Question

In the reaction of sulfur with concentrated sulfuric acid, the oxidised product is X and reduced product is Y. X and Y are respectively:

• A. \( {SO}_3, {SO}_2 \)
• B. \( {SO}_2, {SO}_2 \)
• C. \( {SO}_2, {H}_2{S} \)
• D. \( {SO}_2, {H}_2{O} \)

Hint:

In reactions involving sulfur and sulfuric acid, the products \( {SO}_2 \) and \( {H}_2{S} \) are formed, with sulfur undergoing both oxidation and reduction.

Answer 1

The Correct Option is B

When sulfur reacts with concentrated sulfuric acid, sulfur undergoes oxidation and reduction. The reaction is: \[ {S} + {H}_2{SO}_4 \rightarrow {SO}_2 + {H}_2{S} \] In this reaction:
- Sulfur (S) is oxidized to \( {SO}_2 \), which is the oxidized product (X).
- The sulfur (S) is also reduced to \( {SO}_2 \), which is the reduced product (Y).
Step 1: The sulfur reacts with sulfuric acid to form \( {SO}_2 \) as both the oxidized and reduced product. This is because sulfur in the reaction can simultaneously undergo oxidation and reduction. 
Step 2: Therefore, the correct answer is \( {SO}_2, {SO}_2 \), corresponding to option (2).

Answer 2

To determine the oxidised and reduced products when sulfur reacts with concentrated sulfuric acid, we must consider the redox (reduction-oxidation) changes in the reaction. Sulfuric acid, \(H_2SO_4\), is both a strong oxidizing agent and a strong acid. When sulfur (\(S\)) is added to concentrated sulfuric acid, an oxidation-reduction reaction occurs. In this process, sulfur and sulfuric acid react according to the following equation: \[S + 2H_2SO_4 \rightarrow 3SO_2 + 2H_2O\]  Here, sulfur starts in its elemental state with an oxidation number of 0 and gets oxidized to sulfur dioxide \((SO_2)\), where sulfur has an oxidation state of +4. This transition indicates that sulfur is oxidized. Simultaneously, the sulfur in sulfuric acid \((H_2SO_4)\), which initially has an oxidation state of +6, is reduced to sulfur dioxide \((SO_2)\), with an oxidation state of +4. This shows the reduction process. Therefore, both the oxidised and reduced products are \(SO_2\). Thus, the correct answer to the problem is: \(SO_2, SO_2\)