# What is a disproportionation reaction?

Jun 21, 2016

A reaction in which the same species is both oxidised and reduced.

#### Explanation:

A great example is chlorine mixing with water.

$C {l}_{2} + {H}_{2} O \to H C l + H C l O$

Elemental chlorine ($C {l}_{2}$) has a redox number of 0.
In $H C l$, it has a redox number of -1
And in $H C l O$ it has a redox number of +1

Therefore chlorine has been both oxidised and reduced in the same reaction. This is disproportionation.

Jul 5, 2016

Generally in redox chemical reactions one entity is oxidized and other is reduced. In a disproportionation reaction one entity under goes both oxidation as well as reduction and two different products are formed.

Jul 5, 2016

A disproportionation reaction is when a multiatomic species whose pertinent element has a specific oxidation state gets oxidized and reduced in two separate half-reactions, yielding two other products containing the same pertinent element.

EXAMPLE: MANGANESE OXIDES

A convenient example is ${\text{Mn"_2"O}}_{3}$ becoming ${\text{Mn}}^{2 +}$ and ${\text{MnO}}_{2}$.

From this Pourbaix diagram:

...we see that ${\text{Mn"_2"O}}_{3}$ in acidic pH's has boundary lines that converge, and on either side of the converged line is ${\text{Mn}}^{2 +}$ and ${\text{MnO}}_{2}$.

In a Pourbaix diagram, this indicates a disproportionation reaction (which, by the way, is spontaneous below pH 4).

(Hence, ${\text{MnO}}_{4}^{2 -}$ could also disproportionate into ${\text{MnO}}_{2}$ and ${\text{MnO}}_{4}^{-}$, but in basic pH.)

You could also see it in a Frost diagram, albeit it's a bit harder to identify unless you remember how it works.

In this case, the disproportionation occurs from a "hill" point to the immediate left and right points (e.g. $+ 3 \to \left\{+ 2 , + 4\right\}$).

THE HALF-REACTIONS

We can write the half-reactions like so.

Reduction:

$\stackrel{\textcolor{red}{+ 3}}{{\text{Mn")_2stackrel(color(red)(-2))("O"_3)(s) -> stackrel(color(red)(+2))("Mn}}^{2 +}} \left(a q\right)$

This is a reduction from $\textcolor{red}{+ 3}$ to $\textcolor{red}{+ 2}$. Now we balance these. Balance the manganese first:

$\implies {\text{Mn"_2"O"_3(s) -> 2"Mn}}^{2 +} \left(a q\right)$

Add water to balance the oxygens.

$\implies \text{Mn"_2"O"_3(s) -> 2"Mn"^(2+)(aq) + 3"H"_2"O} \left(l\right)$

Add protons (since we are in acidic pH) to balance the hydrogens.

$\implies \text{Mn"_2"O"_3(s) + 6"H"^(+)(aq) -> 2"Mn"^(2+)(aq) + 3"H"_2"O} \left(l\right)$

Now adding the electrons balances the charge.

$\implies \textcolor{g r e e n}{\text{Mn"_2"O"_3(s) + 6"H"^(+)(aq) + 2e^(-) -> 2"Mn"^(2+)(aq) + 3"H"_2"O} \left(l\right)}$

Oxidation:

${\stackrel{\textcolor{red}{+ 3}}{\text{Mn")_2stackrel(color(red)(-2))("O"_3)(s) -> stackrel(color(red)(+4))("Mn")stackrel(color(red)(-2))("O}}}_{2} \left(s\right)$

This is an oxidation from $\textcolor{red}{+ 3}$ again, but to $\textcolor{red}{+ 4}$. Balance the manganese as before.

$\implies {\text{Mn"_2"O"_3(s) -> 2"MnO}}_{2} \left(s\right)$

Add water to balance the oxygens...

$\implies {\text{Mn"_2"O"_3(s) + "H"_2"O"(l) -> 2"MnO}}_{2} \left(s\right)$

and protons to balance the hydrogens...

$\implies {\text{Mn"_2"O"_3(s) + "H"_2"O"(l) -> 2"MnO"_2(s) + 2"H}}^{+} \left(a q\right)$

and electrons to balance the charge.

$\implies \textcolor{g r e e n}{{\text{Mn"_2"O"_3(s) + "H"_2"O"(l) -> 2"MnO"_2(s) + 2"H}}^{+} \left(a q\right) + 2 {e}^{-}}$

Overall reaction:

And the final result would be:

$\text{Mn"_2"O"_3(s) + cancel(6)^(4)"H"^(+)(aq) + cancel(2e^(-)) -> 2"Mn"^(2+)(aq) + cancel(3)^(2)"H"_2"O} \left(l\right)$
"Mn"_2"O"_3(s) + cancel("H"_2"O"(l)) -> 2"MnO"_2(s) + cancel(2"H"^(+)(aq)) + cancel(2e^(-))
$\text{--------------------------------------------------------------}$
$2 \text{Mn"_2"O"_3(s) + 4"H"^(+)(aq) -> 2"MnO"_2(s) + 2"Mn"^(2+)(aq) + 2"H"_2"O} \left(l\right)$

Finally, cancel out the common multiples.

