What is the oxidation state of S in the sulphur containing product of the following reaction? SO 3 - 2 ( aq…

This is a redox reaction in which Sulphur will oxidised and Bromine will reduce according to the following reaction.

${{\mathrm{SO}}_3^{-2}}_{\left(\mathrm{aq}\right)} + {{\mathrm{Br}}_2}_{\left(\mathrm{l}\right)} + {\mathrm{H}}_2\mathrm{O} \to {{\mathrm{SO}}_4^{-2}}_{\left(\mathrm{aq}\right)} + {\mathrm{Br}}^{-}+ {\mathrm{H}}^{+}$

Balanced reaction will be

${{\mathrm{SO}}_3^{-2}}_{\left(\mathrm{aq}\right)} + {{\mathrm{Br}}_2}_{\left(\mathrm{l}\right)} + {\mathrm{H}}_2\mathrm{O} \to {{\mathrm{SO}}_4^{-2}}_{\left(\mathrm{aq}\right)} + 2{{\mathrm{Br}}^{-}}_{\left(aq\right)}+2{\mathrm{H}}^{+}$

The sum of oxidation states of all elements in a compound is always zero or equal to total charge on ion.

Let the oxidation state of $\mathrm{S}$ in ${\mathrm{SO}}_4^{-2}=\mathrm{x}$

$\therefore \mathrm{x}+\left(-2\times 4\right)=-2$

or $\mathrm{x}+2-8=0$

$\therefore \mathrm{x}=+6$