Step-by-step explanation:
We have a galvanic cell with an aqueous solution of Zn²⁺ and a Zn electrode on one side, and on the other side we have a Cu²⁺ aqueous solution and a Cu electrode.
The reduction potentials of these half-cells are:
Cu²⁺ + 2 e- ----> Cu Potential = 0.34 V
Zn²⁺ + 2 e- ----> Zn Potential = -0.76 V
This means that the Cu²⁺ has a greater reduction potential and it will be our oxidizing agent. And Zn²⁺ will be our reducing agent.
Zn ----> Zn²⁺ + 2 e- is oxidized
Cu²⁺ + 2 e- ----> Cu is reduced
Since the oxidation state of Zn goes from 0 to 2+ it is being oxidized. The oxidation state of Cu goes from 2+ to 0, it is being reduced. And electrons are flowing from Zn to Cu.
Answer: Electrons flow from Zn to Cu.