Electrolysis of Copper (II) Sulfate Solution with graphite electrodes

In the electrolysis of Copper (II) Sulfate solution with inert graphite electrodes, the electrons from the power pack will first go to cathode. I understand what happens at the cathode - the copper ions gets reduced and solidifies.


However, I'm unsure exactly what happens at the anode, specifically with it being made of graphite - as in the redox half equation. I know oxygen forms, but is it a hydroxide ion that gets oxidised, or is it water? Why?


The hydrogen cations then join with the sulfate to make sulfuric acid. From my understanding the sulfate cannot be oxidised because it cannot exist as a neutral atom.


Would anything change if it were a copper electrodes?


Thank you so much.




In the electrolysis of $\ce{CuSO4}$, the following half-equation occurs at the graphite anode : $$\ce{2 H2O -> O2 + 4 H+ + 4 e-}$$ This production of electrons is due to $\ce{H2O}$ molecules and not to $\ce{OH-}$ ions, because $\ce{H2O}$ is more than $\pu{10^8}$ more frequent in solution than $\ce{OH-}$.
If the anode is made of copper, the production of electrons will be due to another half-equation, namely $$\ce{Cu -> Cu^{2+} + 2 e-}$$


 0