Step 1: Note the electrode and the species present.
Platinum electrodes are inert, so they do not dissolve or take part in the reaction. In aqueous \(CuCl_2\) the species are \(Cu^{2+}\), \(Cl^-\), and water (which can give \(H^+\) and \(OH^-\)).
Step 2: Apply the rule for the cathode.
Reduction happens at the cathode and the species easiest to reduce is discharged. \(Cu^{2+}\) is reduced more easily than \(H^+\) from water, so copper metal deposits at the cathode: \(Cu^{2+} + 2e^- \rightarrow Cu\). So no \(H_2\) is released.
Step 3: Apply the rule for the anode.
Oxidation happens at the anode. The competition is between \(Cl^-\) and water (giving \(O_2\)). Although water has a lower standard discharge potential, the overvoltage for \(O_2\) on platinum is high, so chloride is preferentially oxidised.
Step 4: Write the anode reaction.
\(2Cl^- \rightarrow Cl_2 + 2e^-\), so chlorine gas is released at the anode, not oxygen.
Step 5: Conclude.
Copper deposits at the cathode and chlorine gas is released at the anode. The correct statement is that \(Cl_2\) gas is released at the anode.
\[ \boxed{Cl_2\ \text{gas will be released at anode}} \]