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Hydrogen peroxide in its reaction with `KIO_(4)` and `NH_(2)OH` respectively, is acting as aA. reducing agent,oxidising agentB. reducing agent, reducing agentC. oxidising agent,oxidising agentD. oxidising agent, reducing agent |
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Answer» Correct Answer - A `H_(2)O_(2)` reacts with `KIO_(4)` in the following manner: `overset(+7)(KIO_(4))+H_(2)O_(2)rArr overset(+5)(KIO_(3))+H_(2)O+O_(2)` on reaction of `KIO_(4)` with `H_(2)O_(2)`, oxidation state of `I` varies from `+7` to `+5` i.e., decreases. Thus `KIO_(4)` get reduced hence,`H_(2)O_(2)` is reducing agent here. `H_(2)O_(2)` reacts with `NH_(2)OH` in the following manner. `overset(-1)(NH_(2)OH)+H_(2)O_(2)rArr overset(+3)(N_(2)O_(3))+H_(2)O` In this reaction, oxidation state of `N` varies from `-1` to +3 i.e., increases, hence `H_(2)O_(2)` is acting on an oxidising agent here. |
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