Question

An aqueous solution of hydrogen sulphide and sulphur dioxide when mixed together yield:
(A)- sulphur and water
(B)- hydrogen peroxide and sulphur
(C)- sulphur trioxide and water
(D)- hydrogen and sulphurous dioxide

Answer 1

Hint: In the given reaction, having two different oxidation states of the sulfur atom, it may act as both an oxidizing and reducing agent accordingly. It will undergo a redox reaction.

Complete step by step solution:
In the given aqueous solution, the reaction between hydrogen sulphide and sulphur dioxide will undergo a redox reaction.
This is because, in the hydrogen sulphide, with the charge on the hydrogen atom is (+1), then the oxidation state of sulfur will be (-2). Similarly, in sulphur dioxide, with the charge on oxygen atom is (-2), then the oxidation of sulfur is (+4). So, we have the sulfur present in two different states in the reactants.
On reaction, the hydrogen sulphide undergoes oxidation, and the sulphur undergoes reduction to produce sulfur and water, as follows:
Oxidation: ${{S}^{2-}}\to {{S}^{0}}+2{{e}^{-}}$
$2{{S}^{2-}}\to 2{{S}^{0}}+4{{e}^{-}}$ -------- (a)
Reduction: ${{S}^{4+}}+4{{e}^{-}}\to {{S}^{0}}$ -------- (b)
Combining equation (a) and (b), we get,
$2{{S}^{2-}}+{{S}^{4+}}\to 3{{S}^{0}}$
$2{{H}_{2}}S(g)+S{{O}_{2}}(g)\to 3S(s)+2{{H}_{2}}O(l)$
It is seen that the hydrogen sulphide acts as a reducing agent, with the change in the oxidation state of sulfur from (-2) to 0. And the sulphur dioxide acts as an oxidizing agent, with the change in oxidation state of sulfur from (+4) to 0.

Therefore, in the reaction of hydrogen sulphide and sulphur dioxide in aqueous solution, a precipitate of sulfur is obtained. So, the reaction yields option (A)- sulphur and water.

Note: The given reaction is a reverse disproportionation reaction, as the sulphur is found to be in two different oxidation states, that is, (-2) and (+4) in the product, as it undergoes both the oxidation and reduction reaction at the same time.