When ${\mathrm{SO}}_2$ is passed through the acidified ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ solution?

Whenever ${\mathrm{SO}}_2$ is acidified as it passes ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ green solution of ${\mathrm{Cr}}_2{\left({\mathrm{SO}}_4\right)}_3$ is created.

${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7+3{\mathrm{SO}}_2+{\mathrm{H}}_2{\mathrm{SO}}_4\to {\mathrm{K}}_2{\mathrm{SO}}_4+{\mathrm{Cr}}_2{\left({\mathrm{SO}}_4\right)}_3+{\mathrm{H}}_2\mathrm{O}$