Question

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 1

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.

Answer 2

Electrons were moving from the anode to the cathode.

What is ionic equation?

The term ionic equation has to do with those equations that show the oxidation or reduction of ions.

We have to look at the reaction closely to see the oxidized and the reduced species. In this case, we can see that electrons were moving from the anode to the cathode .

Learn more about ionic equation:https://brainly.com/question/15467507

#SPJ5