Question

When copper is reacted with conc. sulphuric acid, which gas is evolved?
(A) ${H_2}$
(B) ${O_2}$
(C) $S{O_2}$
(D) $S{O_3}$

Answer 1

Hint: Sulphuric acid (${H_2}S{O_4}$) can act as a good oxidizing agent. So, in this process, copper will get oxidized. In this way, you can predict the resultant gas produced as a product of reduction of sulphuric acid.

Complete step by step solution:
Note that conc. sulphuric acid is a good oxidizing agent and it can oxidize other species.
-Here, conc. ${H_2}S{O_4}$ will act as an oxidizing agent as well and it will oxidize copper.
-Copper is in zero oxidation state in its metallic form. So, it will get oxidized from zero to +2 oxidation state. So, copper will act as a reducing agent and will reduce sulphuric acid.
-Thus, we can say that copper gets oxidized and sulphuric acid gets reduced during the reaction. So, we can write the reaction as
\[C{u_{(s)}} + 2{H_2}S{O_{4(aq)}} \to CuS{O_{4(aq)}} + S{O_{2(g)}} + 2{H_2}{O_{(aq)}}\]
-So, we can see that sulphur dioxide is the gas that gets formed by this reaction.
-So, as we know that this is a redox reaction, the gas produced can be either sulphur dioxide or it can be sulphur trioxide.
-Now, the oxidation number of sulphur in sulphuric acid is +6. Now, we know that in this reaction, the reduction of sulphuric acid occurs. Now, here hydrogen or oxygen is not reduced. So, there should be a decrease in the oxidation state of sulphur atoms in the product.
-In sulphur trioxide, sulphur is in +6 oxidation state. So, it cannot be a reduction product of sulphuric acid. So, in this way, we can also predict that the product can be sulphur dioxide gas.

Thus, the correct answer is (C).

Note: Remember that oxidizing agent oxidizes other compounds and itself gets reduced during the course of the reaction. Reducing agents get oxidized and reduce other compounds. So, do not get confused between them.