(A) $\mathrm{HOCl}+\mathrm{H}_{2} \mathrm{O}_{2} \rightarrow \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{Cl}^{-}+\mathrm{O}_{2}$
(B) $\mathrm{I}_{2}+\mathrm{H}_{2} \mathrm{O}_{2}+2 \mathrm{OH}^{-} \rightarrow 2 \mathrm{I}^{-}+2 \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}$
(A) $\mathrm{HOCl}+\mathrm{H}_{2} \mathrm{O}_{2} \rightarrow \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{Cl}^{-}+\mathrm{O}_{2}$
In this equation, $\mathrm{H}_{2} \mathrm{O}_{2}$ is reducing chlorine from $+1$ to $-1$.
(B) $\mathrm{I}_{2}+\mathrm{H}_{2} \mathrm{O}_{2}+2 \mathrm{OH}^{-} \rightarrow 2 \mathrm{I}^{-}+2 \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}$
In this equation, $\mathrm{H}_{2} \mathrm{O}_{2}$ is reducing iodine from 0 to $-1$.
Correct Option: , 3
In (A) reduction of HOCl occurs so it will be a oxidising agent hence $\mathrm{H}_{2} \mathrm{O}_{2}$ will be a reducing agent.
In(B) reduction of $\mathrm{I}_{2}$ occurs so it will be a oxidising agent and $\mathrm{H}_{2} \mathrm{O}_{2}$ will be a reducing agent.
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