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If I had some NaCl dissolved in water, it is conductive, so if I were to put two electrodes in the container (powered), what is causing the flow of electrons? Do the NaCl ionic bonds break and the ionic atoms collect at the oppositely charged electrode (how I'd assume electroplating works) with all of the atoms themselves moving to carry current, or are electrons loose in a "sea" like a metallic bond, with some of the atoms being pulled along with the flow?

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  • $\begingroup$ Good question. I'd want to know the answer too. I remember in an exam question my professor asked about the reaction of equivalent concentrations of a strong acid and a strong base; he asked what interesting property could we expect from the water even though it is neutral. The answer was solution conductivity (there are still counterions left in the system). However, we never went over the nature of conductivity and how exactly these electrolytes convey electrons. I know that for there to be current there must be mobile electrons or mobile ions, but what exactly do these ions/electrolytes do? $\endgroup$ – Dissenter Jun 7 '14 at 3:08
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$\ce{NaCl}$ once in water dissociates, producing $\ce{Na+}$ and $\ce{Cl-}$ ions. Note: this process is possible only because immediately after dissociation ions are solvated by water, i.e. water molecules bind to ions

http://en.wikipedia.org/wiki/Solvation#mediaviewer/File:Na%2BH2O.svg

This is releasing energy and increasing effective size of ion, decreasing ion charge density. Once relatively free ions are produced, they can move between electrodes, transferring charge between electrodes. On electrodes ions may discharge taking/giving up electrons, and this is what usually happens with $\ce{Cl-}$ ion. However, if this is not an option, as the expected product is no stable enough in water, water molecule may react instead, producing another ion and either hydrogen or oxygen. This is what happens with $\ce{Na+}$: $\ce{Na}$ is too active to be stable in contact with water, so instead of sodium, hydrogen and $\ce{{}^{-}OH}$ are produced instead, resulting in net reaction (assuming cold solution) $\ce{2NaCl + 2 H2O <=> 2 NaOH + Cl2 + H2}$ with chlorine and hydrogen released on different electrodes.

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