= Solution
Take $V=V_{\rm inside}-V_{\rm outside}$. Equality of <electrochemical potentials> for a monovalent cation gives the <Nernst equilibrium potential>
$$
E=\frac{RT}{F}\ln\frac{c_{\rm outside}}{c_{\rm inside}}.
$$
At $T=293.15\,\mathrm K$, $RT/F=25.26\,\mathrm{mV}$. Therefore
$$
\boxed{E_{\rm Na}=25.26\ln(120/15)=+52.5\,\mathrm{mV},\qquad
E_{\rm K}=25.26\ln(5/100)=-75.7\,\mathrm{mV}.}
$$
The signs refer to the stated electrode convention, not to the direction of current at every voltage. Activities are approximated by <concentrations> and the bulk ionic <concentrations> are treated as fixed during a single spike.
Back to article page
