Solutions A and B are separated by a membrane that is permeable to $\mathrm{Ca}^{2+}$ and impermeable to $\mathrm{Cl}^{-}$. Solution A contains $10 \mathrm{mM} \mathrm{CaCl}_{2}$, and solution $\mathrm{B}$ contains $1 \mathrm{mM}$ $\mathrm{CaCl}_{2}$. Assuming that $2.3 \mathrm{RT} / \mathrm{F}=60 \mathrm{mV}, \mathrm{Ca}^{2+}$ will be at electrochemical equilibrium when

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[IV B]. The membrane is permeable to $\mathrm{Ca}^{2+}$ but impermeable to $\mathrm{Cl}^{-}$. Although there is a concentration gradient across the membrane for both ions, only $\mathrm{Ca}^{2+}$ can diffuse down this gradient. $\mathrm{Ca}^{2+}$ will diffuse from solution $\mathrm{A}$ to solution $\mathrm{B}$, leaving negative charge behind in solution $A$. The magnitude of this voltage can be calculated for electrochemical equilibrium with the Nernst equation as follows: $\mathrm{E}_{\mathrm{Ca}^{2}}$ $=2.3 \mathrm{RT} / \mathrm{zF} \log \mathrm{C}_{\mathrm{A}} / \mathrm{C}_{\mathrm{B}}=60 \mathrm{mV} /+2 \log 10 \mathrm{mM} / 1 \mathrm{mM}=30 \mathrm{mV} \log 10=30 \mathrm{mV}$. The sign is determined with an intuitive approach $-\mathrm{Ca}^{2+}$ diffuses from solution $\mathrm{A}$ to solution $\mathrm{B}$, so solution A develops a negative voltage $(-30 \mathrm{mV})$. Net diffusion of $\mathrm{Ca}^{2+}$ will cease when this voltage is achieved, that is, when the chemical driving force is exactly balanced by the electrical driving force (not when the $\mathrm{Ca}^{2+}$ concentrations of the solutions become equal).

Solutions a and B Are Separated by a Membrane That Ca2+\mathrm{Ca}^{2+}Ca2+

Solutions a and B Are Separated by a Membrane That $\mathrm{Ca}^{2+}$