Write the net ionic equation for the reaction of potassium dichromate$(VI)$,$K_{2}Cr_{2}O_{7}$ with sodium sulphite,$Na_{2}SO_{3}$ in an acidic solution to give chromium$(III)$ ion and the sulphate ion.

(N/A) Step $1:$ The skeletal ionic equation is: $Cr_{2}{O_{7}}^{2-}_{(aq)} + S{O_{3}}^{2-}_{(aq)} \rightarrow Cr^{3+}_{(aq)} + S{O_{4}}^{2-}_{(aq)}$
Step $2:$ Assign oxidation numbers for $Cr$ and $S$: $\mathop {Cr_{2}O_{7}^{2-}}\limits^{+6, -2}_{(aq)} + \mathop {SO_{3}^{2-}}\limits^{+4, -2}_{(aq)} \to \mathop {Cr^{3+}}\limits^{+3}_{(aq)} + \mathop {SO_{4}^{2-}}\limits^{+6, -2}_{(aq)}$. This indicates that the dichromate ion is the oxidant and the sulphite ion is the reductant.
Step $3:$ Calculate the increase and decrease of oxidation number and make them equal. The oxidation state of chromium changes from $+6$ to $+3$ (decrease of $3$ per $Cr$ atom). The oxidation state of sulphur changes from $+4$ to $+6$ (increase of $2$ per $S$ atom). To balance,multiply $Cr$ by $2$ and $S$ by $3$: $Cr_{2}{O_{7}}^{2-}_{(aq)} + 3S{O_{3}}^{2-}_{(aq)} \to 2Cr^{3+}_{(aq)} + 3S{O_{4}}^{2-}_{(aq)}$.
Step $4:$ As the reaction occurs in an acidic medium,balance the charge by adding $8H^{+}$ on the left: $Cr_{2}{O_{7}}^{2-}_{(aq)} + 3S{O_{3}}^{2-}_{(aq)} + 8H^{+}_{(aq)} \to 2Cr^{3+}_{(aq)} + 3S{O_{4}}^{2-}_{(aq)}$.
Step $5:$ Balance the hydrogen atoms by adding $4H_{2}O$ on the right: $Cr_{2}{O_{7}}^{2-}_{(aq)} + 3S{O_{3}}^{2-}_{(aq)} + 8H^{+}_{(aq)} \to 2Cr^{3+}_{(aq)} + 3S{O_{4}}^{2-}_{(aq)} + 4H_{2}O_{(l)}$.