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Does passivation of lead anodes cause them to become semiconductors?

Or even become diodes (eg. a p-n passage)?

How does this work on an atomar level?

This question has been risen by trying to understand an old AC-DC converter alos known as Quecksilberdampfgleichrichter or Mercury-arc valve. Shortly: It turns rotating or alternating current into DC (for street cars traditionally) by that it ignites an arc above a mercury lake (the catode) in an containment. Above that lake are the anodes that hand the arc around as the current rotates.

This requires some place in the anodes to be current conducing in only one direction (diode like behaviour). Otherwise anodes would exchange current between themselves instead of between catode and anodes. It is being said that "Passivation" on those lead anodes in the mercury environment is the reason for that. Therefore: How does passivation (on an atomar or subatomar level) cause the electrons to only move one way, not both?

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  • $\begingroup$ Have you looked at the Wikipedia article? $\endgroup$
    – John Rennie
    Commented Jun 16, 2014 at 11:28
  • $\begingroup$ Yes and either I don't understand it or it is not contained. Neither in the german, nor in the anglais version. $\endgroup$
    – Robetto
    Commented Jun 20, 2014 at 7:58
  • $\begingroup$ The Wikipedia article does not mention passivation, but it appears not to be needed for the operation of the rectifier. The mercury cathode emits electrons easily, the carbon anode does not. Those electrons accelerate towards the anode, ionizing mercury vapor atoms. This creates more electrons, as well as positive ions. The ions accelerate back towards the mercury pool. When the voltage flips sign, the carbon does not emit electrons, so no current will flow. So the rectification is an asymmetry in electron emission from the two different materials. All covered in the article as mentioned. $\endgroup$
    – Jon Custer
    Commented Jun 23, 2014 at 15:36
  • $\begingroup$ You could easily put this as an actual answer. Sounds like the reason, to me. $\endgroup$
    – Robetto
    Commented Jul 2, 2014 at 13:48

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