In a galvanic cell, why do electrons move through the wire from zinc to copper?

Question

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.

Answer 1

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.