In the following reaction, ${\mathrm{SO}}_2$ acts as a reducing agent :

${\mathrm{SO}}_2+{\mathrm{Cl}}_2+2{\mathrm{H}}_2\mathrm{O}\to {\mathrm{H}}_2{\mathrm{SO}}_4+3\mathrm{HCl}$

Find the equivalent weight of ${\mathrm{SO}}_2$.

${\mathrm{SO}}_2+{\mathrm{Cl}}_2+2{\mathrm{H}}_2\mathrm{O}⟶{\mathrm{H}}_2{\mathrm{SO}}_4+3\mathrm{HCl}$ $n\mathit{.}f=2$

S changes oxidation state from + 4 to +6

$\text{ Equivalent weight }=\frac{\text{ Molecular weight }}{\text{ Total change in oxidation number }}$

$\text{ Eq wt. of }{\mathrm{SO}}_2=\frac{\text{ mol. wt of }{\mathrm{SO}}_2}{2}=\frac{64}{2}=32\mathrm{g}$

Thus, the correct answer is 32.