Question

when$S{O}_2$ is passed in a solution of potassium iodate, the oxidation state of iodine change's from :

• A. +5 to 0
• B. +5 to -1
• C. -5 to 0
• D. -7 to -1

Answer 1

The correct option is A

+5 to 0

When $S{O}_2$ gas is passed into a solution of $KI{O}_3$, the products are $I_2$, $H_{2}S{O}_4$ and $K_{2}S{O}_4$.
The balanced equation for the reaction is:

$5\stackrel{+4}{S}{\stackrel{-2}{O}}_2+4{\stackrel{+1}{H}}_2\stackrel{-2}{O}+2\stackrel{+1}{K}\stackrel{+5}{I}{\stackrel{-2}{O}}_3⟶{\stackrel{+1}{K}}_2\stackrel{+6}{S}{\stackrel{-2}{O}}_4+{\stackrel{0}{I}}_2+{\stackrel{+1}{H}}_2\stackrel{+6}{S}{\stackrel{-2}{O}}_4$

As you can see, the $I$ in the $I{O}_3^{-}$ has a +5 oxidation state, which is changed to 0 in $I_2$