# Voltage drop over a cell membrane

Again, a problem from exam preparation:

[A] cell's membrane allows sodium ions to pass through it, but not chlorine ions. The cell is placed in a salty solution with a ten times higher concentration of salt outside than in the cell. As a result, sodium diffuses into the cell. Calculate the voltage difference across the cell membrane as a result.

Well, in equilibrium, the chemical potentials of the solution and the cell should be the same. Let $\mu_A$ be that of the solution and $\mu_B$ that of the cell.

Assuming that the salt is in a dilute solution, I can write its chemical potential as $\mu_A = f(T,P) + kT \ln(n_A)$ where $f(T,P)$ is some function independent of the number of particles, and $n_A$ is the number-concentration of sodium ions in the solution.

I'd assume that for the cell, I have $\mu_B = f(T,P) + kT \ln(n_B) + U$ where now I include the voltage that exists across the cell's membrane.

Problem is, now I have two unknowns: I don't know the final concentration of ions in the cell and I don't know the final voltage of ions in the cell.

Alternative idea but not sure if that makes sense: Osmotic pressure is defined as the pressure you must bring up to prevent osmosis from occurring, and it is also the pressure that finally builds up when you allow osmosis from happening. Could I analogously obtain the final voltage as the voltage that I'd have to apply manually to prevent any diffusion from happening at all?

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When speaking of the ion, generally one uses "chloride", not "chlorine". – Chelonian Feb 17 '14 at 17:54

Okay, if I assume unchanged concentrations, I get $U = kT/e \ln \left( c_A/c_B\right)$, which in the case of $c_A = 10 c_B$ amounts to approximately $60 mV$. Sounds reasonable :) – Lagerbaer Aug 11 '11 at 17:38