Iodine oxidises ${\mathrm{S}}_2{\mathrm{O}}_3^{2-}$ ion to ' $\mathrm{X}$ ', change in oxidation state of sulphur

 Iodine oxidises ${\mathrm{S}}_2{\mathrm{O}}_3^{2-}$ ion to ${\mathrm{S}}_4{\mathrm{O}}_6^{2-}$ ion. The oxidation state of sulphur changes from $+2$ to $+2.5$. Thus there will be a change of $+0.5$ oxidation state overall.