# Could Nernst Equation fail at high Ion concentration diffrence

When I was learning about neuroscience, I came across the Nernst equation and found out that it is only dependent on the quotient of the ion concentration outside compared to the inside. So, now I thought, how is that possible? But, then found out that electrostatic force is much stronger so the effect on the concentration is negligible. But, now my question is: What about if we had a huge difference for example if $$\mathrm{Ion}_\mathrm{out} = \pu{150 mM}$$ and $$\mathrm{Ion}_\mathrm{in} = \pu{0.000015 mM}?$$ So, $$E=\frac{RT}{ZF}\ln(Q),$$ where $$Q$$ is $$Q = \frac{150}{0.000015}.$$ Would the Nernst equation still work, or would diffusion affect it too much. Or is diffusion already counted in, than how and why?

• Be aware that N.E. describes equilibrium. There is no equilibrium when there is (net) diffusion. Commented Sep 4, 2021 at 10:06

Nernst equation works correctly at high dilution. At concentrations higher than about $$0.1$$ M, the concentration must be replaced by the activity. The activity may be considered as a sort of concentration calculated by dividing the number of moles of solute by the volume of the "free" water (water not electrically attached around the ions). This volume may be much smaller than the volume of the solution. And it is also necessary to take into account the probability of recombination of the separated ions in solution.