# 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 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. – andselisk Feb 17 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 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? – andselisk Feb 17 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 at 19:36