On electrolysis of dil. sulphuric acid using platinum ($\mathrm{Pt}$) electrodes, the product obtained at the anode will be:

Solution & Explanation

\textbf{Answer:} (D) Oxygen gas \textbf{Solution:} Dilute $\mathrm{H}{}_2\mathrm{SO}{}_4$ contains $\mathrm{H}{}^{+}$, $\mathrm{OH}{}^{-}$ (from water) and $\mathrm{SO}{}_4{}^{2-}$ ions. At the anode, $\mathrm{OH}{}^{-}$ ions are discharged in preference to $\mathrm{SO}{}_4{}^{2-}$ (the sulphate ion is not oxidised), liberating oxygen: $\ce{2H2O -> O2 + 4H+ + 4e-}$. At the cathode, $\mathrm{H}{}^{+}$ is reduced to hydrogen: $2\mathrm{H}{}^{+}+2\mathrm{e}{}^{-}\to \mathrm{H}{}_2$. Hence the product at the anode is oxygen gas.