To solve this problem, we need to understand the electrolysis of dilute sulphuric acid ($$H$_2$SO$_4$) using Platinum (Pt) electrodes.$\\$During electrolysis, the following reactions occur at the electrodes:$\\$• At the cathode (reduction occurs):$\\ 2$H$^+ + 2e^- \rightarrow$H$_2$(gas)$\\$• At the anode (oxidation occurs):$\\$In dilute sulphuric acid, water is oxidized instead of sulfate ions.$\\ 2$H$_2$O$\rightarrow$O$_2$(gas)$+ 4$H$^+ + 4e^- \\$Explanation:$\\$• The anode reaction involves the oxidation of water to produce oxygen gas.$\\$• Sulfate ions (SO$_4^{2-}$) are not discharged at the anode because the oxidation of water is more favorable.$\\$Therefore, the product obtained at the anode is oxygen gas.$\\$This corresponds to Option 1: Oxygen gas.