Question

Explain the following:

The volume of hydrogen and oxygen formed at the cathode and anode respectively differs during the electrolysis of acidified water using platinum electrodes.

Answer 1

Anode reaction

• The ${\mathrm{OH}}^{-}$and ${{\mathrm{SO}}_4}^{-2}$ both migrate to the anode. and ${\mathrm{OH}}^{-}$ being lower in the electrochemical series is discharged preferentially. it loses one electron to the anode and becomes neutral$\mathrm{OH}$.

${\mathrm{OH}}^{-}-{\mathrm{e}}^{-}\to \mathrm{OH}$

• The combination of $\mathrm{OH}$ forms water with the liberation of oxygen which is given off at the anode.

$4\mathrm{OH}\to 2{\mathrm{H}}_2\mathrm{O}\hspace{0.17em}+\hspace{0.17em}{\mathrm{O}}_2$

• The discharge of ${\mathrm{OH}}^{-}$ disturbs the ionic equilibrium of water and to maintain it, more water ionizes.

${\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathrm{H}}^{+}+{\mathrm{OH}}^{-}$

• Product at the anode: 1 volume of oxygen gas is evolved.

Cathode Reaction

• Hydrogen ions, ${\mathrm{H}}^{+}$ being the only positively charged ions, migrate to the cathode. since the cathode is a reservoir of electrons, ${\mathrm{H}}^{+}$gains an electron and becomes a neutral hydrogen atom

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

• Hydrogen atoms combine to form a molecule and this comes out as a hydrogen gas

$\mathrm{H}+\mathrm{H}\to {\mathrm{H}}_2$

• Since ${{\mathrm{SO}}_4}^{-2}$ ions migrate to the anode, and ${\mathrm{H}}^{+}$ions have been discharged, the concentration of the sulphuric acid will decrease.
• Product at the cathode: 2 volumes of hydrogen gas

Hence, the product formed is hydrogen and oxygen in the ratio of 2:1.