I would like to know what causes the e-field in a Daniell cell.
From a chemistry point of view, I do understand that this is initiated by a redox reaction in which Gibbs free energy is decreased. The elemental zinc anode (in aqueous zinc sulfate) becomes oxidized by losing two electrons. The copper(II) ions (from the aqueous copper sulfate) become reduced by gaining two electrons, thus causing the elemental copper cathode to accumulate in size. The circuit is completed by the addition of a salt bridge that helps balance out the charge of the electrolyte in each of the half cells as the reaction occurs.
However, I have difficulty understanding the process from an electronic point of view --specifically concerning the e-field. I realize that the valence electrons of the elemental zinc (in the 4s orbital) want to be at a lower energy level, and they find that in the 3d10 and 4s1 orbitals of copper. This difference in electric potential gives the cell its voltage.
But how does this potential difference cause an e-field to arise? I suppose it could begin with the spontaneous reaction of elemental zinc to Zn2+ (aq) and two electrons. This would explain the negative pole of the field. But what is drawing the electrons over to the other cell? Is it the positive charge of the Cu2+ in the copper(II) sulfate electrolyte which surrounds the elemental copper cathode? I don't suspect the flow could just be from electron diffusion.
There was a similar question to mine asked a few weeks ago, but the discussion there was a bit fruitless.