8.23 :  Chlorine is used to purify drinking water. Excess of chlorine is harmful. The excess of chlorine is removed by treating with sulphur dioxide. Present a balanced equation for this redox change taking place in water.

The given redox reaction can be represented as:
${{\mathrm{Cl}}_{2}}_{\left(\mathrm{s}\right)}+{\mathrm{SO}}_{2\left(\mathrm{aq}\right)}+{\mathrm{H}}_{2}{\mathrm{O}}_{\left(\mathrm{l}\right)}\to {{\mathrm{Cl}}^{-}}_{\left(\mathrm{aq}\right)}+{{\mathrm{SO}}_{4}^{2-}}_{\left(\mathrm{aq}\right)}$
The oxidation half reaction is:
$\stackrel{+4}{\mathrm{S}}{O}_{2\left(aq\right)}\to \stackrel{+6}{\mathrm{S}}{{\mathrm{O}}_{4}^{2-}}_{\left(\mathrm{aq}\right)}$
The oxidation number is balanced by adding two electrons as:
${\mathrm{SO}}_{2\left(\mathrm{aq}\right)}\to {{\mathrm{SO}}_{4}^{2-}}_{\left(\mathrm{aq}\right)}+2{\mathrm{e}}^{-}$
The charge is balanced by adding 4${\mathrm{H}}^{+}$ ions as:
${\mathrm{SO}}_{2\left(\mathrm{aq}\right)}\to {{\mathrm{SO}}_{4}^{2-}}_{\left(\mathrm{aq}\right)}+4{{\mathrm{H}}^{+}}_{\left(\mathrm{aq}\right)}+2{\mathrm{e}}^{-}$
The O atoms and ${\mathrm{H}}^{+}$ ions are balanced by adding $2{\mathrm{H}}_{2}\mathrm{O}$ molecules as:
The reduction half reaction is:
${\mathrm{Cl}}_{2\left(\mathrm{s}\right)}\to {{\mathrm{Cl}}^{-}}_{\left(\mathrm{aq}\right)}$
The chlorine atoms are balanced as:
$\stackrel{0}{\mathrm{C}}{l}_{2\left(s\right)}\to \stackrel{-1}{\mathrm{C}}{{\mathrm{l}}^{-}}_{\left(\mathrm{aq}\right)}$
The oxidation number is balanced by adding electrons
The balanced chemical equation can be obtained by adding equation (i) and (ii) as:
${\mathrm{Cl}}_{2\left(\mathrm{s}\right)}+{\mathrm{SO}}_{2\left(\mathrm{aq}\right)}+2{\mathrm{H}}_{2}{\mathrm{O}}_{\left(\mathrm{l}\right)}\to 2{{\mathrm{Cl}}^{-}}_{\left(\mathrm{aq}\right)}+{{\mathrm{SO}}_{4}^{2-}}_{\left(\mathrm{aq}\right)}+4{{\mathrm{H}}^{+}}_{\left(\mathrm{aq}\right)}$