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Answer: from the Zn anode to the Cu cathode
Justification:
1) The reaction given is: Zn(s) + Cu₂⁺ (aq) -> Zn²⁺ (aq) +Cu(s)
2) From that, you can see the Zn(s) is losing electrons, since it is being oxidized (from 0 to 2⁺), while Cu²⁺, is gaining electrons, since it is being reduced (from 2⁺ to 0).
3) Then, you can already tell that electrons go from Zn to Cu.
4) The plate where oxidation occurs is called anode, and the plate where reduction occus is called cathode.
So you get that the electrons flow from the anode (Zn) to the cathode (Cu).
Always oxidation occurs at the anode, and reduction occurs at the cathode.
If its sodium it would have 11 electrons
