Sodium-motive force (${\mathcal{E}}_{\mathrm{Na}}=(k_{B}T/e)\ln (c_o/c_i)-V_m$)

Define the displayed quantity as the free energy available per unit positive charge for sodium entry, with $V_m={\psi }_i-{\psi }_o$. Its negative is the signed inward electrochemical cost per charge. A reversal of the transport direction reverses the sign. At equilibrium it vanishes without requiring equal concentrations.

Start at sodium electrochemical equilibrium with $n_0$ internal ions. Exporting $n$ ions changes the membrane potential by $-ne/C$ and the internal number density to $(n_0-n)/V$. Subtracting the initially balanced chemical and electrical contributions gives the displayed incremental cost. An initial zero potential further implies $n_0=c_{o}V$ for an ideal fixed external reservoir. Nonzero initial equilibrium potential permits unequal concentrations, so that latter equality is an additional assumption.