Question

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?

Answer 1

In the electrolysis of Copper (II) Sulfate solution with graphite electrodes, carbon dioxide gas is produced at the anode and copper ions are reduced at the cathode. If copper electrodes were used, the reactions would remain the same.

Step-by-step explanation:

In the electrolysis of Copper (II) Sulfate solution with inert graphite electrodes, the electrons from the power pack will first go to the cathode. At the cathode, copper ions get reduced and solidify. At the anode, the graphite electrode reacts with oxide ions, producing carbon dioxide gas (CO₂).

In the electrolysis of water, which is a different process, the anode reacts with hydroxide ions, producing oxygen gas (O₂).

If copper electrodes were used instead of graphite electrodes, the reactions at the anode and cathode would remain the same. The copper electrode would still produce copper ions (Cu²+) at the anode, and copper metal would still form at the cathode.