I am seriously confuse with the sign of electric field in the electrical double layer. Say zinc is dipped into the zinc sulphate solution. Then why the electrical double layer has positive sign in the electrode side of EDL and negative sign in the electrolyte side of EDL.

I am confused that since zinc has more tendency to dissolve, then zinc should lose two electrons and retain those 2 electrons in the electrode side of the metal-electrolyte interface. This will give eletrode negative sign. Also a layer positive charge of $\ce{Zn^2+}$ develops parallel to electrode. Where I am wrong here?


1 Answer 1


The phrase "retain those 2 electrons" is wrong. Metal surface has positive potential in solution, but this doesn't mean that lost electrons are concentrated somewhere. Electrons don't form layers. You can think of electrons as much more "volatile" things than ions, they may discharge everywhere depending on potential, for example, to reduce Zn ions in solution. So, surface is charged positively and hence first ionic layer is negative.

  • $\begingroup$ Great. I was confusing with the layers of electrons. You cleared major doubts. But what in case of Cu2+? How the electrode surface turns negative even if Cu2+ has to be deposited at the electrode surface as Cu. From where it takes those 2 electrons? $\endgroup$
    – Shivay
    Commented Oct 3, 2016 at 13:07
  • $\begingroup$ As electric charge associated with double layer is typically very small (picofarades), there is not much sense to bother about it in the first place. It may get spread over electrode/solution, or drawed/discharged into external circuitry like the hand of experimenter :), and so on. Btw, metal surface has non-zero potential in the air too. If you absolutely need the source of electrons for mental experiment in absolutely isolated environment, then it may be oxidation of water: H20 --> O2 + H(+) + e(-) $\endgroup$
    – sa7
    Commented Oct 3, 2016 at 14:07

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