Redox and Equivalent ConceptMedium
Question
The number of electrons involved in the reduction of nitrate ion to hydrazine is
Options
A.8
B.7
C.5
D.3
Solution
$2NO_{3}^{-} + 16H^{+} + 14e^{-} \rightarrow N_{2}H_{4} + 6H_{2}O$
∴ Number of e− per NO3− ion =$\frac{14}{2} = 7$.
Method II: Oxidation state of N changes from +5 to –2 and hence, there is a gain of 7e− per NO3− ion.
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