# How do you represent the redox reaction of chlorate ion, ClO_3^(-) with SO_2(g) to give Cl^(-) and SO_4^(2-)?

Oct 14, 2016

$C l {O}_{3}^{-} + 3 S {O}_{2} \left(g\right) + 6 H {O}^{-} \rightarrow 3 S {O}_{4}^{2 -} + C {l}^{-} + 3 {H}_{2} O$

#### Explanation:

Chlorate $\left(C l \left(V\right)\right)$ is reduced to chloride $\left(C {l}^{-} \left(- I\right)\right)$:

$C l {O}_{3}^{-} + 6 {H}^{+} + 6 {e}^{-} \rightarrow C {l}^{-} + 3 {H}_{2} O .$ $\left(i\right)$

And sulfur dioxide, $\left(S \left(I V\right)\right)$, is oxidized to sulfate, $\left(S \left(V I\right)\right)$:

$S {O}_{2} \left(a q\right) + 2 {H}_{2} O \left(l\right) \rightarrow S {O}_{4}^{2 -} + 2 {e}^{-} + 4 {H}^{+} .$ $\left(i i\right)$

Charge and mass are balanced in each equation, so we may cross mulitply to remove the electrons:

$\left(i\right) + 3 \times \left(i i\right)$

$C l {O}_{3}^{-} + 3 S {O}_{2} \left(g\right) + 3 {H}_{2} O \left(a q\right) \rightarrow 3 S {O}_{4}^{2 -} + C {l}^{-} + 6 {H}^{+}$

Are charge and mass balanced here? Do not trust my arithmetic!

On other words, chlorate oxidizes sulfurous acid to sulfate.

Ah, but you specified that the reaction was performed in base. We have the process in acid. What do we do? Easy! Simply add $6 \times O {H}^{-}$ to each side of the equation
(i.e. $H {O}^{-} + {H}^{+} \rightarrow {H}_{2} O$):

$C l {O}_{3}^{-} + 3 S {O}_{2} \left(g\right) + 6 H {O}^{-} \rightarrow 3 S {O}_{4}^{2 -} + C {l}^{-} + 3 {H}_{2} O$

Charge and mass are balanced, the which is absolutely required. If you are unclear of the steps I took and why, ask, and someone will help you.