Sodium-motive force
= Sodium-motive force
{title2=$\mathcal E_{\rm Na}=(k_BT/e)\ln(c_o/c_i)-V_m$}
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.