Hydrogen peroxide (H₂O₂) can act as both an oxidizing agent and a reducing agent because the oxygen atom in H₂O₂ has an oxidation state of -1, which can either decrease to -2 (reduction, acting as oxidizing agent) or increase to 0 (oxidation, acting as reducing agent). Let's analyze its behavior in the given reactions:

Reaction with KIO₄ (Potassium periodate):

KIO₄ is a strong oxidizing agent. In the reaction with H₂O₂, periodate (IO₄^-) is reduced to iodate (IO₃^-). The half-reaction is:

$I{O}_4{}^{-}+2H^{+}+2e^{-}\to I{O}_3{}^{-}+H_{2}O$

H₂O₂ provides the electrons for this reduction. The oxidation half-reaction for H₂O₂ is:

$H_2{O}_2\to O_2+2H^{+}+2e^{-}$

Here, the oxidation state of oxygen increases from -1 in H₂O₂ to 0 in O₂. Since H₂O₂ is losing electrons, it acts as a reducing agent.

Reaction with NH₂OH (Hydroxylamine):

NH₂OH can act as a reducing agent. In the reaction with H₂O₂, hydroxylamine is oxidized. The half-reaction for oxidation of NH₂OH is:

$2N{H}_{2}OH\to N_{2}O+H_{2}O+4H^{+}+4e^{-}$

H₂O₂ accepts these electrons and is reduced. The reduction half-reaction for H₂O₂ is:

$H_2{O}_2+2H^{+}+2e^{-}\to 2H_{2}O$

Here, the oxidation state of oxygen decreases from -1 in H₂O₂ to -2 in H₂O. Since H₂O₂ is gaining electrons, it acts as an oxidizing agent.

Conclusion:

In the reaction with KIO₄, H₂O₂ acts as a reducing agent, and in the reaction with NH₂OH, it acts as an oxidizing agent. Therefore, the correct option is:

reducing agent, oxidising agent

Related Topics & Formulae:

Oxidation-Reduction (Redox) Reactions: Reactions involving transfer of electrons. The species that loses electrons is oxidized (reducing agent), and the species that gains electrons is reduced (oxidizing agent).

Hydrogen Peroxide: A versatile molecule that can act as both oxidizing and reducing agent due to the intermediate oxidation state (-1) of oxygen.

Half-Reaction Method: Useful for balancing redox reactions by separating the oxidation and reduction processes.