Ionic Flux

To derive the Nernst-Planck equation, we must start from the molar equation for electrochemical potential:

${\displaystyle {\begin{aligned}{\bar {\mu ))&=\mu ^{\ominus }+RT\ln {a}+zFE,\qquad {\text{ where ))R=8.314{\text{ )){\text{J)){\text{ K))^{-1}{\text{ mol))^{-1},\ F=96\ 485{\text{ )){\text{ C)){\text{ mol))^{-1}\\F_{\text{x))&=-{\partial {\bar {\mu )) \over \partial x}\\&=-\left({RT \over a}\right){\partial a \over \partial x}-zF{\partial E \over \partial x}\\J&=vc=uF_{\text{x))c\\&=-\left({uRTc \over a}\right){\partial a \over \partial x}-uzFc{\partial E \over \partial x}\\D({\text{diffusivity)))&=uRT\\\therefore J&=-\left({Dc \over a}\right){\partial a \over \partial x}-uzFc{\partial E \over \partial x}\\\end{aligned))}$

Activity (a) is more accurate than using concentration (c), as it takes into account electrostatic forces of attraction within the solution. Activities and concentrations are related by the following equation:

${\displaystyle a=\gamma c}$, where γ is the activity coefficient.

The more dilute a solution is, the less significant these forces become, and the activity approaches the actual concentration.

${\displaystyle \lim _{\gamma \to 1}{a}=c}$

If we assume the solution is dilute and the chemical species has no charge (i.e. z = 0), then the equation simplifies into Fick's first law of diffusion:

${\displaystyle J=-D{\partial c \over \partial x))$

Resting Membrane Potential of a Neuron

The resting membrane potential (V_m) can be calculating using the Goldman-Hodgkin-Katz equation, where M represents cations and A are anions:

${\displaystyle V_{\text{m))={RT \over F}\ln {\left({\frac {\displaystyle \sum _{i}P_{\text{M)){\text{M))_((\text{out)),i}^{+}+\displaystyle \sum _{j}P_{\text{A)){\text{A))_((\text{in)),j}^{-)){\displaystyle \sum _{i}P_{\text{M)){\text{M))_((\text{in)),i}^{+}+\displaystyle \sum _{j}P_{\text{A)){\text{A))_((\text{out)),j}^{-))}\right)))$

The two main ions which contribute to the resting membrane potential in neurons are sodium (Na⁺) and potassium (K⁺).

For sodium:

${\displaystyle {\begin{aligned}J&=-uRT{dc \over dx}-uFc{E \over \delta }\\{dc \over J+uFc{E \over \delta ))&=-{dx \over uRT}\\\int _{c_{\text{out))}^{c_{\text{in))}{dc \over J+uFc{E \over \delta ))&=-\int _{0}^{\delta }{dx \over uRT}\\{\delta \over uFE}\ln {\left({J+uFc_{\text{in)){E \over \delta } \over J+uFc_{\text{out)){E \over \delta ))\right)}&=-{\delta \over uRT}\\{J+uFc_{\text{in)){E \over \delta } \over J+uFc_{\text{out)){E \over \delta ))&=\exp {\left(-{FE \over RT}\right)}\\J&=uF{E \over \delta }\left[{\frac {c_{\text{out))\exp {\left(-{FE \over RT}\right)}-c_{\text{in))}{1-\exp {\left(-{FE \over RT}\right)))}\right]\\P&=u{RT \over \delta }\\J&=P{FE \over RT}\left[{\frac {c_{\text{out))\exp {\left(-{FE \over RT}\right)}-c_{\text{in))}{1-\exp {\left(-{FE \over RT}\right)))}\right]\\\end{aligned))}$

When a given neuron is at rest, there is no net ionic flux across the membrane:

${\displaystyle {\begin{aligned}I_{\text{K))+I_{\text{Na))+I_{\text{Cl))&=0\\&=F(J_{\text{K))+J_{\text{Na))-J_{\text{Cl)))\\\end{aligned))}$

This flux terms in brackets yields the following:

