Question

When sulphur dioxide is passed in an acidified $K_2{Cr}_2{O}_7$ solution, the oxidation state of sulphur is changed from:

• A. $+4$ to $0$
• B. $+4$ to $+2$
• C. $+4$ to $+6$
• D. $+6$ to $+4$

Answer 1

The correct option is C $+4$ to $+6$

When sulphur dioxide is passed in an acidified $K_{2}C{r}_2{O}_7$ solution,$S{O}_2$ conerts to $S{O}_4^{2-}$ in the following reaction:

$3S{O}_2+C{r}_2{O}_7^{2-}\to 3S{O}_4^{2-}+2C{r}^{3+}+H_{2}O$

So, oxidation state of $S$ in $S{O}_2$ is $+4$ it changes to $+6$ in $S{O}_4^{2-}$

Hence, option $C$ is correct.