Question

In the reaction, $2H_{2}S+S{O}_2\to 3S+2H_{2}O$

• A. $H_{2}S$ has been oxidized
• B. $S{O}_2$ has been oxidized
• C. $H_{2}S$ has been reduced
• D. $S{O}_2$ has been reduced

Answer 1

Answer: (A), (D)

The correct options are
A $H_{2}S$ has been oxidized
D $S{O}_2$ has been reduced

${2H}_2\stackrel{-2}{S}+{\stackrel{+4}{SO}}_2⟶3\stackrel{0}{S}+2H_{2}O$

Here, $H_2\stackrel{-2}{S}$ is oxidized to $\stackrel{0}{S}$

and ${\stackrel{+4}{SO}}_2$ is reduced to $\stackrel{0}{S}$.

Therefore, $\left(A\right)$ and $\left(D\right)$ are correct answer.