Question

A solution of a salt with concentrated $H_{2}S{O}_4$ acid produced violet colour vapours which turns starch paste blue. The salt may be:

• A. Chloride
• B. Nitrate
• C. Bromide
• D. Iodide

Answer 1

Answer: (D)

Iodide

Iodide is a stronger reducing agent than bromide, and it is oxidized to iodine by the sulfuric acid:

$2I^{-}\to I_2+2e^{-}$

The reduction of the sulfuric acid is more complicated than with bromide. Iodide is powerful enough to reduce it in three steps:

sulfuric acid to sulfur dioxide (sulfur oxidation state = +4)

sulfur dioxide to elemental sulfur (oxidation state = 0)

sulfur to hydrogen sulfide (sulfur oxidation state = -2).

The most abundant product is hydrogen sulfide. The half-equation for its formation is as follows:

$H_{2}S{O}_4+8H^{+}+8e\to H_{2}S+4H_{2}O$

Combining these two half-equations gives the following net ionic equation:

$H_{2}S{O}_4+8H^{+}+8I^{-}\to 4I_2+H_{2}S+4H_{2}O$

This is confirmed by a trace of steamy fumes of hydrogen iodide, and a large amount of iodine. The reaction is exothermic: purple iodine vapor is formed, with dark gray solid iodine condensing around the top of the reaction vessel. There is also a red color where the iodine comes into contact with solid iodide salts. The red color is due to the $I^{-3}$ ion formed by reaction between$I_2$ molecules and $I^{-}$ ions. Hydrogen sulfide gas can be detected by its "rotten egg" smell, but this gas is intensely poisonous.

Hence, the Correct option is $D$