Question

$1$ mole of $HN{O}_3$ absorbs $8$ moles of electrons in a redox reaction. The reduced product of this reaction is:

• A. $1$ mole $N_2$
• B. $1$ mole $N_{2}O$
• C. $1$ mole $N{O}_2$
• D. $1$ mole $N{H}_3$

Answer 1

The correct option is D $1$ mole $N{H}_3$

Oxidation number of $N$ in $HN{O}_3$ is $+5$ and oxidation number of $N_2$ is $O,N_{2}O$ is $+1,N{O}_2$ is $+4$ and $N{H}_3$ is $-3$. So each change in oxidation number needs $1$ mole of electrons per mole of product.

(a) Needs $10$ moles ($5$ electrons per $N$, times $2$ to get $N_2$)

(b) Change in oxidation number is $4$, so $4$ electrons per $N$. Therefore $4\times 2$ for two $N$ atoms i.e $N_2$. Hence, $8$ electrons.

(c) Change in Oxidation state=$1$, so $1$ mole of electrons to change to $+4$

(d) Change in Oxidation state=$8$, so $8$ moles of electron.