# Observation of sodium sulfite solid added to a little alkaline potassium permanganate?

Jun 5, 2018

Well what did you see? The colour would dissipate certainly, but was there a brown precipitate? Did it go green..?

#### Explanation:

In acidic medium sulfite would be oxidized to sulfate, and permanganate reduced to $M {n}^{2 +}$

$S {O}_{3}^{2 -} + {H}_{2} O \left(l\right) \rightarrow S {O}_{4}^{2 -} + 2 {H}^{+} + 2 {e}^{-}$ $\left(i\right)$

And of course permanganate reduced to $M {n}^{2 +}$...

$M n {O}_{4}^{-} + 8 {H}^{+} + 5 {e}^{-} \rightarrow M {n}^{2 +} + 4 {H}_{2} O$ $\left(i i\right)$

But this is in an ACID medium. In a basic medium, which was specified by the question, more likely the reduction product would be manganate ion, $M n {O}_{4}^{2 -}$, $M n \left(V I +\right)$, OR $M n {O}_{2} \left(s\right)$...i.e. a hydrous oxide of $M n \left(I V +\right)$...

For manganate ion....

${\underbrace{M n {O}_{4}^{-}}}_{\text{red" +e^(-) rarr underbrace(MnO_4^(2-))_"green}}$

...or MORE LIKELY.....

$M n {O}_{4}^{-} + 2 {H}_{2} O + 3 {e}^{-} \rightarrow {\underbrace{M n {O}_{2} \left(s\right)}}_{\text{brown solid}} \downarrow + 4 H {O}^{-}$ $\left(i i i\right)$

I am inclined to think that $\left(i i i\right)$ is the reduction that occurred. But what do I know? I did not do the experiment. And so we cross multiply...

$3 \times \left(i\right) + 2 \times \left(i i i\right)$

$3 S {O}_{3}^{2 -} + 2 M n {O}_{4}^{-} + {H}_{2} O \left(l\right) \rightarrow 2 M n {O}_{2} \left(s\right) + 3 S {O}_{4}^{2 -} + 2 H {O}^{-}$

The which is balanced with respect to mass and charge... Whew.

See here for another example of this process...