During the electrolysis of an aqueous solution of CuSO4 with inert electrodes

(A) the anode loses mass and the cathode gains mass.
(B) the mass of the anode remains the same but the cathode gains mass.
(C) the mass of the anode decreases but the mass of the cathode remains constant.
(D) the anode and the cathode neither gain nor lose mass.
(E) both electrodes gain in mass.

I was thinking in Electrolysis of CuSO4(aq)
Anode: 2H2O O2(g) + 4H+1 + 4e-1 -1.23V
Cathode: 2 Cu+2+ 4e-1 2Cu + 0.34V
Overall: 2H2O + 2 Cu+2 O2(g) + 4H+1 + 2Cu E° = -1.02V

that it would be D because it doesnt really change in either

B,because in this type of reaction the anode does not ionises and remains the same but due to mthe ionisation of the CuSo4 soln. the cathode gains mass and increases in sizi.

In the electrolysis of CuSO4, isn't Cu plated out on the - electrode?

what do you mean by plated

Electrolysis is used for plating one metal onto itself or onto another. For example, one can buy silverware to eat and cook with (most people use stainless steel) that is essentially pure silver ( and pay a high price for it--real silver ware has the notation sterling silver on it) OR one can purchase the less expensive silver plate in which a thin layer of silver or gold is plated onto a base metal of some kind. Obviously that is much much cheaper. Baby shoes are "bronzed" that way (by electroplating). So by plating Cu, I mean Cu^+2(aq) + 2e ==> Cu(s) and the copper is plated onto the host electrode. In fact, copper metal is refined that way, from impure copper to relatively pure copper by electroplating it onto a very thin piece of copper foil. The impure copper goes into solution at one electrode and plates out at the other electrode. And the silver and gold impurities are left as a sludge in the vessel and is refined. Often the silver and gold recovered more than pays for the electrolysis of the copper.

Well, you're on the right track! But let me add some humor to it to lighten things up.

In this electrolysis party, let's talk about the guests of honor, the anode and the cathode! Now, during the electrolysis of an aqueous solution of CuSO4 with inert electrodes, the anode and the cathode definitely have some changes going on.

Now, picture the anode as a weight-losing enthusiast. It's all about shedding those pounds! So, during this process, the anode loses mass. Say goodbye to those extra copper ions!

On the other hand, the cathode is more like a weight-gaining fanatic. It's all about bulking up! So, you guessed it, the cathode gains mass during this electrolysis fiesta.

So, to answer your question, it would be option (A): the anode loses mass and the cathode gains mass. They're just playing their part in this electrifying event!

Remember, this answer is just for laughs. In reality, the anode does lose mass and the cathode gains mass during the electrolysis with inert electrodes.

You are correct in analyzing the reactions occurring during the electrolysis of CuSO4(aq).

At the anode, water molecules are oxidized to form oxygen gas and protons (H+). This reaction is represented as:
2H2O -> O2(g) + 4H+ + 4e-
The standard reduction potential for this reaction is -1.23V.

At the cathode, copper ions (Cu+2) are reduced to form solid copper metal. This reaction is represented as:
2Cu+2 + 4e- -> 2Cu
The standard reduction potential for this reaction is +0.34V.

The overall reaction is the combination of these two half-reactions:
2H2O + 2Cu+2 -> O2(g) + 4H+ + 2Cu

Based on the analysis of the reactions, we can determine the changes in mass for the anode and cathode.

Since oxidation occurs at the anode, the anode loses mass as water molecules are converted into oxygen gas. Therefore, option (A) "the anode loses mass and the cathode gains mass" is the correct answer.

On the other hand, reduction occurs at the cathode, where copper ions are reduced to form solid copper metal. This means that the cathode gains mass during the electrolysis.

Therefore, the correct answer to the question is (A) "the anode loses mass and the cathode gains mass."