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.
I think it’s 7.41 because you count up all the atoms and find out how many are x (the large grey ones) and you do 2/27 x 100 which gives you 7.41 :) (sorry if i counted wrong it’s kinda hard)
C, to make sure the design works as expected.
A prototype is first, typical model of the said product. Hope this helps!
