I understand that electroplating involves using an anode (copper in this case) and an electrolytic solution containing Cu(II) ions with some sulfuric acid. Why exactly does the anode need to be copper if you want to plate something with copper; couldn't something like a graphite electrode be used instead, as the $\ce{Cu^2+}$ ions can be provided by the solution?
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$\begingroup$ Yes, the anode can be a different metal. It probably is shown as the same metal so as the metal from solution would "bind" with the anode material, and so that the reaction would be reversible. $\endgroup$– MaxWCommented Dec 20, 2015 at 6:19
1 Answer
You are right in saying that the electroplating could equally well be carried out with any electrode. Just offhand though, I could think of a couple of reasons why you would want the anode to be made of copper.
Choosing copper as the anode makes the overall cell potential equal to zero. The reaction at the anode is $\ce{Cu -> Cu^2+ + 2e-}$ and the reaction at the cathode is $\ce{Cu^2+ + 2e- -> Cu}$ - they are identical but one is flipped, so $E^\circ_{\mathrm{cell}} = 0~\mathrm{V}$. Compare this to if you used graphite - the reaction at the anode would then be $\ce{2H2O -> O2 + 4H+ + 4e-}$ and you would have $E^\circ_{\mathrm{cell}} = -0.892~\mathrm{V}$, meaning you would need more energy to run the reaction.
More importantly, it also ensures that the concentration of $\ce{Cu^2+}$ ions in solution remains constant. If you use a graphite electrode, the solution can initially provide the $\ce{Cu^2+}$ ions required for electroplating, but as the concentration of $\ce{Cu^2+}$ decreases, $E_{\ce{Cu^2+}/\ce{Cu}}$ becomes smaller (by the Nernst equation) and it becomes harder and harder to reduce $\ce{Cu^2+}$ at the cathode.