Question

On addition of conc. ${{H}_{2}}S{{O}_{4}}$ to a chloride salt, colorless fumes are evolved but in case of iodide salt, violet fumes come out. This is because:
(A) ${{H}_{2}}S{{O}_{4}}$ reduces HI to ${{I}_{2}}$
(B) HI is a violet color
(C) HI gets oxidized to ${{I}_{2}}$
(D) HI changes to $HI{{O}_{3}}$

Answer 1

Hint: In terms of electrons transfer, removing electrons from molecules is known as oxidation, and adding electrons the same is reduction. When a reaction both oxidation and reduction take place is named as redox reactions. In the case of salt analysis at laboratory requirements, these redox reactions are more useful to identify the cation or anion in the given salt sample by changing the color solution or precipitate of color fumes.

Complete answer:
Step-1: Chloride salt when treated with a conc. of sulphuric acid (${{H}_{2}}S{{O}_{4}}$) gives colorless gas due to HCl gas. According to that, the reaction follows,
\[NaCl+{{H}_{2}}S{{O}_{4}}\to \underset{Colourless}{\mathop{HCl}}\,+N{{a}_{2}}S{{O}_{4}}\]

Step-2: Hydrogen iodide is an iodide salt which is a stronger reducing agent than sulphuric acid (${{H}_{2}}S{{O}_{4}}$). So ${{H}_{2}}S{{O}_{4}}$ to $S{{O}_{2}}$ and $HI$ to ${{I}_{2}}$ , due to the release of iodine gas violet fumes are observed. Hence, $HI$ oxidized to ${{I}_{2}}$ . the oxidation number of Iodine in $HI$ is -1 and in ${{I}_{2}}$ is 0 respectively, hence iodine gets oxidized from -1 to 0.
\[2NaI+{{H}_{2}}S{{O}_{4}}\to N{{a}_{2}}S{{O}_{4}}+2HI\]

\[2HI+{{H}_{2}}S{{O}_{4}}\to S{{O}_{2}}+\underset{Violet}{\mathop{{{I}_{2}}}}\,+{{2H}_{2}}O\]

Therefore the correct answer is HI gets oxidized to ${{I}_{2}}$- option C.

Note: In the salt analysis, conc. ${{H}_{2}}S{{O}_{4}}$ acts as the best reducing agent to identify the anions in the given salt sample. Regarding organic chemistry analysis, determine the unknown given compound structure conc. ${{H}_{2}}S{{O}_{4}}$ used as a dehydrated agent. In the volumetric analysis by using conc. ${{H}_{2}}S{{O}_{4}}$, easily prepared to dilute ${{H}_{2}}S{{O}_{4}}$ for volumetric titrations.