Question

Write down separate equations for the reactions which take place at the two platinum electrodes when dilute ${\mathrm{H}}_2{\mathrm{SO}}_4$, which is electrolysed.

Answer 1

Electrolysis of dilute Sulphuric acid ${\mathrm{H}}_2{\mathrm{SO}}_4$ takes place when electricity is passed through the solution of sulphuric acid. The aqueous solution contains ions from both sulphuric acid and water.

Part 1: Electrolysis of Sulphuric acid:

$\underset{\mathrm{Sulphuric}\mathrm{acid}}{{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)}\to \underset{\mathrm{Sulphate}\mathrm{ion}}{{\mathrm{SO}}_4^{2-}\left(\mathrm{aq}\right)}+\underset{\mathrm{Hydrogen}\mathrm{ion}}{2{\mathrm{H}}^{+}\left(\mathrm{aq}\right)}$

Part 2: Dissociation of water$\left({\mathrm{H}}_2\mathrm{O}\right)$:

$\underset{\mathrm{Water}}{{\mathrm{H}}_2\mathrm{O}(\mathrm{l}})\to \underset{\mathrm{Hydrogen}\mathrm{ion}}{{\mathrm{H}}^{+}\left(\mathrm{aq}\right)}+\underset{\mathrm{Hydroxide}\mathrm{ion}}{{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)}$

Part 3: The reaction taking place at the cathode is:

$\underset{\mathrm{Hydrogen}\mathrm{ion}}{2{\mathrm{H}}^{+}\left(\mathrm{aq}\right)}+2{\mathrm{e}}^{-}\to \underset{\mathrm{Hydrogen}}{{\mathrm{H}}_2\left(\mathrm{g}\right)}$

Part 4: The reaction taking place at the anode is:

$\underset{\mathrm{Hydroxide}\mathrm{ion}}{4{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)}\to \underset{\mathrm{Water}}{2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}+\underset{\mathrm{Oxygen}}{{\mathrm{O}}_2\left(\mathrm{g}\right)}+4{\mathrm{e}}^{-}$