Passive transport

Passive transport is net movement down the transported substance's electrochemical gradient without coupling that movement to another source of free energy. Facilitated diffusion is passive transport through an ion channel or a passive carrier protein. Providing such a route can speed passage, but does not turn an uphill transfer into a downhill one. For an uncharged solute under the ideal conditions above, net passive transport runs from higher to lower concentration; for an ion, the membrane potential also matters. See Cooper, The Cell: transport of small molecules.

In the figure, stipulate eight neutral solute tracers in A and two in B, with equal fixed volumes, identical surroundings, and no imposed electric potential difference. The concentration ratio is $c_{\mathrm{A}}/c_{\mathrm{B}}=8/2=4$, so A-to-B transfer has $\Delta \tilde{\mu }=-RT\ln 4\simeq -1.38629RT$, or $-3.458\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ at the assumed temperature $T=300\mathrm{K}$. If this solute can cross, thermodynamic equilibrium has equal expected concentrations: five tracers per side on average. Individual crossings occur in both directions; an observed ten-particle snapshot need not contain exactly five on each side. These stipulated counts illustrate a driving force, not a measured flow rate or a prediction of the time to equilibrate.

For a separate illustrative ion calculation, keep $c_{\mathrm{A}}/c_{\mathrm{B}}=4$, but assign $z=+1$ and $V_{\mathrm{A}}-V_{\mathrm{B}}=-60\mathrm{mV}$ at $300\mathrm{K}$. Using $R\simeq 8.31446\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$ and $F\simeq 96485.33\mathrm{C}\mathrm{mo}{\mathrm{l}}^{-1}$, B-to-A transfer gives $RT\ln 4-0.060F\simeq 3.458-5.789=-2.331\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. Thus an appropriate open ion channel can permit passive transport toward the higher concentration: the electrical contribution outweighs the opposing chemical potential contribution. This stipulated membrane potential is not inferred from the tracer counts.