$(a) H_{2} O_{2} + O_{3} ⟶ H_{2} O + 2 O_{2} (b) H_{2} O_{2} + A g_{2} O ⟶ 2 A g + H_{2} O + O_{2}$
Role of hydrogen peroxide in the above reactions is respectively

Oxidizing in (a) and (b)

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Reducing in (a) and oxidizing in (b)

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Reducing in (a) and (b)

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Oxidizing in (a) and reducing in (b)

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Solution

The correct option is C Reducing in (a) and (b)
In the reaction,
$(a) H_{2} O_{2} + {0 O}_{3} ⟶ H_{2} O + 2 O_{2}$
Since, $H_{2} O_{2}$ reduces $O_{3}$ into $O^{2 -}$ ion, thus it behaves as a reducing agent. Further, in the reaction,
$(b) H_{2} O_{2} + A g_{2} O ⟶ 2 A g + H_{2} O + O_{2}$
Here, $H_{2} O_{2}$ reduces $A g_{2} O$ into metallic silver $[A g]$ (as oxidation number is reducing from +1 to 0). Thus, $H_{2} O_{2}$ behaves as a reducing agent.
Hence, option (c) is correct.

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Similar questions

(I) H_{2} O_{2} + O_{3} \to H_{2} O + 2 O_{2}

(I I) H_{2} O_{2} + A g_{2} O \to 2 A g + H_{2} O + O_{2}

H_{2} O_{2} + O_{3} \to H_{2} O + 2 O_{2}

H_{2} O_{2} + A g_{2} O \to 2 A g + H_{2} O + O_{2}

(I) H_{2} O_{2} + O_{3} \to H_{2} O + 2 O_{2}

(I I) H_{2} O_{2} + A g_{2} O \to 2 A g + H_{2} O + O_{2}

(i) H_{2} O_{2} + O_{3} \to H_{2} O + 2 O_{2} (i i) H_{2} O_{2} + A g_{2} O \to 2 A g + H_{2} O + O_{2}

H_{2} O_{2}

A g_{2} O + H_{2} O_{2} \to 2 A g + H_{2} O + O_{2}

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