Are the plates of a battery really charged? In a zinc/copper Daniell cell correct me if I am wrong :

*

*Zinc has 2 valence electrons. So it wants to get rid of them. To do so it sends them to the copper which needs 2 to complete its valence shell.


*There needs to be a wire between the zinc and the copper for this reaction to happen.


*So technically the plates are not charged. It's just the charges flowing out that create the electric field.
TLDR : Are the plates of a battery more like a capacitor with excess charges on the plates? Or do they simply throw in and out electrons near their terminals and the individual plates of zinc and copper are neutral?
My confusion is this : I understand that the zinc wants to get rid of electrons, and the copper wants more electrons, but : The zinc and copper atom are "neutral". it's only the defecit of electrons on the conductor near the positive terminal and the excess of electrons near the negative terminal, That for me would make an electric field .
Or Maybe it's the "wanting to get rid" and the "wanting to get more" electrons that create an electric field, if it's indeed that please confirm !
Thanks !
 A: I think that most confusion about batteries comes from ignoring the electrolyte. For example:

Zinc has 2 valence electrons. So it wants to get rid of them. To do so it sends them to the copper which needs 2 to complete its valence shell.

The actual reaction is between the metallic zinc and the dissolved zinc. Zinc wants to get rid of two electrons and it does so by becoming an ion and going into solution. 
This reaction is energetically favorable and can occur even if there is a small electric field opposing it at the surface of the electrode. However, the reaction products near the electrode surface, zinc ions and electrons, are highly charged and quickly produce a strongly opposing field which overcomes the energetic favorability and halts the reaction. For the reaction to proceed the reaction products must be removed from the region near the electrode surface. 
The electrons can be removed from the electrode surface by transport through the wire, and the ions can be removed from the surface by transport through the fluid. The transport of electrons requires the complementary reaction at the copper electrode, and the transport of the ions requires a complementary transport of the solute ion in the electrolyte. Understanding the electrolyte is essential for understanding batteries, and is the usual neglected piece. 

There needs to be a wire between the zinc and the copper for this reaction to happen.

The purpose of the wire is not to make the reaction happen. The reaction is energetically favorable, so it briefly happens regardless. The purpose of the wire is to remove the reaction products so it can continue to happen. 
Note that in this process not all of the excess electrons on the plate are normally removed. As soon as a few are removed the reaction proceeds and replenishes those few. The reaction thus proceeds at the rate that the products are removed from the immediate vicinity of the electrode surface. In abnormal situations, like a short circuit, a substantial fraction of the excess charges at the surface can be depleted and the current is limited by the reaction kinetics. This manifests as an “internal resistance” for the cell. 
A: Metals of electrodes in the open circuit undergo these  electrochemical reactions:
$$\require{mhchem} \ce{Zn(s) <=> Zn^2+(aq) + 2 e-}$$
$$\ce{Cu(s) <=> Cu^2+(aq) + 2 e-}$$
The reaction comes to chemical equilibrium when the electrode potential of metal M reaches the particular equilibrium value, given by the Nernst equation.
$$E_M=E^\circ_M + \frac{RT}{nF}\ln a_M^{n+}$$
Zinc ions have higher tendency to dissolve in solution, compared to copper, leading to more positive potential on the copper electrode.
When the circuit is closed, the potential difference causes the current flowing.
As consequence, both electrode potentials are disbalanced from their equilibrium values.
Zinc electrode potential gets higher, what leads to dissolving of zinc and pumping electrons to the circuit.
Copper electrode potential gets lower, what leads to copper ion reduction to metsl and draining electrons from the circuit.
Within the cell, a ion flow forms between electrodes, via a diaphragm or ionic bridge, following potential and concentration gradients.
About capacitors, there are at least 3 type of capacitance:


*

*bulk metal of electrodes

*dielectric layer of ions adsorbed at electrodes

*solution itself due bulk volume charge displacement.
A: Both Cu and Zn are dissolving from the respective metal pieces and return back at some rate. In order to be dissolved in water they need to be ions (because water is a polar solvent). So who has a greater tendency to dissolve? The element that's easier to ionize. In this case it's Zn because it has lower electronegativity and can give away its electrons (to Zn electrode) easier.
So a bit more of Zn will be dissolved than Cu and thus Zn electrode will have some build up of electrons. That will create an electric field in the conductor which will increase the density of electrons nearby which in turn will create an electric field down the wire.
Though this would halt very quickly because electrons will quickly build up on the other end. For this reason you also need a salt bridge. AFAIU, it will either contain additional ions itself that would dissolve in water or simply allow Zn- to flow to the Cu cell and vice versa.
PS: I'm not sure what role SO4 plays in this though, but it seems like it has even more affinity to Zn than just water which makes the dissolution better.
A: 
TLDR : Are the plates of a battery more like a capacitor with excess
  charges on the plates?

Yes. This is why a (charged) cell has a non-zero open-circuit (no external circuit) voltage across. Despite the fact that there is no external circuit through which charge can flow, the reactions of the plates with the electrolyte result in one plate having a deficit of electrons (positively charged) and the other plate having an excess of electrons (negatively charged).
Essentially, it is this separation of charge (and the associated electric field) that 'halts' the (net) chemical reactions at the plates and produces the constant open-circuit voltage. 
If electrons are allowed to flow through an external circuit (load) from the negative plate to the positive plate, the reactions proceed and the cell discharges.
In the case of a rechargeable cell, an external source can move electrons from the positive plate to the negative plate essentially reversing the chemical reactions at the plates and the cell charges.

(From the comments)

OK but when we connect both terminals with a circuit wouldn't all the
  excess charge simply flow in the wire and terminals become neutral
  again ?

Unlike a capacitor, the energy stored in a cell is in the form of chemical energy and not electrostatic energy. Only a tiny fraction of the stored chemical energy is needed to establish the open-circuit voltage.
Like a capacitor, the plates have equal and opposite electric charge but unlike a capacitor, chemical reactions 'try to' maintain the charge separation on the plates even as excess electrons flow from the negative plate, through the load, an onto the positive plate.
Of course, there is a limit and cells have a so-called short-circuit current that is the maximum current that can be supplied when the plates are connected together with a wire (I should point out that this is generally not a safe test to perform).
There is of course, lots more to all of this, and I do believe there's some good Q & A here on this subject.
