If I had a negatively charged body , say an electret , and i put it in a container of NaCl solution . Will the positive ions of sodium be attracted to it ? and why? If no, why do the positive ions attract to the cathode during electrolysis?
$\begingroup$
$\endgroup$
1
asked Oct 1, 2012 at 20:08
-
$\begingroup$ Hello Abdelrahman, Good Question. But short answer... Take a basic view upon Electrolysis and other related links thereby..! $\endgroup$– Waffle's Crazy PeanutOct 2, 2012 at 14:37
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
Yes, and this is a very important effect in synthetic polyelectrolytes and in biological macromolecules.
The usual model is to treat the added charge as an imobile object with a given charge. The dissolved ions are then treated as mobile charges. The mobile ions with the opposite charge will be attracted to the fixed charge, and repelled from each other. The mobile ions with the same charge as the fixed charge will be repelled from it, and each other. The end result of that is a continuous charge distribution throughout the liquid. The electrostatic potential is the sum of the potential from the fixed ion with the potential due to the sea of mobile ions. You can treat the mobile ions statistical mechanically: they follow a Boltzmann distribution with energies equal to their charge times the electrostatic potential at their location in the liquid. Put those two terms together to get the Poisson-Boltzmann equation, a partial differential equation for the electrostatic potential in the liquid. In general that is difficult to solve, and it's usually solved numerically. For some special problems, it is possible to solve the PB equation analytically. The usual approximation is that the electrostatic energy is much less than the thermal energy $kT$, in which case you can linearize the PB equation to get the Debye-Huckel equation.
The Debye-Huckel equation is actually a bit instructive qualitatively. In the absence of anything else, the electrostatic potential falls off as $V(r) \propto 1/r$. In the presence of the mobile ions, the electrostatic falls off as $V(r) \propto e^{-\kappa r}/r$. $\kappa^{-1}$ is caled the Debye screening length, and is proportional to (among other things) the square root of the ionic strength of the solution.
answered Oct 2, 2012 at 14:30