(A) HOCl + H_2O_2 rightarrow H_3O^{+} + Cl^{ | vedclass

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(C) In equation $(A)$: $HOCl + H_2O_2 \rightarrow H_3O^{+} + Cl^{-} + O_2$. The oxidation state of $Cl$ changes from $+1$ in $HOCl$ to $-1$ in $Cl^{-}

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$H_2O_2$ acts as a reducing agent in equations $(A)$ and $(B)$.

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(C) In equation $(A)$: $HOCl + H_2O_2 \rightarrow H_3O^{+} + Cl^{-} + O_2$. The oxidation state of $Cl$ changes from $+1$ in $HOCl$ to $-1$ in $Cl^{-}$. Thus,$HOCl$ is reduced and $H_2O_2$ acts as a reducing agent. In equation $(B)$: $I_2 + H_2O_2 + 2 OH^{-} \rightarrow 2 I^{-} + 2 H_2O + O_2$. The oxidation state of $I$ changes from $0$ in $I_2$ to $-1$ in $I^{-}$. Thus,$I_2$ is reduced and $H_2O_2$ acts as a reducing agent. Therefore,in both equations $(A)$ and $(B)$,$H_2O_2$ acts as a reducing agent.

$(A) \ HOCl + H_2O_2 \rightarrow H_3O^{+} + Cl^{-} + O_2$
$(B) \ I_2 + H_2O_2 + 2 OH^{-} \rightarrow 2 I^{-} + 2 H_2O + O_2$
Choose the correct option.

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$(A) \ HOCl + H_2O_2 \rightarrow H_3O^{+} + Cl^{-} + O_2$ $(B) \ I_2 + H_2O_2 + 2 OH^{-} \rightarrow 2 I^{-} + 2 H_2O + O_2$ Choose the correct option.