= Solution
Suppose initially $c_1>c_2$. Positive <ions> initially move by <diffusion> from side 1 to side 2. Since negative <counterions> cannot cross, side 1 acquires a negative interfacial charge and side 2 a positive one. The resulting electrical <force> opposes further positive-ion transfer. The two contributions balance when the <electrochemical potentials> are equal; individual <ions> still cross in both directions, but there is no net <electric current> of the permeant <chemical species>.
A simple <capacitor> model makes the feedback quantitative. If $n$ moles of monovalent positive <ions> move from initially neutral side 1 to side 2, the charge on side 1 is $-Fn$, hence $V=-Fn/C$. The free-energy stationarity condition is
$$
RT\ln\frac{c_1^0-n/\mathcal V}{c_2^0+n/\mathcal V}-\frac{F^2n}{C}=0.
$$
As $n$ increases, both the <concentration> imbalance and the <voltage> drive toward reversal diminish the original <diffusion> tendency. The equilibrium lies before the bulk <concentrations> become equal. This demonstrates why a nonzero <Nernst potential> can coexist with very little change in bulk composition and approximate bulk <electroneutrality>. A <concentration> difference alone does not maintain an electrical <voltage> if both <counterions> can freely follow each other; \b[<selective permeability>] is essential to this model.
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