Question:
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.
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.
Show Hint
Compare the oxidation number of sulphur before and after the reaction.
Updated On: Nov 12, 2026
- \(\text{H}_{2}\text{S}\)
- \(\text{SO}_{2}\)
- \(\text{H}\)
- both \(\text{H}_{2}\text{S}\) and \(\text{SO}_{2}\)
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept
Oxidation means loss of electrons, or an increase in oxidation number.
Step 2: Oxidation numbers of sulphur
In \(\text{H}_{2}\text{S}\), S is \(-2\). In \(\text{SO}_{2}\), S is \(+4\). In elemental S, it is 0.
Step 3: Result
In \(\text{H}_{2}\text{S}\), sulphur goes from \(-2\) to 0, so it loses electrons and is oxidised.
In \(\text{SO}_{2}\), sulphur goes from \(+4\) to 0, so it is reduced. Hydrogen stays at \(+1\).
Final Answer:
Hydrogen sulphide is the substance that is oxidised. \[ \boxed{\text{(A) }\text{H}_{2}\text{S}} \]
Oxidation means loss of electrons, or an increase in oxidation number.
Step 2: Oxidation numbers of sulphur
In \(\text{H}_{2}\text{S}\), S is \(-2\). In \(\text{SO}_{2}\), S is \(+4\). In elemental S, it is 0.
Step 3: Result
In \(\text{H}_{2}\text{S}\), sulphur goes from \(-2\) to 0, so it loses electrons and is oxidised.
In \(\text{SO}_{2}\), sulphur goes from \(+4\) to 0, so it is reduced. Hydrogen stays at \(+1\).
Final Answer:
Hydrogen sulphide is the substance that is oxidised. \[ \boxed{\text{(A) }\text{H}_{2}\text{S}} \]
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