Question

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

• A. $Pb{S}_{\left(s\right)}+4H_2{O}_{2\left(aq\right)}\stackrel{H^{+}}{\to }PbS{O}_{4\left(s\right)}+H_{2}O\left(l\right)$
• B. $HOCl+H_2{O}_2\stackrel{H^{+}}{\to }{H}_3{O}^{+}+C{l}^{-1}+O_2$
• C. $M{n}^{2+}+H_2{O}_2\stackrel{O{H}^{-}}{\to }M{n}^{4+}+2O{H}^{-}$
• D. $2F{e}^{2+}+H_2{O}_2\stackrel{O{H}^{-}}{\to }2F{e}^{3+}+2O{H}^{-}$

Answer 1

The correct option is B $HOCl+H_2{O}_2\stackrel{H^{+}}{\to }{H}_3{O}^{+}+C{l}^{-1}+O_2$

solution:

$HOCl+H_2{O}_2\to H_3{O}^{+}+C{l}^{-}+O_2$

Here, the $H_2{O}_2$ act as the reducing agent, as the $\left(Cl\right)$ 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.

hence the correct option B