Question

Predict the product of electrolysis in the following:

A dilute solution of H₂SO₄ with platinum electrodes

Answer 1

H₂SO₄ ionizes in aqueous solutions to give H⁺ and `"SO"_4^(2-)` ions.

\[\ce{H_2SO_{4(aq)} -> 2H+_{ (aq)} + SO^{2-}_{4(aq)}}\]

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.

\[\ce{2H+_{ (aq)} + 2e- -> H_{2(g)}}\]; E° = 0.0 V

\[\ce{2H_2O_{ (aq)} + 2e- -> H_{2(g)} + 2OH-_{(aq)}}\]; 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.