# How would MnO_4^(-), and SO_2(g) interact?

Mar 2, 2017

We would assume that the permanganate ion, which is a potent oxidant...............

#### Explanation:

We would assume that the permanganate ion, which is potent oxidant, oxidizes the $S {O}_{2} \left(g\right)$ up to $S \left(V I +\right)$. Let's assume such oxidation:

$\text{Oxidation rxn (i):}$

$S {O}_{2} \left(g\right) + 2 {H}_{2} O \left(l\right) \rightarrow S {O}_{4}^{2 -} + 4 {H}^{+} + 2 {e}^{-}$

$\text{Reduction rxn (ii):}$

$M n {O}_{4}^{-} + 8 {H}^{+} + 5 {e}^{-} \rightarrow M {n}^{2 +} + 4 {H}_{2} O$

$\text{Overall rxn:}$ $=$ $5 \times \left(i\right) + 2 \times \left(i i\right)$ $=$

$2 M n {O}_{4}^{-} + 5 S {O}_{2} \left(g\right) + 2 {H}_{2} O \left(l\right) \rightarrow 2 M {n}^{2 +} + 5 S {O}_{4}^{2 -} + 4 {H}^{+}$

I would recheck this question. They should give you a bit of context (I am guessing that you are an undergrad). Namely, that the reaction proceeds with the disappearance of the deep purple colour of permanganate to give the almost colourless $M {n}^{2 +}$ ion. The redox equation is as far as I know stoichiometrically balanced; i.e. garbage out equals garbage in, including electronic charge.