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

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

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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.


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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 #"Mn"_2"O"_3# becoming #"Mn"^(2+)# and #"MnO"_2#.

From this Pourbaix diagram:

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...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.

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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.

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#=> 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))#