${\displaystyle {P_{\text{K)){FV_{\text{m)) \over RT}\left[{\frac {\left[{\text{K))^{+}\right]_{\text{out))\exp {\left(-{FV_{\text{m)) \over RT}\right)}-\left[{\text{K))^{+}\right]_{\text{in))}{1-\exp {\left(-{FV_{\text{m)) \over RT}\right)))}\right]}+{P_{\text{Na)){FV_{\text{m)) \over RT}\left[{\frac {\left[{\text{Na))^{+}\right]_{\text{out))\exp {\left(-{FV_{\text{m)) \over RT}\right)}-\left[{\text{Na))^{+}\right]_{\text{in))}{1-\exp {\left(-{FV_{\text{m)) \over RT}\right)))}\right]}+{P_{\text{Cl)){FV_{\text{m)) \over RT}\left[{\frac {\left[{\text{Cl))^{-}\right]_{\text{in))\exp {\left(-{FV_{\text{m)) \over RT}\right)}-\left[{\text{Cl))^{-}\right]_{\text{out))}{1-\exp {\left(-{FV_{\text{m)) \over RT}\right)))}\right]))$

This means that:

${\displaystyle {\begin{aligned}\exp {\left(-{FV_{\text{m)) \over RT}\right)}\left(P_{\text{K))\left[{\text{K))^{+}\right]_{\text{out))+P_{\text{Na))\left[{\text{Na))^{+}\right]_{\text{out))+P_{\text{Cl))\left[{\text{Cl))^{-}\right]_{\text{in))\right)&=P_{\text{K))\left[{\text{K))^{+}\right]_{\text{in))+P_{\text{Na))\left[{\text{Na))^{+}\right]_{\text{in))+P_{\text{Cl))\left[{\text{Cl))^{+}\right]_{\text{out))\\-{FV_{\text{m)) \over RT}&=\ln {\left({\frac {P_{\text{K))\left[{\text{K))^{+}\right]_{\text{in))+P_{\text{Na))\left[{\text{Na))^{+}\right]_{\text{in))+P_{\text{Cl))\left[{\text{Cl))^{+}\right]_{\text{out))}{P_{\text{K))\left[{\text{K))^{+}\right]_{\text{out))+P_{\text{Na))\left[{\text{Na))^{+}\right]_{\text{out))+P_{\text{Cl))\left[{\text{Cl))^{-}\right]_{\text{in))))\right)}\\\therefore V_{\text{m))&={RT \over F}\ln {\left({\frac {P_{\text{K))\left[{\text{K))^{+}\right]_{\text{out))+P_{\text{Na))\left[{\text{Na))^{+}\right]_{\text{out))+P_{\text{Cl))\left[{\text{Cl))^{-}\right]_{\text{in))}{P_{\text{K))\left[{\text{K))^{+}\right]_{\text{in))+P_{\text{Na))\left[{\text{Na))^{+}\right]_{\text{in))+P_{\text{Cl))\left[{\text{Cl))^{+}\right]_{\text{out))))\right)}\\\end{aligned))}$

A typical neuron has a resting V_m of -70 mV. As such, chloride, whose equilibrium potential is roughly the same, does not contribute significantly to the resting V_m, and can be omitted:

${\displaystyle {\begin{aligned}V_{\text{m))&\approx E_{\text{Cl))\\\therefore V_{\text{m))&={RT \over F}\ln {\left({\frac {\left[{\text{K))^{+}\right]_{\text{out))+\alpha \left[{\text{Na))^{+}\right]_{\text{out))}{\left[{\text{K))^{+}\right]_{\text{in))+\alpha \left[{\text{Na))^{+}\right]_{\text{in))))\right)},\qquad {\text{where ))\alpha ={\frac {P_{\text{Na))}{P_{\text{K))))\\\end{aligned))}$

This equation can be refined by taking into account the effect of the sodium/potassium pump, which counteracts sodium and potassium leak and is vital in maintaining a constant resting V_m.

${\displaystyle V_{\text{m))={RT \over F}\ln {\left({\frac {\left[{\text{K))^{+}\right]_{\text{out))+{\dfrac {\alpha }{r))\left[{\text{Na))^{+}\right]_{\text{out))}{\left[{\text{K))^{+}\right]_{\text{in))+{\dfrac {\alpha }{r))\left[{\text{Na))^{+}\right]_{\text{in))))\right)},\qquad {\text{where ))r={\frac {J_{\text{Na))}{J_{\text{K))))={\frac {3}{2))}$