Hydrogen peroxide in its reaction with KIO_4 | vedclass

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(A) In the reaction with $KIO_4$: $KIO_4 + H_2O_2 \rightarrow KIO_3 + H_2O + O_2$. Here,the oxidation state of $I$ in $KIO_4$ is $+7$ and in $KIO_3$ i

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reducing agent,oxidising agent

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(A) In the reaction with $KIO_4$: $KIO_4 + H_2O_2 \rightarrow KIO_3 + H_2O + O_2$. Here,the oxidation state of $I$ in $KIO_4$ is $+7$ and in $KIO_3$ is $+5$. Since $I$ is reduced,$H_2O_2$ acts as a reducing agent.In the reaction with $NH_2OH$: $2NH_2OH + H_2O_2 \rightarrow N_2 + 4H_2O$. Here,the oxidation state of $N$ in $NH_2OH$ is $-1$ and in $N_2$ is $0$. Since $N$ is oxidized,$H_2O_2$ acts as an oxidising agent.

Hydrogen peroxide in its reaction with $KIO_4$ and $NH_2OH$ respectively,is acting as a

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