Suppose I were to electrolyse some molten lead(II) bromide. According to what I have been taught, the lead ions travel towards the cathode and are reduced at it, while the bromide ions are oxidised at the anode. The movement of the ions creates a charge imbalance (there will be more lead ions near the cathode and vice versa) in the molten fluid.

Suppose I were to let this run for a little while in a vacuum (and siphoning off the $\ce{Br2}$ as soon as it is produced) and non-conducting container. Now suppose I was to quickly put an impermeable non-conducting plate between the cathode and the anode and at the same time remove both the cathode and anode, so separating the molten $\ce{PbBr2}$ into two halves. This will stop the electrolysis but there is already a charge imbalance between the two halves (e.g. the half which originally had the cathode has more $\ce{Pb^2+}$ than $\ce{Br-}$) which has no way to correct itself.

If I cooled down both halves of the mixture now so that they both solidified there would be excess ions on both side which wouldn't be able to recombine. It might even be possible to remove the re-formed $\ce{PbBr2}$ by some method and have just the uncombined ions left on each side. My question is whether this is even possible, and if so what would happen next.

  • $\begingroup$ There's no need to "thank in advance". If you do get an answer, upvote it instead of saying thanks. If it completely solved your problem, accept it. $\endgroup$ – Pritt says Reinstate Monica Jul 9 '17 at 13:44
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    $\begingroup$ There's not that much of a charge imbalance. The ions move in the melt to quickly restore imbalances. $\endgroup$ – Zhe Jul 9 '17 at 13:47
  • $\begingroup$ @PrittBalagopal There's no appreciable excess of ions when the barrier would be put in place. Electrolysis doesn't separate the charges. They are fluid in the melt and adjust to establish a charge balance. $\endgroup$ – Zhe Jul 9 '17 at 16:21
  • $\begingroup$ @Zhe I don't know whether I am missing something but in a similar experiment I saw with dissolved KMnO4 on a piece of paper all the purple colour moved towards the anode so an appreciable difference of ions was present in that case, I don't know whether it would be the same in this case or different. $\endgroup$ – Hadi K says thanks to Monica Jul 9 '17 at 17:10
  • $\begingroup$ I'm pretty sure this is a misconception. You just can't build up excess charge to any appreciable extent. Electrostatic forces are really strong, and you'll rapidly reach equilibrium... $\endgroup$ – Zhe Jul 9 '17 at 17:19

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