Iodide reduces sulfuric acid due to stronger reducing power

Steps:

• Concentrated H2SO4 protonates both Cl- and I- to form HX (HCl or HI) and NaHSO4.
• HCl is stable and escapes as gas without further reaction.
• HI, however, reduces H2SO4 because I- is easily oxidized to I2.
• This reduction produces SO2 and H2S alongside HI and NaHSO4.

Why C is correct:

• Iodide ions are better reducing agents than chloride ions, as I- has lower standard reduction potential (E° for I2/2I- is +0.54 V vs. +1.36 V for Cl2/2Cl-), favoring oxidation of I- by H2SO4.

Why the others are wrong:

• A: No displacement of iodide by chloride occurs; reactions are separate.
• B: Volatility difference exists but does not explain extra products like SO2 and H2S.
• D: Dehydration is not primary; reduction by I- drives the unique products.

Final answer: C