Question

Predict the products of electrolysis in each of the following.

A dilute solution of $H_{2}S{O}_4$ with platinum electrodes.

Answer 1

The dissociation of sulphuric acid gives protons and sulphate ions.
At cathode, either hydrogen ions or water molecules can be reduced. Since protons have higher reduction potential than water, hydrogen ions are reduced to hydrogen gas.
At anode, either sulphate ions or water molecule can get oxidized. Since, during oxidation of sulphate ions, more bonds are broken than oxidation of water molecules, sulphate ions have lower oxidation potential than water. Hence, water is oxidized at the anode to liberate oxygen molecules.

$H_{2}S{O}_4\to 2H^{+}+S{O}_4^{-}$

$H_{2}O\to H^{+}+O{H}^{-}$

Reaction at cathode;

$2H^{+}+2e^{-}\to H_2$

Reaction at anode;

Due to platinum electrode, self ionization of water will take place.

$H_{2}O\to 2H^{+}+\frac{1}{2}{O}_2+2e^{-}$

Hence, hydrogen gas will generate at cathode and oxygen gas will generate at anode.