partial y} \mathbf{j}+ \frac{\partial c_{n}}{\partial z} \mathbf{k} \mbox{.}

where phi is the flux of the concentration due to pure diffusion, c is the concentration, and D is the diffusion coefficient.

In one dimensional steady-state, this equation turns into

\vec{\varphi}_{n(D)} = -D_{n} \frac{\mathrm{d}c_{n}}{\mathrm{d}x} \mbox{,}

\mbox{where } c_{n} [=] \mbox{mol}/\mbox{L} = \mbox{M} \mbox{, } \vec{\varphi}_{n} [=] \mbox{mol}/(\mbox{sm}^{2}) \mbox{, and } D_{n} [=] \mbox{m}^{2}/\mbox{s}.

The diffusion coefficient relies on the lipid solubility of the species being examined. Now, let's add in the flux due to external force

\vec{\varphi}_{n(F)} = c_{n} v_{n}, \mbox{ where } v_{n} [=] \mbox{ m/s is the drift velocity of the species}.

We can substitute

v = u_{n} f, \mbox{ where } u_{n} [=] \mbox{(m mol)/(N s) is the molar mechanical mobility of the species}.

Now we add the two fluxes together

\vec{\varphi}_{n} = \vec{\varphi}_{n(D)} + \vec{\varphi}_{n(F)} = -D_{n} \frac{\mathrm{d}c_{n}}{\mathrm{d}x} + u_{n} f c_{n}(x,t).

In the case of ions in a neuron, the force is applied via an electric field.

\varepsilon = -\frac{\mathrm{d}\psi}{\mathrm{d}x}, \mbox{ where } \psi [=] V.

From here, we convert the force per charge of voltage to force per mole by multiplying by zF, where z is the charge per molecule and F is the Faraday constant (about 9.65x10^4 C/mol). Doing this gives us

f = \varepsilon z_{n} F = -z_{n} F \frac{\mathrm{d}\psi}{\mathrm{d}x}.

Therefore, our sum of fluxes is now

\vec{\varphi}_{n} = -D \frac{\mathrm{d}c_{n}}{\mathrm{d}x} -u_{n} z_{n} F c_{n}(x,t) \frac{\mathrm{d}\psi}{\mathrm{d}x}.

Granted, this equation only takes into account an electric field of one-dimension and a concentration gradient of one-dimension. To make this more useful, we will assume the membrane is of width d and that the flux across the membrane is zero.

\vec{\varphi}_{n} = 0 \Rightarrow u_{n} z_{n} F c_{n}(x) \frac{\mathrm{d}\psi(x)}{\mathrm{d}x} = -D \frac{\mathrm{d}c_{n}(x)}{\mathrm{d}x}

The diffusion coefficient is related to molar molecular mobility by the relationship

D_{n} = u_{n} R T,

which leads us to

R T \frac{\mathrm{d}c_{n}(x)}{\mathrm{d}x} = -z_{n} F c_{n}(x) \frac{\mathrm{d}\psi(x)}{\mathrm{d}x} \Rightarrow R T \frac{1}{c_{n}(x)} \frac{\mathrm{d}c_{n}(x)}{\mathrm{d}x} = -z_{n} F \frac{\mathrm{d}\psi(x)}{\mathrm{d}x}

This is equivalent to

\frac{R T}{z_{n} F} \frac{\mathrm{d}[\ln c_{n}(x)]}{\mathrm{d}x} = \frac{\mathrm{d}\psi(x)}{\mathrm{d}x}

Almost done! If we then integrate both sides of the equation over the interval [0,d],

\frac{R T}{z_{n} F} [\ln c_{n}(d) - \ln c_{n}(0)] = \frac{R T}{z_{n} F}\ln \frac{c_{n}(d)}{c_{n}(0)} = \psi(0)-\psi(d).

Assuming both c(x) and psi(x) are continuous at the boundaries, this leaves us with

V_{n} = \frac{R T}{z_{n} F} \ln \frac{c_{n}^{o}}{c_{n}^{i}},

or the Nernst Equation!

Posted September 8th, 2010 in Neuroscience, Science.