Deriving goldman equation

Question

So I've been trying to follow the derivation of the Goldman-Hodgkin-Katz equation on Wikipedia (https://en.wikipedia.org/w/index.php?title=Goldman_equation ) - I can follow it through after it sets up the initial model.

However, I'm having trouble using the Einstein-Stokes relationship to derive the initial model in the first place.

There is the flux from diffusion down the chemical gradient, $-D_{A}\frac{d[A]}{dx}$. The negative sign is there because I'm imagining the ions being "lost" from the side I'm considering.

Then we have the electrical field pushing ions the other way. This won't depend upon the rate of change of concentration because the electric field is uniform and so there's effectively a constant force on each of the particles. The flow will instead depend upon the constant velocity that these particles will be pushed at. So the flow will be $[A]v_d$ - the moles per unit volume being multiplied by the speed in the direction towards the membrane will give me the number of moles per unit area passing through the membrane per second.

So all I need to do is work out $v_d$ I'm guessing. So I use the electrical mobility equation, $D_A=\frac{\mu_{q}k_BT}{q}$.

$q$ will be the charge, so $nzF$ (where $z$ is the valance, $nz$ gives the moles of electrons and multiplying by $F$ gives the charge).

$\mu_q$ will be $\frac{v_d}{E} = \frac{v_d L}{V_m}$

$k_B = \frac{R}{N_A}$ where $N_A$ is Avogadro's number.

So $D_A=v_d \frac{L}{V_m} \frac{RT}{zF} \frac{1}{N_A n}$, thus $D_A \frac{V_m}{L} \frac{zF}{RT} N_A n=v_d$

Putting this back in to get the ion flux, $D_A \frac{V_m}{L} \frac{zF}{RT} N_A n [A]$.

Adding my two fluxes together to get the total flux, I get: $j_a = -D_A(\frac{d[A]}{dx} - \frac{V_m}{L} \frac{zF}{RT} N_A n [A] )$ - this is not quite what the Wikipedia article has - a stray $n N_A$ has crept in, somehow!

Please help me by pointing out where I'm making a mistake!

Answer 1

The mistake is in the charge $q$, that you have put in:

$ D_A=\frac{\mu_{q}k_BT}{q} $

$q$ is simply the particle charge (and not the whole charge, as you've unfortunately considered) and should then be expressed as:

$q=\frac{zF}{N_{A}}$

with $F$ the Faraday constant (charge / mole ), ${N_{A}}$ the Avogadro's number (number of particle / mole) and $z$ the charge valence.

${N_{A}}$ will finally cancel out when you divide Boltzmann's constant $k_B$ by the the charge $q$, in the definition of diffusion coefficient. That will allow you to recover the Goldman equation found on Wikipedia.

By checking all that through an always useful dimension analysis for $D_A$ and using the defintions you have given in your question for ($\mu_q$ = $\frac{v_d}{E} = \frac{v_d L}{V_m}$ and $k_B = \frac{R}{N_A}$):

$ \frac{\mu_{q}k_BT}{q}=\frac{m/s\,\times m}{V}\times \frac{J/(mol\times K)\,\times K}{{N_A}\times C/mol}\times{N_A}=\frac{m^2}{s} $

you find the correct units for a diffusion coefficient.