Consider the following statements:Statement-I | vedclass

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(B) Statement-$I$ is correct: $H_2O_2$ has an oxidation state of $-1$ for oxygen. It can be reduced to $H_2O$ (oxidation state $-2$) or oxidized to $O

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Statement-$I$ is correct,but statement-$II$ is not correct

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(B) Statement-$I$ is correct: $H_2O_2$ has an oxidation state of $-1$ for oxygen. It can be reduced to $H_2O$ (oxidation state $-2$) or oxidized to $O_2$ (oxidation state $0$),allowing it to act as both an oxidizing and a reducing agent in acidic and basic media.Statement-$II$ is incorrect: $10 \ V \ H_2O_2$ means $1 \ mL$ of $H_2O_2$ solution gives $10 \ mL$ of $O_2$ at $STP$.The decomposition reaction is: $2H_2O_2 ightarrow 2H_2O + O_2$.$2 	imes 34 \ g$ of $H_2O_2$ produces $22400 \ mL$ of $O_2$ at $STP$.So,$22400 \ mL \ O_2$ is produced by $68 \ g \ H_2O_2$.$10 \ mL \ O_2$ is produced by $(68 	imes 10) / 22400 \approx 0.03036 \ g \ H_2O_2$.Since this is in $1 \ mL$ of solution,the concentration is $0.03036 \ g/mL$,which is $3.03\% \ (w/v)$. The statement claims $6\%$,which is incorrect.

Consider the following statements:
Statement-$I$: $H_2O_2$ acts as an oxidising as well as a reducing agent in both acidic and basic media.
Statement-$II$: $10 \ V \ H_2O_2$ sample means it contains $3.03\% \ (w/v) \ H_2O_2$.

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Consider the following statements:Statement-$I$: $H_2O_2$ acts as an oxidising as well as a reducing agent in both acidic and basic media.Statement-$II$: $10 \ V \ H_2O_2$ sample means it contains $3.03\% \ (w/v) \ H_2O_2$.