Question

Consider the following reactions,
(I) $Zn+dil.H_{2}S{O}_4\to ZnS{O}_4+H_2$
(II) $Zn+conc.H_{2}S{O}_4\to ZnS{O}_4+S{O}_2+H_{2}O$
Oxidising agents in $I$ and $II$ are _______ respectively

• A. $H^{+},S{O}_4^{2-}$
• B. $S{O}_4^{2-},H^{+}$
• C. $H^{+},H^{+}$
• D. $S{O}_4^{2-},S{O}_4^{-2}$

Answer 1

The correct option is A $H^{+},S{O}_4^{2-}$

The oxidizing agent is the species that oxidizes the other species and itself gets reduced.

In reaction $\left(I\right)$ $H^{+}$ is reduced to $H_2,$

While in reaction $II$ $S{O}_4^{2-}$ is reduced to $S{O}_2$.

Hence, $H^{+}$ and $S{O}_4^{2-}$ is reduced to $H_2$ and $S{O}_2.$

Hence, $H^{+}$ and $S{O}_4^{2-}$ are oxidizing agents in $I$ and $II$ respectively.