# Why this redox equation occurs on basic medium?

I have the following redox equation that I need to balance:

$$\ce{MnO4-(aq) + SO3^2-(aq) → MnO2(s) + SO4^2-(aq)}$$

The problem says that it happens on basic medium, but I don't see the $$\ce{OH-}$$ which I usually use to identify the basic medium.

How do I identify that this reaction happens on basic medium? Is it something that the problem should clarify?

You are given a net ionic equation, which is the reduced form of a complete ionic equation: that's why you see neither $$\ce{H2O}$$ nor $$\ce{OH-}$$. You can deduce that the medium should be slightly alkaline a priori only if you:

• either know that manganate $$\ce{MnO4^2-}$$ formed in alkaline medium hydrolyses when there is a lack of reducing agent (here, sulfite $$\ce{SO3^2-}$$) to manganese(IV) oxide $$\ce{MnO2}$$ and that $$\ce{MnO2}$$ is also formed near neutral $$\mathrm{pH}$$ values;
• or you compose the redox reaction as you were asked to and notice that two $$\ce{OH-}$$ are produced:

\begin{align} \ce{\overset{+7}{Mn}O4- + 2 H2O + 3 e- &→ \overset{+4}{Mn}O2 + 4 OH-} & |\cdot 2 \tag{red}\\ \ce{\overset{+4}{S}O3- + 2 OH- &→ \overset{+6}{S}O4^2- + H2O + 2 e-} & |\cdot 3 \tag{ox}\\ \hline \ce{2\overset{+7}{Mn}O4- + 3\overset{+4}{S}O3- + H2O &→ 2\overset{+4}{Mn}O2 + 3\overset{+6}{S}O4^2- + 2 OH-} \tag{redox} \end{align}

As for why your problem states the redox reaction takes place in basic medium, it's just an extra hint assisting you in balancing the oxidation and reduction half-reactions (number of oxygens in oxides and oxoanions, for instance), but you could've balance the equation all right even if there were no information about the alkalinity of the medium.

• RE: "but you could've balance the equation all right even if there were no information about the alkalinity of the medium." -- There is a point here which has not been clearly expressed. In basic solution $\ce{MnO4^- -> MnO2}$, but in acidic solution $\ce{MnO4^- -> Mn^{+2}}$.
– MaxW
Feb 17, 2019 at 19:00
• @MaxW What do you mean? OP is already given the product and it's $\ce{MnO2}$, not $\ce{Mn^2+}$. The rest can be deduced from this very fact. Feb 17, 2019 at 19:02
• That is my point. Since the reduction product is $\ce{MnO2}$ the solution must be basic. If the solution were acidic then the reduction product would be $\ce{Mn^{+2}}$.
– MaxW
Feb 17, 2019 at 19:05
• @MaxW I see, and if the medium were strongly alkaline, then permanganate were reduced to manganate. But what's the point of listing all possibilities unrelated to the terms of OP's question? Feb 17, 2019 at 19:09
• Your English is very very good. I studied French as a second language but forgot most since I never really used it outside of class. So like most Americans I know English and Greek. If it isn't English, it is Greek to me. ;-)
– MaxW
Feb 17, 2019 at 19:36