# Balance this reaction? "H"_ ((aq))^(+) + "MnO"_ (4(aq))^(2-) -> "MnO"_ ((aq))^(-) + "MnO"_ (2(s)) + "H"_ 2"O"_ ((l))

Jan 27, 2018

Here's what's going on here.

#### Explanation:

You're actually dealing with a disproportionation reaction here. In a disproportionation reaction, the same element undergoes both oxidation and reduction.

In this case, manganese(VI) is reduced to manganese(IV) and oxidized to manganese(VII).

stackrel(color(blue)(+6))("Mn")"O"_ (4(aq))^(2-) -> stackrel(color(blue)(+7))("Mn")"O"_ (4(aq))^(-) + stackrel(color(blue)(+4))("Mn")"O"_ (2(s))

The reduction half-reaction looks like this

stackrel(color(blue)(+6))("Mn")"O"_ (4(aq))^(2-) + 2"e"^(-) -> stackrel(color(blue)(+4))("Mn")"O"_ (2(s))

Here each atom of manganese takes in $2$ electrons, which is why the oxidation number of manganese goes from $\textcolor{b l u e}{+ 6}$ on the reactants' side to $\textcolor{b l u e}{+ 4}$ on the products' side.

To balance the atoms of oxygen, use the fact that this reaction takes place in an acidic medium and add water molecules to the side that needs oxygen and protons, ${\text{H}}^{+}$, to the side that needs hydrogen.

$4 {\text{H"_ ((aq))^(+) + stackrel(color(blue)(+6))("Mn")"O"_ (4(aq))^(2-) + 2"e"^(-) -> stackrel(color(blue)(+4))("Mn")"O"_ (2(s)) + 2"H"_ 2"O}}_{\left(l\right)}$

Notice that the half-reaction is balanced in terms of charge because you have

$4 \times \left(1 +\right) + \left(2 -\right) + 2 \times \left(1 -\right) = 0 + 0$

$\textcolor{w h i t e}{a}$
The oxidation half-reaction looks like this

stackrel(color(blue)(+6))("Mn")"O"_ (4(aq))^(2-) -> stackrel(color(blue)(+7))("Mn")"O"_ (4(aq))^(-) + "e"^(-)

This time, each atom of manganese loses $1$ electron, which is why the oxidation number of manganese goes from $\textcolor{b l u e}{+ 6}$ on the reactants' side to $\textcolor{b l u e}{+ 7}$ on the products' side.

The atoms of oxygen are already balanced, so you don't need to use water molecules and protons. Once again, the half-reaction is balanced in terms of charge because you have

$\left(2 -\right) = \left(1 -\right) + \left(1 -\right)$

So, you know that the balanced half-reactions look like this

$\left\{\begin{matrix}4 {\text{H"_ ((aq))^(+) + stackrel(color(blue)(+6))("Mn")"O"_ (4(aq))^(2-) + 2"e"^(-) -> stackrel(color(blue)(+4))("Mn")"O"_ (2(s)) + 2"H"_ 2"O"_ ((l)) \\ color(white)(aaaaaaaaaaaaaa)stackrel(color(blue)(+6))("Mn")"O"_ (4(aq))^(2-) -> stackrel(color(blue)(+7))("Mn")"O"_ (4(aq))^(-) + "e}}^{-}\end{matrix}\right.$

In every redox reaction, the number of electrons lost in the oxidation half-reaction must be equal to the number of electrons gained in the reduction half-reaction, so multiply the oxidation half-reaction by $2$ to get

$\left\{\begin{matrix}4 {\text{H"_ ((aq))^(+) + stackrel(color(blue)(+6))("Mn")"O"_ (4(aq))^(2-) + 2"e"^(-) -> stackrel(color(blue)(+4))("Mn")"O"_ (2(s)) + 2"H"_ 2"O"_ ((l)) \\ color(white)(aaaaaaaaaaaaa)2stackrel(color(blue)(+6))("Mn")"O"_ (4(aq))^(2-) -> 2stackrel(color(blue)(+7))("Mn")"O"_ (4(aq))^(-) + 2"e}}^{-}\end{matrix}\right.$

Add the two half-reactions to find the balanced chemical equation that describes this disproportionation reaction.

$\left\{\begin{matrix}4 {\text{H"_ ((aq))^(+) + stackrel(color(blue)(+6))("Mn")"O"_ (4(aq))^(2-) + 2"e"^(-) -> stackrel(color(blue)(+4))("Mn")"O"_ (2(s)) + 2"H"_ 2"O"_ ((l)) \\ color(white)(aaaaaaaaaaaaa)2stackrel(color(blue)(+6))("Mn")"O"_ (4(aq))^(2-) -> 2stackrel(color(blue)(+7))("Mn")"O"_ (4(aq))^(-) + 2"e}}^{-}\end{matrix}\right.$
$\frac{\textcolor{w h i t e}{a}}{\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a}}$
$4 {\text{H"_ ((aq))^(+) + ["MnO"_ (4(aq))^(2-) + 2"MnO"_ (4(aq))^(2-)] + color(red)(cancel(color(black)(2"e"^(-)))) -> 2"MnO"_ (4(aq))^(-) + "MnO"_ (2(s)) + color(red)(cancel(color(black)(2"e"^(-)))) + 2"H"_ 2"O}}_{\left(l\right)}$

You will end up with

$4 {\text{H"_ ((aq))^(+) + 3"MnO"_ (4(aq))^(2-) -> 2"MnO"_ (4(aq))^(-) + "MnO"_ (2(s)) + 2"H"_ 2"O}}_{\left(l\right)}$