= Nernst equilibrium potential
{c}
{title2=$E_{\mathrm{ion}}=\frac{RT}{zF}\ln\!\left(\frac{a_{\mathrm{outside}}}{a_{\mathrm{inside}}}\right)$}
= Nernst potential
{c}
{synonym}
The <Nernst equilibrium potential> is the inside-minus-outside <membrane potential> at which one <ion> species has equal <electrochemical potential> on both sides of a <cell membrane>. At uniform <temperature> $T$, using the same reference <chemical potential> on both sides, $z\neq0$ is the signed <ion> charge number, $R$ the <molar gas constant>, $F$ the <Faraday constant>, and $a$ the dimensionless <thermodynamic activity> of that <ion>. For an ideal dilute <chemical solution>, the activity ratio becomes the <concentration> ratio. See https://www.ncbi.nlm.nih.gov/books/NBK11102/[Purves et al.: the forces that create membrane potentials and the Nernst equation].
The short derivation is to set the outside-to-inside <electrochemical potential> change from <membrane transport> to zero: $RT\ln(a_{\mathrm{inside}}/a_{\mathrm{outside}})+zFV=0$, and solve for $V$. An open pathway carrying only this species by <passive transport> then has this <reversal potential>. A balance of several nonzero <ion> <electric currents> does not establish this equilibrium condition for each species.
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