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Ionic Flux

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

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

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:

$a=\gamma c$, 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.

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

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:

$J=-D{\partial c \over \partial x}$

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:

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

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

For sodium:

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

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

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

This flux terms in brackets yields the following:

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

This means that:

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

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:

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

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.

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