Question

What is meant by 'disproportionation'? Give two examples of disproportionation
reaction in aqueous solution.
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Answer 1

Answer:

Chlorine gas undergoes disproportionation in the alkaline conditions to provide chlorate and chloride ions. Two examples of the disproportionation reaction in the aqueous solution are Reduction - 12Cl2(g)+e−→Cl− (i)Oxidation 12Cl2(g)+6HO−→ClO−3+5e−+3H2O. (ii) And 5 × (i)+(ii) gives - 3Cl2(g)+6HO−→5Cl−+ClO−3+3H2O.

Answer 2

DISPROPOTIONATION REACTION:

• Disproportionation reaction is a special type of chemical reaction in which an element in one oxidation state is oxidized and reduced simulateneously.

• Disproportionation reaction always contains an element which can exist in three oxidation states which is favourable for disproportionation reaction.

• The element in reactant side will always possess a particular element with intermediate oxidation state.

• Higher and lower oxidation states of that element will occur in product side.

EXAMPLE 1:

$\sf P_4(s) +3OH^-(aq)+3H_2O(l)\longrightarrow PH_3(g)+3H_2PO_2^-(aq)$

$\tt Here\ Oxidation\ state\ of\ P\ in\ P_4 = 0$

$\tt Oxidation\ state\ of\ P\ in\ PH_3 = -3$

$\tt Oxidation\ state\ of\ P\ in\ H_2PO_2^-= +1$

EXAMPLE 2:

$\sf S_8(s) +12OH^-(aq) \longrightarrow 4S^{-2}(aq) + 2S_2O_3^{2-}(aq)+6H_2O(l)$

$\tt Here \ Oxidation\ state\ of \ S \ in \ S_8 = 0$

$\tt Oxidation\ state\ of \ S \ in \ S^{-2} = - 2$

$\tt Oxidation\ state\ of \ S \ in \ S_2O_3^{2-} = + 2$