Given the balanced ionic equation representing the reaction in an operating voltaic cell: zn (s) + cu2 + (aq) → zn2 + (aq) + cu (s) the flow of electrons through the external circuit in this cell is from the

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.

It might be through the (AQ) equation, I'm not sure, but it's what I came up with.