H _ 2 O _ 2 acts as a reducing agent in

Here, the ${H}_{2}{O}_{2}$ act as the reducing agent, as the $Cl$ has +1 oxidation state of in $NaOCl$ and after reaction with the ${H}_{2}{O}_{2}$, in the presence of the acid, it reduces the $Cl$ in $NaOCl$ to ${Cl}^{-}$ 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.