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Chemistry · Ch 7 — Redox Reactions

Disproportionation reactions

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Disproportionation reactions

Disproportionation reactions are a special type of redox reaction where the same element is simultaneously oxidised and reduced. One substance acts as both the oxidising agent and the reducing agent.

For disproportionation to occur, the reacting substance must contain an element that can exist in at least three oxidation states. The element starts in an intermediate oxidation state and produces both a higher and a lower oxidation state.

Example — Decomposition of hydrogen peroxide:

2H2+1O2−1(aq)→2H2+1O−2(l)+O20(g)(7.45) 2\overset{+1}{H_2}\overset{-1}{O_2}(aq) \rightarrow 2\overset{+1}{H_2}\overset{-2}{O}(l) + \overset{0}{O_2}(g) \qquad(7.45)

Oxygen in H2_2O2_2 is in the –1 oxidation state. In the products, some oxygen is reduced to –2 (in H2_2O) and some is oxidised to 0 (in O2_2).

Disproportionation in alkaline medium — Phosphorus:

P40(s)+3OH−(aq)+3H2O(l)→P−3H3(g)+3H2+1PO2−−2(aq)(7.46) \overset{0}{P_4}(s) + 3OH^-(aq) + 3H_2O(l) \rightarrow \overset{-3}{P}H_3(g) + 3\overset{+1}{H_2}P\overset{-2}{O_2^-}(aq) \qquad(7.46)

Phosphorus (0) is reduced to –3 (in PH3_3) and oxidised to +1 (in H2_2PO2−_2^-).

Disproportionation in alkaline medium — Sulphur:

S80(s)+12OH−(aq)→4S2−−2(aq)+2S2+2O32−−2(aq)+6H2O(l)(7.47) \overset{0}{S_8}(s) + 12OH^-(aq) \rightarrow 4\overset{-2}{S^{2-}}(aq) + 2\overset{+2}{S_2}\overset{-2}{O_3^{2-}}(aq) + 6H_2O(l) \qquad(7.47)

Sulphur (0) is reduced to –2 (in S2−^{2-}) and oxidised to +2 (in S2_2O32−_3^{2-}).

Disproportionation in alkaline medium — Chlorine (formation of household bleach):

Cl20(g)+2OH−(aq)→Cl+1O−(aq)+Cl−−1(aq)+H2O(l)(7.48) \overset{0}{Cl_2}(g) + 2OH^-(aq) \rightarrow \overset{+1}{Cl}O^-(aq) + \overset{-1}{Cl^-}(aq) + H_2O(l) \qquad(7.48)

Chlorine (0) is oxidised to +1 (in ClO−^-) and reduced to –1 (in Cl−^-). The hypochlorite ion (ClO−^-) is the active bleaching agent — it oxidises coloured stains to colourless compounds.

Watch out

Bromine and iodine follow the same pattern as chlorine in reaction (7.48), but fluorine does not. Fluorine is the most electronegative element and cannot exhibit positive oxidation states. When fluorine reacts with alkali, a different reaction occurs:

2F2(g)+2OH−(aq)→2F−(aq)+OF2(g)+H2O(l)(7.49)2F_2(g) + 2OH^-(aq) \rightarrow 2F^-(aq) + OF_2(g) + H_2O(l) \qquad(7.49) …