When $\text{SO}_2$ gas is passed through an acidified solution of potassium dichromate ($\text{K}_2\text{Cr}_2\text{O}_7$), it acts as a reducing agent. It reduces the orange dichromate ion ($\text{Cr}_2\text{O}_7^{2-}$) to green chromium(III) ion ($\text{Cr}^{3+}$), while itself being oxidised to sulphate ion ($\text{SO}_4^{2-}$). The overall reaction is: $\text{K}_2\text{Cr}_2\text{O}_7 + \text{H}_2\text{SO}_4 + 3\text{SO}_2 \rightarrow \text{K}_2\text{SO}_4 + \text{Cr}_2(\text{SO}_4)_3 + \text{H}_2\text{O}$. Therefore, green chromium(III) sulphate, $\text{Cr}_2(\text{SO}_4)_3$, is formed.