In the reaction with potassium periodate (KIO₄) and hydroxylamine (NH₂OH), hydrogen peroxide (H₂O₂) exhibits different roles :

(i) When hydrogen peroxide reacts with potassium periodate (KIO₄), it forms potassium iodate (KIO₃).

The balanced chemical reaction is :

$$ \mathrm{H}_2 \mathrm{O}_2 + \underset{\substack{\text{Potassium}\\ \text{periodate}}}{\mathrm{KIO}_4} \longrightarrow \underset{\begin{array}{c}\text{Potassium}\\ \text{iodate}\end{array}}{\mathrm{KIO}_3} + \mathrm{H}_2 \mathrm{O} + \mathrm{O}_2 $$

Oxidation state of oxygen in hydrogen peroxide :

(a) In H₂O₂ :

$$ \begin{aligned} & 2x + 2 = 0 \\\\ & x = -1 \end{aligned} $$

(b) In O₂ :

$$ x = 0 $$

The oxidation state of oxygen increases from -1 to 0, indicating that H₂O₂ is oxidized and acts as a reducing agent for KIO₄.

(iii) When hydrogen peroxide reacts with hydroxylamine (NH₂OH), it produces nitric acid (HNO₃) and water (H₂O).

$$ 3 \mathrm{H}_2 \mathrm{O}_2 + \mathrm{NH}_2 \mathrm{OH} \longrightarrow \mathrm{HNO}_3 + 4 \mathrm{H}_2 \mathrm{O} $$

Oxidation state of oxygen :

(a) In H₂O₂ :

$$ \begin{aligned} & 2x + 2 = 0 \\\\ & x = -1 \end{aligned} $$

(b) In H₂O :

$$ 2 + x = 0 $$

$$ x = -2 $$

The oxidation state of oxygen decreases from -1 to -2, meaning H₂O₂ is reduced and acts as an oxidizing agent for NH₂OH.