Why electrons move to Cu in galvanic cells?

I'm student who learning about chemistry, and I got problem about understanding why electorns move to Cu in galvanic cells. Let's think the reason for electrons going to Cu is that the standard reduction potential of Cu is high. It seems strange that Cu does not have a Cu2 + ion to be reduced, but Cu takes electrons. How about lead? it is a bit strange too that the electrons follow the lead always.

• Electrons follow neither copper neither lead. They follow voltage gradient. Galvanic connection of electrodes disbalance their equilibrium potential, causing reduction of $\ce{Cu^2+}$ on copper electrode. – Poutnik Jun 16 at 6:07
• Yeah, Poutnik is right. If you're wondering where from did the $\ce{Cu^2+}$ come, it is present in the electrolyte solution. – William R. Ebenezer Jun 16 at 11:27
• With great reluctance, I downvoted this. I will reverse my vote if the question is much improved (and assuming I am allowed to do so). The following (relatively deep) reference directly addresses why those electrons move as they do: K. Schmidt-Rohr, "How Batteries Store and Release Energy: Explaining Basic Electrochemistry", Journal of Chemical Education, 95 (2018) 1801-1810. I am currently reading all the galvanic cell questions and answers here and looking for a clean and factually correct answer to which I can link. If I find one, I will put the link in a comment here. – Ed V Jun 19 at 2:29

In the galvanic cell you are talking about, the copper is the inactive electrode. The activity, the electron production, comes from the more active metal. When a neutral metal atom falls into solution, it will leave is electrons behind:

$$\ce{M -> M^2+ + 2 e-}$$

But those electrons are just a few of the many that are left behind by all the $$\ce{M}$$ atoms falling into solution. They repel one another, pushing all the way to the copper electrode, where they may find an exit. If the solution contains an easily reducible metal ion, or a $$\ce{H+}$$ ion, the ion may pick up an electron (or two):

$$\ce{Z^2+ + 2 e- -> Z}$$

or

$$\ce{H+ + e- -> H}$$

followed by

$$\ce{H + H -> H2}$$