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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.

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  • $\begingroup$ 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, 2017 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, 2017 at 13:47
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    $\begingroup$ @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, 2017 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 Khan
    Jul 9, 2017 at 17:10
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    $\begingroup$ 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, 2017 at 17:19

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An electret is a dielectric (non-conducting) material that has a semi-permanent electric charge or polarization, the electrostatic equivalent of a permanent magnet. Electrets were made by first melting a polymer or wax containing polar molecules, and then allowing it to re-solidify in a powerful electrostatic field (several kilovolts/cm). The polar molecules in the dielectric align to the direction of the electrostatic field, producing a dipole with a permanent electrostatic bias when the melt solidifies.

Although electrets are only in a metastable state, those fashioned from very low leakage materials can retain excess charge or polarization for many years. The Japanese used microphones polarized by electrets in WWII.

https://en.wikipedia.org/wiki/Electret

So, using the electret as an example, the likely charge imbalance in the solidified lead bromide melt would be very small because the electric field across the molten salt is nowhere near kilovolts per centimeter. It might be measurable, but very small, and probably not very long-lived.

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