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

Question

The number of electrons involved in the reduction of one nitrate ion to hydrazine is (A) 0 (B) 2 (C) 6 (D) 5

Tardigrade Admin · Accepted Answer

Oxidation state of N in NO-3 is +5 Oxidation state of N in N2H4 is -2 So number of electrons involved in the reduction of one nitrate ion to hydrazine is 6

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

$0$

$2$

$6$

$5$

Oxidation state of $N$ in $N O_{3}^{-}$ is $+ 5$ Oxidation state of $N$ in $N_{2} H_{4}$ is $- 2$ So number of electrons involved in the reduction of one nitrate ion to hydrazine is $6$

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

0

2

6

5

Oxidation state of N in NO3−​ is +5 Oxidation state of N in N2​H4​ is −2 So number of electrons involved in the reduction of one nitrate ion to hydrazine is 6

$0$

$2$

$6$

$5$

Oxidation state of $N$ in $N O_{3}^{-}$ is $+ 5$ Oxidation state of $N$ in $N_{2} H_{4}$ is $- 2$ So number of electrons involved in the reduction of one nitrate ion to hydrazine is $6$