Question

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

Answer 1

`ClO_3^(-)+3SO_2(g) +6HO^(-) rarr 3SO_4^(2-) + Cl^(-) +3H_2O`

Explanation:

Chlorate `(Cl(V))` is reduced to chloride `(Cl^(-)(-I))`:

`ClO_3^(-)+6H^+ +6e^(-) rarr Cl^(-) + 3H_2O.` `(i)`

And sulfur dioxide, `(S(IV))`, is oxidized to sulfate, `(S(VI))`:

`SO_2(aq) +2H_2O(l) rarr SO_4^(2-) + 2e^(-) + 4H^(+).` `(ii)`

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

`(i)+3xx(ii)`

`ClO_3^(-)+3SO_2(g) +3H_2O(aq) rarr 3SO_4^(2-) + Cl^(-)+ 6H^+`

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 `6xxOH^-` to each side of the equation
(i.e. `HO^(-)+H^(+) rarr H_2O`):

`ClO_3^(-)+3SO_2(g) +6HO^(-) rarr 3SO_4^(2-) + Cl^(-) +3H_2O`

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.