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Q.

$H_{2} O_{2}$ oxidises HCl to ${Cl}_{2}; H_{2} O_{2}$ reduces HOCl to chloride. The other products formed are respectively

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a

$(H_{2} O + O_{2}) and H_{2} O$

b

$H_{2} O and (H_{2} O + O_{2})$

c

$(H_{2} O + O_{2}) and (H_{2} O + {OH}^{-})$

d

$H_{2} O and (H_{2} O + O_{2})$

answer is D.

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Detailed Solution

Oxidation of HCl : - $2 H C l + H_{2} O \to C l_{2} + 2 H_{2} O$

Reduction of HOCl : - $H O C l + H_{2} O_{2} \to H C l + H_{2} O + O_{2}$

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