Q104 / 200 · MHT CET - Medical 2026 (25 April Shift 1) · Chemistry
In a reaction,
$2\text{H}_2\text{S} + \text{SO}_2 \rightarrow 2\text{H}_2\text{O} + 3\text{S}$
Identify the substance oxidized.
$2\text{H}_2\text{S} + \text{SO}_2 \rightarrow 2\text{H}_2\text{O} + 3\text{S}$
Identify the substance oxidized.
- Correct answerA$\text{H}_2\text{S}$
- B$\text{SO}_2$
- C$\text{H}$
- Dboth $\text{H}_2\text{S}$ and $\text{SO}_2$
Solution
The given reaction is $2\text{H}_2\text{S} + \text{SO}_2 \rightarrow 2\text{H}_2\text{O} + 3\text{S}$.
The oxidation state of sulphur in $\text{H}_2\text{S}$ is $-2$.
The oxidation state of sulphur in $\text{SO}_2$ is $+4$.
The oxidation state of elemental sulphur ($\text{S}$) is $0$.
The oxidation state of sulphur increases from $-2$ in $\text{H}_2\text{S}$ to $0$ in $\text{S}$. An increase in oxidation state corresponds to oxidation.
The oxidation state of sulphur decreases from $+4$ in $\text{SO}_2$ to $0$ in $\text{S}$. A decrease in oxidation state corresponds to reduction.
Thus, $\text{H}_2\text{S}$ is oxidized and $\text{SO}_2$ is reduced.
Answer: $\text{H}_2\text{S}$
The oxidation state of sulphur in $\text{H}_2\text{S}$ is $-2$.
The oxidation state of sulphur in $\text{SO}_2$ is $+4$.
The oxidation state of elemental sulphur ($\text{S}$) is $0$.
The oxidation state of sulphur increases from $-2$ in $\text{H}_2\text{S}$ to $0$ in $\text{S}$. An increase in oxidation state corresponds to oxidation.
The oxidation state of sulphur decreases from $+4$ in $\text{SO}_2$ to $0$ in $\text{S}$. A decrease in oxidation state corresponds to reduction.
Thus, $\text{H}_2\text{S}$ is oxidized and $\text{SO}_2$ is reduced.
Answer: $\text{H}_2\text{S}$
Hence, the correct answer is option: 'A'.