= Solution
Use signed <ion> valence $z$ and define $V=\psi_1-\psi_2$. The molar <electrochemical potential> in side $i$ of an ideal dilute solution is
$$
\widetilde\mu_i=\mu^\circ+RT\ln(c_i/c^\circ)+zF\psi_i.
$$
Transfer of an infinitesimal amount $dn$ from side 1 to side 2 changes the total <Gibbs free energy> by $(\widetilde\mu_2-\widetilde\mu_1)dn$. At equilibrium this derivative is zero, so
$$
RT\ln(c_1/c_2)+zF(\psi_1-\psi_2)=0,
\qquad\boxed{V=\frac{RT}{zF}\ln\frac{c_2}{c_1}.}
$$
For a cell with side 1 inside, this is $E_i=(RT/z_iF)\ln(c_{i,\rm out}/c_{i,\rm in})$. The <concentrations> are \b[equilibrium free-ion <concentrations>], not automatically their exact initial values. For nonideal solutions the ratio must be a ratio of <thermodynamic activities>. The <temperature> and solvent must be common to both reservoirs, and an <ion> must actually be able to exchange for this equilibrium condition to be dynamically established.
Equal compartment volumes are unnecessary for deriving the <voltage> law, but help specify number conservation if <concentration> changes are appreciable. For equal <volume> $\mathcal V$ and a monovalent cation transfer $n$ from side 1 to 2, $c_1=c_1^0-n/\mathcal V$, $c_2=c_2^0+n/\mathcal V$. Combined with the <membrane capacitance>, this gives an implicit finite-reservoir equilibrium rather than treating both <concentrations> as externally clamped.
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