Predict the products of electrolysis in the following:
A dilute solution of H2SO4 with platinum electrodes
H2SO4 ionizes in aqueous solutions to give H+ and `SO_4^(2-)` ions.
`H_2SO_(4(aq)) -> 2H_((aq))^+ + SO_(4(aq))^(2-)`
On electrolysis, either of H+ ions or H2O molecules can get reduced at the cathode. But the reduction potential of H+ ions is higher than that of H2O molecules.
`2H_((aq))^+ + 2e^(-) -> H_(2(g)); E^@ = 0.0 V`
`2H_2O_((aq)) + 2e^(-) -> H_(2(g)) + 2OH_((aq))^(-) ; E^@ = -0.83 V`
Hence, at the cathode, H+ ions are reduced to liberate H2 gas.
On the other hand, at the anode, either of `SO_4^(2-)`ions or H2O molecules can get oxidized. But the oxidation of `SO_4^(2-)` involves breaking of more bonds than that of H2O molecules. Hence, `SO_4^(2-)` ions have a lower oxidation potential than H2O. Thus, H2O is oxidized at the anode to liberate O2 molecules