= Selective potassium equilibrium
{title2=$E_{\mathrm K}=\frac{RT}{F}\ln(c_{\mathrm{out}}/c_{\mathrm{in}})$}
For an ideal dilute <chemical solution>, moving one <mole> of a species with signed <ion> charge number $z$ from outside to inside changes its <electrochemical potential> by
$$
\Delta\widetilde\mu=RT\ln\!\left(\frac{c_{\mathrm{in}}}{c_{\mathrm{out}}}\right)+zFV_m.
$$
Here $R$ is the <molar gas constant>, $T$ the uniform absolute <temperature>, $F$ the <Faraday constant>, and $c$ the <concentration>. This assumes the same reference <chemical potential> on both sides. At <thermodynamic equilibrium>, $\Delta\widetilde\mu=0$, hence
$$
E_{\mathrm{ion}}=\frac{RT}{zF}\ln\!\left(\frac{c_{\mathrm{out}}}{c_{\mathrm{in}}}\right).
$$
The <Nernst equilibrium potential> uses <thermodynamic activities> in the general case; replacing them by <concentrations> is an ideal-solution approximation.
Choose an original passive model: only $\mathrm K^+$ permeates ($z=+1$), other <ions> cannot cross, there are no pumps or imposed <electric currents>, and there is no net <water> transfer. Stipulate reservoir <concentrations> $c_{\mathrm{in}}=140\,\mathrm{mM}$ and $c_{\mathrm{out}}=5\,\mathrm{mM}$, with $T=310\,\mathrm K$, $R=8.314\,\mathrm{J\,mol^{-1}\,K^{-1}}$ and $F=96485\,\mathrm{C\,mol^{-1}}$. Treat reservoir <concentrations> as effectively unchanged by charging. Then
$$
E_{\mathrm K}=\frac{8.314\times310}{96485}\ln\!\left(\frac5{140}\right)\,\mathrm V
\approx-0.08901\,\mathrm V=-89.01\,\mathrm{mV}.
$$
The <concentration> ratio is dimensionless; $\mathrm{J/C}=\mathrm V$ supplies the voltage unit. At $V_m=0$, $\Delta\widetilde\mu>0$ for inward transport, so outward transport is favored. At $V_m=E_{\mathrm K}$ it vanishes. A more negative $V_m$ favors inward transport. These sign checks explain the result physically.
Keeping the other chosen inputs fixed, doubling $c_{\mathrm{out}}$ to $10\,\mathrm{mM}$ raises $E_{\mathrm K}$ by $(RT/F)\ln2\approx18.52\,\mathrm{mV}$. Equal inside and outside <concentrations> give $E_{\mathrm K}=0$; reversing their ratio reverses its sign. In finite reservoirs, transport changes both <concentrations>; the exact final <thermodynamic equilibrium> must also satisfy material conservation. The fixed-reservoir calculation is useful when only a small fraction of the available <ions> moves.
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