In alkaline solution, $\text{MnO}_4^-$ ions oxidise $\text{SO}_3^{2-}$ ions to form $\text{SO}_4^{2-}$ ions. At the same time, the $\text{MnO}_4^-$ ions are reduced to $\text{MnO}_2$. What is the ratio of the two ions in the balanced chemical equation for this reaction?
- A$\text{MnO}_4^- : \text{SO}_3^{2-} = 2 : 3$
- B$\text{MnO}_4^- : \text{SO}_3^{2-} = 3 : 2$
- C$\text{MnO}_4^- : \text{SO}_3^{2-} = 4 : 7$
- D$\text{MnO}_4^- : \text{SO}_3^{2-} = 7 : 4$