An ion separated unequally across a membrane experiences both a chemical driving force and an electrical driving force. For one ion species, there is a particular membrane voltage at which those two contributions exactly balance. This voltage is the ion's equilibrium potential.

Define the membrane potential as

$$V_m=\psi_{\rm in}-\psi_{\rm out}.$$

For an ion with charge number $z$, electrochemical equilibrium requires zero free-energy change for moving the ion across the membrane. Rearranging that condition gives the Nernst equation

$$\boxed{E_{\rm ion}=\frac{RT}{zF}\ln!\left(\frac{a_{\rm out}}{a_{\rm in}}\right)},$$

where $E_{\rm ion}$ is the membrane potential at equilibrium for that ion, $R$ is the gas constant, $T$ is absolute temperature, $F$ is the Faraday constant, and $a_{\rm in}$ and $a_{\rm out}$ are ion activities inside and outside. In dilute solutions, activities are often approximated by concentrations.

Example: potassium

Suppose at body temperature

$$[\mathrm{K^+}]{\rm in}=140\ \mathrm{mM},\qquad [\mathrm{K^+}]{\rm out}=5\ \mathrm{mM}.$$

For $z=+1$ and $RT/F\approx26.7\ \mathrm{mV}$,

$$E_K\approx26.7\ln!\left(\frac{5}{140}\right)\mathrm{mV}\approx-89\ \mathrm{mV}.$$

The concentration gradient favors K$^+$ leaving the cell, while a sufficiently negative interior attracts K$^+$ inward. Near $-89$ mV, those tendencies balance.

Equilibrium potential is ion-specific

Different ions generally have different concentration ratios and charges, so they have different equilibrium potentials. A membrane can therefore be at equilibrium for K$^+$ while Na$^+$ still experiences a strong electrochemical driving force.

Opening a channel does not force the membrane voltage to jump instantly to that ion's equilibrium potential. Instead, ion flow tends to move $V_m$ toward the equilibrium potential of the ions that the channel conducts.

The equilibrium potential is therefore a quantitative bridge between ion concentration gradients and electrical signaling.