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(B) The reduction half-reaction for the conversion of nitrate ion $(NO_{3}^{-})$ to hydrazine $(N_{2}H_{4})$ is determined by balancing the oxidation

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(B) The reduction half-reaction for the conversion of nitrate ion $(NO_{3}^{-})$ to hydrazine $(N_{2}H_{4})$ is determined by balancing the oxidation states.In $NO_{3}^{-}$,the oxidation state of $N$ is $+5$.In $N_{2}H_{4}$,the oxidation state of $N$ is $-2$.The change in oxidation state per nitrogen atom is $5 - (-2) = 7$.Therefore,for one mole of $NO_{3}^{-}$ to be reduced to half a mole of $N_{2}H_{4}$ (which contains one mole of $N$ atoms),$7$ moles of electrons are required.$NO_{3}^{-} + 7e^{-} + 6H^{+} \longrightarrow \frac{1}{2} N_{2}H_{4} + 3H_{2}O$.

How many moles of electrons are needed to convert one mole of nitrate ion to hydrazine?

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