Question

What is a disproportionation reaction?

Answer 1

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

Explanation:

A great example is chlorine mixing with water.

`Cl_2 + H_2O -> HCl + HClO`

Elemental chlorine (`Cl_2`) has a redox number of 0.
In `HCl`, it has a redox number of -1
And in `HClO` it has a redox number of +1

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

Answer 2

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.

Answer 3

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.

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EXAMPLE: MANGANESE OXIDES

A convenient example is `"Mn"_2"O"_3` becoming `"Mn"^(2+)` and `"MnO"_2`.

From this Pourbaix diagram:

...we see that `"Mn"_2"O"_3` in acidic pH's has boundary lines that converge, and on either side of the converged line is `"Mn"^(2+)` and `"MnO"_2`.

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

(Hence, `"MnO"_4^(2-)` could also disproportionate into `"MnO"_2` and `"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 -> {+2,+4}`).

THE HALF-REACTIONS

We can write the half-reactions like so.

Reduction:

`stackrel(color(red)(+3))("Mn")_2stackrel(color(red)(-2))("O"_3)(s) -> stackrel(color(red)(+2))("Mn"^(2+))(aq)`

This is a reduction from `color(red)(+3)` to `color(red)(+2)`. Now we balance these. Balance the manganese first:

`=> "Mn"_2"O"_3(s) -> 2"Mn"^(2+)(aq)`

Add water to balance the oxygens.

`=> "Mn"_2"O"_3(s) -> 2"Mn"^(2+)(aq) + 3"H"_2"O"(l)`

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

`=> "Mn"_2"O"_3(s) + 6"H"^(+)(aq) -> 2"Mn"^(2+)(aq) + 3"H"_2"O"(l)`

Now adding the electrons balances the charge.

`=> color(green)("Mn"_2"O"_3(s) + 6"H"^(+)(aq) + 2e^(-) -> 2"Mn"^(2+)(aq) + 3"H"_2"O"(l))`

Oxidation:

`stackrel(color(red)(+3))("Mn")_2stackrel(color(red)(-2))("O"_3)(s) -> stackrel(color(red)(+4))("Mn")stackrel(color(red)(-2))("O")_2(s)`

This is an oxidation from `color(red)(+3)` again, but to `color(red)(+4)`. Balance the manganese as before.

`=> "Mn"_2"O"_3(s) -> 2"MnO"_2(s)`

Add water to balance the oxygens...

`=> "Mn"_2"O"_3(s) + "H"_2"O"(l) -> 2"MnO"_2(s)`

and protons to balance the hydrogens...

`=> "Mn"_2"O"_3(s) + "H"_2"O"(l) -> 2"MnO"_2(s) + 2"H"^(+)(aq)`

and electrons to balance the charge.

`=> color(green)("Mn"_2"O"_3(s) + "H"_2"O"(l) -> 2"MnO"_2(s) + 2"H"^(+)(aq) + 2e^(-))`

Overall reaction:

And the final result would be:

`"Mn"_2"O"_3(s) + cancel(6)^(4)"H"^(+)(aq) + cancel(2e^(-)) -> 2"Mn"^(2+)(aq) + cancel(3)^(2)"H"_2"O"(l)`
`"Mn"_2"O"_3(s) + cancel("H"_2"O"(l)) -> 2"MnO"_2(s) + cancel(2"H"^(+)(aq)) + cancel(2e^(-))`
`"--------------------------------------------------------------"`
`2"Mn"_2"O"_3(s) + 4"H"^(+)(aq) -> 2"MnO"_2(s) + 2"Mn"^(2+)(aq) + 2"H"_2"O"(l)`

Finally, cancel out the common multiples.

`=> color(blue)("Mn"_2"O"_3(s) + 2"H"^(+)(aq) -> "MnO"_2(s) + "Mn"^(2+)(aq) + "H"_2"O"(l))`

Answer 4

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 `"chloride"` and `"chlorate ions"`.

`"Reduction"`
`1/2Cl_2(g) + e^(-) rarr Cl^(-)` `(i)`

`"Oxidation"`
`1/2Cl_2(g) + 6HO^(-) rarr ClO_3^(-)+ 5e^(-) +3H_2O` `(ii)`

And `5xx(i) + (ii)` gives:

`3Cl_2(g) + 6HO^(-) rarr 5Cl^(-) + ClO_3^(-)+3H_2O`

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.

Answer 5

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 `"pH"`s against the redox potential at that `"pH"`. Higher up on the diagram, we have species with higher oxidation states. For instance, manganese in `"Mn"` has a `0` oxidation state, while `"MnO"_4^(-)` has a manganese oxidation state of `+7`.

A disproportionation reaction could occur starting from `"MnO"_4^(2-)(aq)` in `"pH 13"` solution to form `"MnO"_4^(-)(aq)` and `"MnO"_2(s)`. Visually, we branch off into the two species to the upper left and lower left with respect to `"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 `"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:

`2e^(-) + 4"H"^(+)(aq) + stackrel(color(blue)(+6))("Mn")"O"_4^(2-)(aq) -> stackrel(color(blue)(+4))("Mn")"O"_2(s) + 2"H"_2"O"(l)`

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

Finally, we are in basic solution, so we add `"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:

`bb(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) + 4"OH"^(-)(aq))`