I. , H_2O_2 + O_3 rightarrow H_2O + 2O_2II. , | vedclass

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(C) In reaction $I$: The oxidation state of oxygen in $H_2O_2$ is $-1$. It changes to $0$ in $O_2$. Since the oxidation state increases,$H_2O_2$ acts

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reducing in $I$ and $II$

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(C) In reaction $I$: The oxidation state of oxygen in $H_2O_2$ is $-1$. It changes to $0$ in $O_2$. Since the oxidation state increases,$H_2O_2$ acts as a reducing agent.In reaction $II$: The oxidation state of oxygen in $H_2O_2$ is $-1$. It changes to $0$ in $O_2$. Since the oxidation state increases,$H_2O_2$ acts as a reducing agent.Therefore,$H_2O_2$ acts as a reducing agent in both reactions.

$I. \, H_2O_2 + O_3 \rightarrow H_2O + 2O_2$
$II. \, H_2O_2 + Ag_2O \rightarrow 2Ag + H_2O + O_2$
The role of hydrogen peroxide in the above reactions is respectively:

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$I. \, H_2O_2 + O_3 \rightarrow H_2O + 2O_2$$II. \, H_2O_2 + Ag_2O \rightarrow 2Ag + H_2O + O_2$The role of hydrogen peroxide in the above reactions is respectively: