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Name: Max
Status: student
Grade: 9-12
Location: NY
Date: March 2008

How can fully hydrated Na+ cations or fully hydrated Cl- anions conduct electricity in water? When a Na+ or Cl- ions are dislodged from the crystal, those ions become immediately surrounded by water molecules in the form of a primary hydration shell. The (+) Na ions are surrounded by water molecules having their (-) ends facing toward the ion. Likewise, the (-) Cl ions become hydrated with the (+) end of the water molecules facing the (-) ion. The ions are each solvated by 6-8 water molecules. I assume this makes the solivation compounds (Na+&H2Os, Cl-&H2Os) electro-statically neutral to their surroundings and that is how they float around in solution in water. On the other hands, many texts state that Na+ and Cl- float around "freely" in water, and are attracted to cathode and anode. They do not mention the hydration effect when discussing electrical conductivity of salt water. Questions: a.) How can a fully hydrated Na+ or Cl- conduct electricity? b.) Do the Na+ or Cl- ions get rid of the H2O molecules around them? c.) If so, where in the process and how does it happen?

Your picture of the hydration shell of the ions is basically ok, but you should keep in mind that it is not something rigid, but that it is fluid and changing. Additionally, it does not simply terminate at one contact shell, but the effect of the charge is felt at greater distances from the charge, though of course at lesser intensity.

You are mistaken to think that the ion's hydration shell "makes the solvation compounds (Na+&H2Os, Cl-&H2Os) electro-statically neutral." It does not. Think about the structure you described: the negative Cl- ion is surrounded by water molecules with their positive ends pointing inward. This means that their negative ends are pointing outward. Any object outside the hydration shell will encounter a ball of negative charges. The charge is still there; it is not in any way neutralized. It is simply effectively "spread out" over a larger surface area. Again, the hydration shell does not neutralize the charge of the ion. It effectively increases the ion's size, surface area, and mass, but it does not affect its charge.

So, electrical current can be carried through a solution of hydrated ions by movement of the hydrated ions. No problem.

Richard Barrans, Ph.D., M.Ed.
Department of Physics and Astronomy
University of Wyoming

Hi Max-

The hydrated ions are not neutral. The cluster as a whole has the same charge as the ion alone. It's like: (-1) + 8*(0) = (-1), a simple fact. The difference is in the imaginary sphere which encloses this charged object. The sphere enclosing all the un-balanced charge of a hydration cluster is required to be larger than the sphere enclosing a naked ion. That makes the charge-density lower, and thus the electrostatic self-energy is lower, which is what motivates the hydration cluster to form around a naked ion.

Imagine n~10 point-charges, each having 1/n of an electron charge. These start infinitely far apart and then are pushed towards each other, doing work against their mutual electrostatic repulsion. At the point where they are just within one's imaginary enclosing sphere the further progress is allowed to stop. The energy used to get to this point is larger if the sphere is smaller. If you've taken calculus, you might be able to do an integral of a uniformly-charged hollow spherical shell being compressed from infinity to a given size. If not, you can get a representative idea by using as few as two charges and basic language or a spreadsheet on a computer. You may have seen formulas for such potential energy in physics class. Just as the repulsive force goes as inverse-square of distance, the potential energy goes as the inverse of the sphere's radius or diameter.

In response to electric fields, two motions happen.

(1) Whole hydration-clusters migrate slowly, baffled as they are by constant molecular collisions around them. This means the action is viscous; the net speed is proportional to the electric force dragging them. Large species migrate slower than small ones, for a given force. In a sense, the published mobility numbers might tell you the size of the cluster typical for that ion.

(2) "Exchange" happens, clusters breaking up and reforming randomly as part of thermal motion. An Na+ with 6 H2O's might sluff off 2 H2O's in the "rearward" direction, immediately picking up 2 new ones in the "forwards" direction. If this happens often enough, and if the directions are biased by the E-field instead of random, the ion will seem to be going somewhat faster than its typical hydration cluster can migrate.

The extreme example of exchange is the H+ ion. It can be seen as:

a) the hopping of an electron-absence (a "hole") from H-atom of one water-molecule, thru the oxygen, to an opposing H-atom (remember it's usually H3O+, so there are 2 opposing H's) which then might quickly travel the very short distance to the neighboring oxygen it had previously been H-bonded to, easily repeating the process, or

b) transfer of a naked proton from H3O+ onto a neighboring H2O, which then needs only to swivel in place 180 degrees to repeat the process and make net progress.

The massive cluster around it need not migrate at all, though I am sure it does anyway. The hole-hopping and/or proton exchanges go considerably faster than the cluster drift.

So H+ ion is a few times more conductive in water than any other ion. Likewise OH- is an unusually fast ion too.

The measured value of ion-mobility can be seen as some mathematical combination of (1) and (2).

Jim Swenson

Wow! Some question. But there is a relatively simple answer. A water molecule has an overall charge of nil. No matter how many of them you pile up around a Na+ ion, the overall charge on the group will be +. Similarly, no matter how many water molecules surround a Cl- ions, the overall group will still have a - charge.

The ions would become something like .. (NaH12O6)+ and (ClH12O6)- On the other hand ... There are some who believe that in some solutions, the charge on the salt ions may result in some ionization of the water, so that you have Na+OH- and H+Cl- If these neutral pairs are formed, there would be a significant reduction in the conductivity of the solution, but this does not appear to be the case with most solutions. If there is ionization of the water, the bonding between the salt ions and the water ions must be weak. The conductivity of ionic solutions suggests very strongly that whatever the actual structure of the components, the overall charges are relatively free to migrate in the solution. Further, the accumulation of the expected element at the electrodes is strong evidence that we are on the right track - Copper is deposited from copper sulphate at the negative electrode.

Good question - thanks

Nigel Skelton


I like that you are thinking about the process deeply and not simply accepting what you are told.

While the solution may be neutral (there are an equal number of Na(+) and Cl(-) ions in the water) the charges of each individual solvated ion is not reduced to zero by the solvating water molecules. Remember that the water molecules are themselves neutral even though they may have a partially positive end and a partially negative end. The surrounding solvating water molecules may dampen or partially mask the charge of the ion -making it more stable- but they do not completely neutralize the charge. Therefore, we can still expect that the ions will travel through the salt bridge either because of the attraction to the electrodes, or as an effect of the charge build up due to the separated oxidation-reduction reaction and the transport of electrons through the wire.

Greg (Roberto Gregorius)

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