Question

In a reaction, 4 moles of electrons are transferred to 1 mole of $HN{O_3}$ when acted as an oxidant. The possible reduction product is:
(A) $\dfrac{1}{2}$ mole of ${N_2}$
(B) $\dfrac{1}{2}$ mole of ${N_2}O$
(C) 1 mole of $N{O_2}$
(D) 1 mole of $N{H_3}$

Answer 1

To find the oxidation number of an atom in $HN{O_3}$, the formula is:
Overall charge on $HN{O_3}$ = Oxidation number of H + Oxidation number of N + 3(Oxidation number of O)

Complete step by step solution:
Here, we can see that reduction of $HN{O_3}$ is done by some reducing agent. So, here $HN{O_3}$ oxidizes the other compound and so that $HN{O_3}$ is called an oxidizing agent here.
-We are given that 4 moles of electrons are transferred to 1 mole of $HN{O_3}$ in the process of reduction. So, let’s first find the oxidation number of N in $HN{O_3}$.
We know that for any compound,
Overall charge = Sum of oxidation number of all atoms
Overall charge on $HN{O_3}$ = Oxidation number of H + Oxidation number of N + 3(Oxidation number of O)
0 = 1 + Oxidation number of N + 3(-2)
0 = 1 + Oxidation number of N – 6
Oxidation number of N = 6-1 = +5
Here, we are given that four electrons are consumed by one mole of $HN{O_3}$ during reduction.
So, if four moles of electrons are consumed by a nitrogen atom, then the resulting product will have nitrogen in the +1 oxidation state. This can be explained by the following reaction.
${N^{ + 5}} + 4{e^ - } \to {N^{ + 1}}$
Let’s find the oxidation number of nitrogen in given products.
For ${N_2}$:
Overall charge = 2(Oxidation number of N)
0 = 2(Oxidation number of N)
Oxidation number of N = 0
For ${N_2}O$ :
Overall charge = 2(Oxidation number of N) + Oxidation number of O
0 = 2(Oxidation number of N) + (-2)
Oxidation number of N = $\dfrac{{ + 2}}{2} = + 1$
For $N{O_2}$ :
Overall charge = Oxidation number of N + 2(Oxidation number of O)
0 = Oxidation number of N + 2(-2)
Oxidation number of N = +4
For $N{H_3}$ :
Overall charge = Oxidation number of N + 3(Oxidation number of H)
0 = Oxidation number of N + 3(+1)
Oxidation number of N = -3
So, a half mole of ${N_2}O$ should be a resultant product of this reaction which can be given by:
$HN{O_3} + 4{e^ - } \to \dfrac{1}{2}{N_2}O$

Therefore, the correct answer is (B).

Note: Note that here we will not obtain 1 mole of ${N_2}O$ because there is only one nitrogen atom present in the $HN{O_3}$. So, one mole of $HN{O_3}$ will only give $\dfrac{1}{2}$ mole of ${N_2}O$ in the reduction reaction.