Question

Identify reductant in following reaction.
$\mathrm{H_2S} + \mathrm{NO_2} \rightarrow \mathrm{H_2O} + \mathrm{NO} + \mathrm{S}$

• A. $\mathrm{H_2S}$
• B. $\mathrm{NO_2}$
• C. $\mathrm{NO}$
• D. $\mathrm{S}$

Hint:

A reductant must always be chosen from the reactant side of the equation, which instantly rules out products like $\mathrm{NO}$ and $\mathrm{S}$. Since sulfur goes from a negative oxidation state ($-2$) up to zero ($0$) by shedding electrons, $\mathrm{H_2S}$ is clearly the oxidized agent!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The task is to pinpoint the reducing agent (reductant) in the reaction: H₂S + NO₂ → H₂O + NO + S.

Step 2: Key Formula or Approach:
A reductant is the species that gets oxidized during the process. Oxidation corresponds to a rise in oxidation number caused by electron loss.

Step 3: Detailed Explanation:
Track the oxidation states of the elements across the equation. For sulfur in H₂S: H is +1, making S equal to -2. Elemental sulfur on the product side has an oxidation number of 0. Moving from -2 to 0 signifies an increase, meaning H₂S loses electrons and undergoes oxidation, thereby functioning as the reductant. For nitrogen in NO₂: O is -2 each, so N is +4. In NO, O is -2, giving N a value of +2. This decrease indicates NO₂ gains electrons and acts as the oxidant.

Step 4: Final Answer:
The reductant is H₂S, matching option (A).