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$I) H_{2} O_{2} + {Cl}_{2} \to 2 HCl + O_{2}; II) H_{2} O_{2} + 2 NaOH \to 2 {Na}_{2} O_{2} + 2 H_{2} O; Role of H_{2} O_{2} in the above reactions is$

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Oxidizing agent in (I) and reducing agent in (II)

Reducing agent in (I) and oxidizing agent in (II)

Reducing agent in (I) and acid in (II)

Oxidizing in (I) and acid in (II)

answer is C.

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Detailed Solution

$H_{2} O_{2}$ acts as reducing agent in (I); It reduces ${Cl}_{2} to {Cl}^{-}$

$H_{2} O_{2}$ acts as an acid in (II) ; It reacts with sodium hydroxide to form a salt sodium peroxide ;

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