$\implies \textcolor{b l u e}{\text{Mn"_2"O"_3(s) + 2"H"^(+)(aq) -> "MnO"_2(s) + "Mn"^(2+)(aq) + "H"_2"O} \left(l\right)}$

Oct 15, 2016

A disproportionation reaction is a redox reaction in which one species undergoes BOTH oxidation and reduction.

#### Explanation:

This is best illustrated by an actual example. Chlorine gas is known to undergo disproportionation in alkaline conditions to give $\text{chloride}$ and $\text{chlorate ions}$.

$\text{Reduction}$
$\frac{1}{2} C {l}_{2} \left(g\right) + {e}^{-} \rightarrow C {l}^{-}$ $\left(i\right)$

$\text{Oxidation}$
$\frac{1}{2} C {l}_{2} \left(g\right) + 6 H {O}^{-} \rightarrow C l {O}_{3}^{-} + 5 {e}^{-} + 3 {H}_{2} O$ $\left(i i\right)$

And $5 \times \left(i\right) + \left(i i\right)$ gives:

$3 C {l}_{2} \left(g\right) + 6 H {O}^{-} \rightarrow 5 C {l}^{-} + C l {O}_{3}^{-} + 3 {H}_{2} O$

Which is balanced with respect to mass and charge, as is always required. Clearly zerovalent chlorine gas has undergone reduction to chloride ion, and oxidation to chlorate ion; i.e. a disproportionation reaction.

Jun 19, 2017

It is when a species reacts to simultaneously produce one oxidized and one reduced species.

For instance, consider the following Pourbaix diagram of manganese:

This diagram graphs the species present in aqueous solution at various $\text{pH}$s against the redox potential at that $\text{pH}$. Higher up on the diagram, we have species with higher oxidation states. For instance, manganese in $\text{Mn}$ has a $0$ oxidation state, while ${\text{MnO}}_{4}^{-}$ has a manganese oxidation state of $+ 7$.

A disproportionation reaction could occur starting from ${\text{MnO}}_{4}^{2 -} \left(a q\right)$ in $\text{pH 13}$ solution to form ${\text{MnO}}_{4}^{-} \left(a q\right)$ and ${\text{MnO}}_{2} \left(s\right)$. Visually, we branch off into the two species to the upper left and lower left with respect to ${\text{MnO}}_{4}^{2 -}$.

The skeletal half-reactions are (with only manganese oxidation states labeled):

stackrel(color(blue)(+6))("Mn")"O"_4^(2-)(aq) -> stackrel(color(blue)(+4))("Mn")"O"_2(s)
(reduction)

stackrel(color(blue)(+6))("Mn")"O"_4^(2-)(aq) -> stackrel(color(blue)(+7))("Mn")"O"_4^(-)(aq)
(oxidation)

You can see that the same species got oxidized and reduced at the same time.

As the reaction is at $\text{pH 13}$, we shall balance these in base as usual using the method shown in more detail here.

The resultant two balanced half-reactions are:

$2 {e}^{-} + 4 \text{H"^(+)(aq) + stackrel(color(blue)(+6))("Mn")"O"_4^(2-)(aq) -> stackrel(color(blue)(+4))("Mn")"O"_2(s) + 2"H"_2"O} \left(l\right)$

stackrel(color(blue)(+6))("Mn")"O"_4^(2-)(aq) -> stackrel(color(blue)(+7))("Mn")"O"_4^(-)(aq) + e^(-)

So, we add them to get:

$\cancel{2 {e}^{-}} + 4 \text{H"^(+)(aq) + stackrel(color(blue)(+6))("Mn")"O"_4^(2-)(aq) -> stackrel(color(blue)(+4))("Mn")"O"_2(s) + 2"H"_2"O} \left(l\right)$
2(stackrel(color(blue)(+6))("Mn")"O"_4^(2-)(aq) -> stackrel(color(blue)(+7))("Mn")"O"_4^(-)(aq) + cancel(e^(-)))
$\text{-----------------------------------------------------------------------}$
$4 \text{H"^(+)(aq) + 3stackrel(color(blue)(+6))("Mn")"O"_4^(2-)(aq) -> stackrel(color(blue)(+4))("Mn")"O"_2(s) + 2stackrel(color(blue)(+7))("Mn")"O"_4^(-)(aq) + 2"H"_2"O} \left(l\right)$

Finally, we are in basic solution, so we add ${\text{OH}}^{-}$ to both sides to make sure the solution is basic.

cancel(4"H"^(+)(aq) + 4"OH"^(-)(aq))^(2"H"_2"O"(l)) + 3stackrel(color(blue)(+6))("Mn")"O"_4^(2-)(aq) -> stackrel(color(blue)(+4))("Mn")"O"_2(s) + 2stackrel(color(blue)(+7))("Mn")"O"_4^(-)(aq) + cancel(2"H"_2"O"(l)) + 4"OH"^(-)(aq)

As a result, we cancel out some of the waters to get:

$\boldsymbol{2 {\text{H"_2"O"(l) + 3stackrel(color(blue)(+6))("Mn")"O"_4^(2-)(aq) -> stackrel(color(blue)(+4))("Mn")"O"_2(s) + 2stackrel(color(blue)(+7))("Mn")"O"_4^(-)(aq) + 4"OH}}^{-} \left(a q\right)}$