When dilute sulphuric acid (H\(_2\)SO\(_4\)) undergoes electrolysis, the solution includes these ions:
\[\n \text{H\(_2\)O} \rightleftharpoons \text{H}^+ + \text{OH}^-, \quad \text{H\(_2\)SO\(_4\)} \rightarrow 2\text{H}^+ + \text{SO}_4^{2-}\n \]\
At the cathode (reduction):
- H\(^+\) ions are reduced, forming H\(_2\) gas:
\[\n 2\text{H}^+ + 2e^- \rightarrow \text{H}_2(g)\n \]\
At the anode (oxidation):
- OH\(^-\) ions (from water) are oxidized instead of SO\(_4^{2-}\):
\[\n 4\text{OH}^- \rightarrow 2\text{H}_2\text{O} + \text{O}_2(g) + 4e^-\n \]\
Therefore, oxygen gas (O\(_2\)) is produced at the anode.