How many moles of electrons are needed to con | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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?

Similar Questions

The coefficients $x, y, p, q,$ and $r$ in the following balanced equation are respectively:
$xMnO_4^{2-}{_{\text{(aq)}}} + yH_2O_{\text{(l)}} \rightarrow pMnO_2{_{\text{(s)}}} + qMnO_4^{-}{_{\text{(aq)}}} + rOH^{-}{_{\text{(aq)}}}$

Which of the following is a redox reaction?

In a weak alkaline medium,iodide is changed into which of the following compounds by $KMnO_4$?

What are $x, y$ and $z$ respectively in the following reaction?
$xFe^{2+}{_{\text{(aq)}}} + yH^{+}{_{\text{(aq)}}} + zCr_2O_7^{2-}{_{\text{(aq)}}} \rightarrow \frac{y}{2}H_2O_{\text{(l)}} + xFe^{3+}{_{\text{(aq)}}} + 2zCr^{3+}{_{\text{(aq)}}}$

Justify that the reaction: $2 Na ( s ) + H_2 ( g ) \rightarrow 2 NaH ( s )$ is a redox change.

Vedclass Test Series

Exam Paper Generator

Online Exam Module