Question

In which of the following reactions, $H_2{O}_2$ acts as a reducing reagent ?

• A. $HOCl+H_2{O}_2\stackrel{H^{+}}{\to }{H}_3{O}^{+}+C{l}^{-}+O_2$
• B. $M{n}^{2+}+H_2{O}_2\stackrel{O{H}^{-}}{\to }M{n}^{4+}+2O{H}^{-}$
• C. $2F{e}^{2+}+H_2{O}_2\stackrel{O{H}^{-}}{\to }2F{e}^{3+}+2O{H}^{-}$
• D. $PbS\left(s\right)+4H_2{O}_2\left(aq\right)\stackrel{H^{+}}{\to }PbS{O}_4\left(s\right)+4H_{2}O\left(l\right)$

Answer 1

The correct option is A $HOCl+H_2{O}_2\stackrel{H^{+}}{\to }{H}_3{O}^{+}+C{l}^{-}+O_2$

$HOCl+H_2{O}_2\stackrel{H^{+}}{\to }{H}_3{O}^{+}+C{l}^{-}+O_2$

Here, the $H_2{O}_2$ act as the reducing agent, as the $Cl$ has $+1$ oxidation state of in $HOCl$ and after reaction with the $H_2{O}_2$ , in the presence of the acid, it reduces the $Cl$ in $HOCl$ to $C{l}^{-}$ having -1 oxidation state.

So, as $Cl$ oxidation state decreases from the +1 to -1 oxidation state, hence the $H_2{O}_2$ is acting as the reducing agent.