Question

During the electrolysis of acidified water, the reaction taking place at anode is

• A. $2{\text{H}}_3{\text{O}}^{+}\text{(aq)}+4e^{-}\to {\text{H}}_2\text{(g)}+2{\text{O}}_2\text{(g)}$
• B. ${\text{H}}_2{\text{SO}}_4\text{(l)}+{\text{H}}_2\text{O (l)}\to 2{\text{H}}_3{\text{O}}^{+}\text{(aq)}+{\text{SO}}_4^{2-}$
• C. $2{\text{H}}_2\text{O (l)}\to {\text{O}}_2\text{(g)}+4{\text{H}}^{+}\text{(aq)}+4e^{-}$
• D. $2{\text{H}}_3{\text{O}}^{+}\text{(aq)}+2e^{-}\to {\text{H}}_2\text{(g)}+2{\text{H}}_2\text{O (l)}$

Answer 1

The correct option is C $2{\text{H}}_2\text{O (l)}\to {\text{O}}_2\text{(g)}+4{\text{H}}^{+}\text{(aq)}+4e^{-}$

Pure water does not ionize to a greater extent and therefore does not allow passage of electricity. A little of dilute sulphuric acid is mixed with water, which when ionizes gives ${\text{H}}^{+}$ ions. These ${\text{H}}^{+}$ ions combine with water and form ${\text{H}}_3{\text{O}}^{+}$ ions.

${\text{H}}_2{\text{SO}}_4\text{(l)}+{\text{H}}_2\text{O (l)}\to 2{\text{H}}_3{\text{O}}^{+}\text{(aq)}+{\text{SO}}_4^{2-}$

The reaction occurring at the cathode is:

$2{\text{H}}_3{\text{O}}^{+}\text{(aq)}+2e^{-}\to {\text{H}}_2\text{(g)}+2{\text{H}}_2\text{O (l)}$

The ions moving towards positive electrode is ${\text{SO}}_4^{2-}$. On comparing ${\text{SO}}_4^{2-}$ with water, water has a higher tendency to get oxidised.

The reaction occurring at the anode is:

$2{\text{H}}_2\text{O (l)}\to {\text{O}}_2\text{(g)}+4{\text{H}}^{+}\text{(aq)}+4e^{-}$

Hydrogen gas is obtained at the cathode and oxygen gas is obtained at the anode.