The number of moles of `KMnO_(4)` reduced by `1 "mol of" KI` in alkaline medium isA. oneB. twoC. fiveD. one-fifthe

The number of moles of `KMnO_(4)` reduced by `1 "mol of" KI` in alkali

1.	The number of moles of `KMnO_(4)` reduced by `1 "mol of" KI` in alkaline medium isA. oneB. twoC. fiveD. one-fifthe
Answer» Correct Answer - B In alkaline medium, ` KMnO_4` oxidixes iodide to iodate and gets reduced to manganese dioxide : `overset(+7)(M)nO_(4)^(-) rarr overset(+4)(M)n O_(2)` `overset(-1)(I^(-)) rarr overset(+5)(I)O_(3)^(-)` Thus, 1 mol of `KMnO_4` contains `3` equivalents (because its oxidation number decreases nu `3` unts : ` +7 rarr +4`). On the other hand, 1 mol of `KI` contains `6` equivalents (because oxidation number of iodine increases ny `6` units, ` -1 rarr +5`) According to the law of equivalents, we need `6` equivalents (2 mol) of `KMnO_4` to react with `6` requivalents (1 mol) of ` KI`. Alternatively, we can get the answer by means fo chemical equations : `{:(2KMnO(4)+H_(2)O rarr 2KOH+2MnO_(2)+3[O]),(KI+3[O] rarr KIO_(3)),(bar(2KMnO_(4) +KI +H_(2)O rarr 2KOH +2MnO_(2) +KIO_(3))):}`

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The number of moles of `KMnO_(4)` reduced by `1 "mol of" KI` in alkaline medium isA. oneB. twoC. fiveD. one-fifthe

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