Solid sodium iodide reacts with concentrated sulfuric acid. Which product shows iodide reducing the sulfur most strongly?
Answer and explanation
Correct answer: D. Hydrogen sulfide, H2S
- Step 1: Iodide is the strongest reducing agent of the halides. Step 2. It reduces sulfur from +6 in H2SO4 to -2 in H2S (also forming iodine and sulfur). Answer: Hydrogen sulfide, H2S. Why not the others: SO2 (+4) is formed from bromide; HCl is not formed from iodide; SO3 has sulfur at +6, so no reduction.
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