Question

A solution of $S{O}_2$ in water reacts with $H_{2}S$ precipitating sulphur. Here $S{O}_2$ acts as:

• A. Oxidising agent
• B. Reducing agent
• C. Acid
• D. Catalyst

Answer 1

Answer: (A)

Oxidising agent

$8\stackrel{+4}{S}\stackrel{-2}{O_2}\left(aq\right)+16\stackrel{+1}{H_2}\stackrel{-2}{S}\left(aq\right)⟶3\stackrel{0}{S_8}\downarrow +16\stackrel{-1}{H_2}\stackrel{-2}{O}\left(l\right)$

The above reaction is a disproportion reaction. We're dealing with a disproportionation reaction, which a redox reaction in which the same chemical species is both oxidized and reduced.

Here $H_{2}S$ acts oxidising agent.

Hence option A is correct.