Question

In a reaction 4 moles of electrons are transferred to one mole of ${HNO}_3$ when acted as an oxidant. The possible reduction product is?

• A. (1/2) mole of $N_2$
• B. (1/2) mole of $N_{2}O$
• C. 1 mole of ${NO}_2$
• D. 1 mole of $N{H}_3$

Answer 1

Answer: (B)

(1/2) mole of $N_{2}O$

$\left(1/2\right)$ mole of $N_{2}O$

Oxidation state of nitrogen in $HN{O}_3$

Let the oxidation state of nitrogen in $HN{O}_3$ be x.

$\therefore 1+x+(3\times -2)=0$

$\Rightarrow x-5=0$

$\Rightarrow x=5$

Hence, Oxidation state of nitrogen in $HN{O}_3=+5$

As given that 4 moles of electrons are transferred to one mole of $HN{O}_3$ i.e., the oxidation no. of nitrogen in reduction product becomes +1.

Hence, the possible reduction product is $\left(1/2\right)$ mole of $N_{2}O$ as the oxidation no. of nitrogen is +1 in $N_{2}O$.