Ok let's take a look at a standard reduction potential table
Cu+2 + 2e- --> Cu(s) E = 0.337 (stronger oxidizing agent)
Zn+2 + 2e- --> Zn(s) E= -0.763 (stronger reducing agent)
Cu is a stronger oxidizing agent which means that in the process of the chemical reaction it will become reduced by gaining electrons from an outside source (in this case the Zn plate.
Similarly, Zn is a stronger reducing agent and in the process of a chemical reaction it will become oxidized (loose electrons) to Cu.
Honestly I have a bit of difficulty in explaining away the answer to D.
It would seem to me from an intuitive standpoint that the Zn would also react with the Sulfuric Acid in solution, but since it is touching the Cu, plate directly then maybe this prevents that rxn.
Zn(s) + H2SO4 -->ZnSO4(aq) + H2(g) (these are the bubbles you would see if the
plates were not touching I guess)
Ah, this would not occur because Cu is the stronger oxidizing agent than H+ therefore, the electrons would go towards the Cu plate to reduce the copper.
Hope that helps.