The number of electrons involved in the reduction of one nitrate ion to hydrazine is :

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Solution

The correct option is D 7
Given, $N O_{3}^{-} ⟶ N_{2} H_{4}$
The balanced reaction is as
$2 N O_{3}^{-} + 16 H^{+} + 14 e^{-} \to N_{2} H_{4} + 6 H_{2} O$
$∴$ Two nitrate ions when reduced involved $= 14 e^{-}$
$∴$ One nitrate ion will involve $= \frac{14}{2} e^{-} = 7 e^{-}$

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