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I understand that there is aqueous $\ce{Cu^2+}$ in this scenario which pulls on the electrons from the zinc causing the electrons to move through the wire, but how does the electronegativity of $\ce{Cu^2+}$ pull on the electrons from the $\ce{Zn}$ when they aren't "connected".

What I mean by that is the $\ce{Zn}$ metal and the wire are only connected to the $\ce{Cu}$ metal and not the aqueous $\ce{Cu^2+}$ solution. This would make more sense to me if the $\ce{Zn}$ metal was connected to $\ce{Cu^2+}$ solution via a wire rather than being connected to the $\ce{Cu}$ metal.

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    $\begingroup$ The wire is made of a conducting material, usually copper! So the wire and the Cu electrode are all part of the same conducting system. $\endgroup$ – Withnail May 17 '19 at 1:16
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    $\begingroup$ $\ce{Cu^2+}$ ions in the bulk solution don't react. Only if they are in contact with the electrode will they be able to gain electrons. So in that sense the Zn metal is connected with the $\ce{Cu^2+}$ ions via conducting metal. $\endgroup$ – Karsten Theis May 17 '19 at 1:33
  • $\begingroup$ For an extremely thorough discussion of the findamentals, see this paper: K. Schmidt-Rohr, "How Batteries Store and Release Energy: Explaining Basic Electrochemistry", J. Chem. Ed., 95 (10) (2018) 1801-1810. The Zn and Cu Daniell cell is addressed at great length (way too long for an answer here) and the metal cohesive energy is the source of more than 75% of the energy in the Daniell cell (p. 1805.). $\endgroup$ – Ed V Jun 17 '20 at 13:31
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If we consider the classical Daniel cell $$\ce{Zn|Zn^2+||Cu^2+|Cu|}\text{(wiring to Zn)}$$

then due higher tendency of zinc ions to leave the metal, getting hydrated in solution, the zinc electrode obtains more negative electrostatic potential due extra electrons, compared to cupper electrode.

As electrodes are galvanically connected, electrons flow from more negative potential of the $\ce{Zn}$ electrode toward more positive $\ce{Cu}$ electrode, increasing potential of the former and decreasing the potential of the latter.

The $\ce{Cu}$ electrode, with the potential lower then its equilibrium potential, reduces $\ce{Cu^2+}$ to $\ce{Cu}$.

The $\ce{Zn}$ electrode gets the higher than equilibrium potential, as electrodes leave it by the wire, so more $\ce{Zn^2+}$ ions are transferred to $\ce{Zn^2+}$ salt solution, to approach the equilibrium potential.

Ions in both solutions, connected by diaphragm or salt bridge, migrate along the potential and concentration gradients, caused by unequal potentials and electrode reactions.

The whole cycle is closed.

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I've also searched multiple places for the answer and came across several similar questions. What causes electrons to move from zinc to copper? Why do Electrons leave the Zinc in a Galvanic Cell

However, none of the answers inside seems to be on point or understand the intention of the people who asked the question.

I've thought about the problem and have an explanation myself. Why does the electron leave zinc and flow into copper? It doesn't make much sense since both sides are neutral and where does the attraction force come from then?

The answer I believe lies in the way that a piece of solid zinc reacts with copper sulfate solution. Copper sulfate is also neutral then why the reaction happens when you throw in a piece of zinc? I think it's because although copper sulfate solution is overall neutral, they are not locally. The copper ion close to zinc will show as a positive charge locally and attract the electron.

When it comes to the battery, the copper ion in the solution will also attract the electrons in the copper sheet which makes the sheet have a polarity even though neutral overall. When a wire is connected from zinc to copper, this polarity will then polarize the wire and propagate all the way to the zinc end. So zinc now feels the attraction from the wire on their end the electron will flow.

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