sblock Elements 1 Question 25
25. Hydrogen peroxide in its reaction with $\mathrm{KIO}_4$ and $\mathrm{NH}_2 \mathrm{OH}$ respectively, is acting as a
(2014 Adv.)
(a) reducing agent, oxidising agent
(b) reducing agent, reducing agent
(c) oxidising agent, oxidising agent
(d) oxidising agent, reducing agent
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Answer:
Correct Answer: 25. (a)
Solution: This problem can be solved by using concept of oxidant and reductant.
Oxidant: Oxidant increases the oxidation number of the species with which it is reacted.
Reductant: Reductant decreases the oxidation number of the species with which it is reacted.
$\mathrm{H}_2 \mathrm{O}_2$ reacts with $\mathrm{KIO}_4$ in the following manner:
$$ \stackrel{+7}{\mathrm{KIO}_4}+\mathrm{H}_2 \mathrm{O}_2 \longrightarrow \stackrel{+5}{\mathrm{KIO}_3}+\mathrm{H}_2 \mathrm{O}+\mathrm{O}_2 $$
On reaction of $\mathrm{KIO}_4$ with $\mathrm{H}_2 \mathrm{O}_2$, oxidation state of I varies from +7 to +5 , i.e. decreases. Thus, $\mathrm{KIO}_4$ gets reduced hence, $\mathrm{H}_2 \mathrm{O}_2$ is a reducing agent here.
With $\mathrm{NH}_2 \mathrm{OH}$, it given following reaction:
$$ \stackrel{-1}{\mathrm{~N}} \mathrm{H}_2 \mathrm{OH}+\mathrm{H}_2 \mathrm{O}_2 \longrightarrow \stackrel{+3}{\mathrm{~N}_2} \mathrm{O}_3+\mathrm{H}_2 \mathrm{O} $$
In the above reaction, oxidation state of N varies from -1 to +3 . Here, oxidation number increases, hence $\mathrm{H}_2 \mathrm{O}_2$ is acting as an oxidising agent here.
Hence, (a) is the correct choice.
BETA
