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An equal volume of a reducing agent is titrated separately with 1 M KMnO_(4) in acid , neutral and alkaline media . The volumes of KMnO_(4)required are 20 mL in acid , 33.4 mL in neutral and 100 mL in alkaline media . Find out the oxidation state of manganesein eachreduction product . Givethe balanced equations , for all the three half reactions . Find out the volume of 1 M K_(2)Cr_(2)O_(7) consumed , if the same volume of the reducing agent is titrated in acid medium . |
Answer» Solution :Given that :  where `x_(1),x_(2) and x_(3)`are the oxidation STATES of Mn in the product in acidic , NEUTRAL and alkalinemedia respectively .Since equal volumes of the reducingagentis usedin each titration , ` :. ` ,.e of reducing agent = m.e of `KMnO_(4)` in acidic medium = m.e of `KMnO_(4)`in neutral medium = m.e of `KMnO_(4)`in ALKALINE medium or `1xx (7-x_(1)) xx 20 = 1 xx ( 7 - x_(2)) xx 33.4 ` ` = 1xx ( 7 - x_(3)) x 100` [ m.e = `N xx V` (mL) , N = M `xx` change in On] or ` (7-x_(1))/5 = (7-x_(2))/3 = (7-x_(3))/1 ` On inspection , we see that the equality EXISTS for `x_(1)= +2 , x_(2) = +4 and x_(3) = + 6 " as " x_(1),x_(2) and x_(3)` can never be greater than 7 . The balanced chemicalequations of all the three half reactions are `MnO_(4)^(-) +8H^(+) +5e to Mn^(2+) +4H_(2)O` (acidic medium ) `MnO_(4)^(-) 2H_(2)O+3e to MnO_(2)+4OH^(-)`(neutral medium ) `{:(MnO_(4)^(-)+e, to,MnO_(4)^(2-)),(,,+ 6):}`(alkaline medium) Further , in acidic medium , ` {:(Cr_(2)O_(7)^(2-) to ,2Cr^(3+),,"Change in ON =6"),(+12,+6,):}` ` :. ` normalityof `K_(2)Cr_(2)O_(7)` solution ` = 1 xx 6 =N ` Let the VOLUME of `K_(2)Cr_(2)O_(7)` solution be v mL . ` :. ` m.e of `K_(2)Cr_(2)O_(7)` = m.eof `KMnO_(4)` in acidic medium ` 6 xx v = 5 xx 20 ` ( normality of `KMnO_(4) ` = 5 N) ` v = 16.67 mL ` |
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