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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?

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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.

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  • $\begingroup$ 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 $\ce{MnO4^- -> Mn^{+2}}$. $\endgroup$
    – MaxW
    Commented Feb 17, 2019 at 19:00
  • $\begingroup$ @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. $\endgroup$
    – andselisk ♦
    Commented Feb 17, 2019 at 19:02
  • $\begingroup$ 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 $\ce{Mn^{+2}}$. $\endgroup$
    – MaxW
    Commented Feb 17, 2019 at 19:05
  • $\begingroup$ @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? $\endgroup$
    – andselisk ♦
    Commented Feb 17, 2019 at 19:09
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    $\begingroup$ 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. ;-) $\endgroup$
    – MaxW
    Commented Feb 17, 2019 at 19:36

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