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

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

You are watching: In a disproportionation reaction, the disproportionate substance


*

Generally in redox chemistry reactions one entity is oxidized and other is reduced. In a disproportionation reaction one reality under go both oxidation as well as reduction and two different assets are formed.


*

A disproportionation reaction is when a multiatomic species whose pertinent aspect has a specific oxidation state gets oxidized and diminished in two different half-reactions, yielding 2 other commodities containing the exact same pertinent element.

EXAMPLE: MANGANESE OXIDES

A convenient instance is #"Mn"_2"O"_3# coming to be #"Mn"^(2+)# and also #"MnO"_2#.

From this Pourbaix diagram:

*

...we view that #"Mn"_2"O"_3# in acidic pH"s has boundary lines that converge, and also on either side of the converged heat is #"Mn"^(2+)# and #"MnO"_2#.

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

(Hence, #"MnO"_4^(2-)# could also disproportionate right into #"MnO"_2# and also #"MnO"_4^(-)#, yet in an easy pH.)

You could additionally see that in a Frost diagram, albeit it"s a little harder to identify unless you remember how it works.

*

In this case, the disproportionation occurs from a "hill" allude to the instant left and also right points (e.g. #+3 -> +2,+4#).

THE HALF-REACTIONS

We have the right to 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 palliation from #color(red)(+3)# come #color(red)(+2)#. Currently 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 room in acidic pH) come 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 indigenous #color(red)(+3)# again, yet to #color(red)(+4)#. Balance the manganese as before.

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

Add water come 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 come balance the charge.

See more: Pouring Water On A Drowning Man Gospel, The Five Singing Sons

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

Overall reaction:

And the final an outcome 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 typical multiples.

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