Question

The number of electrons required to balance the following equation is.
$N{O_3}^ - + 4{H^ + } + {e^ - }\xrightarrow{{}}2{H_2}O + NO$
A)$5$
B)$4$
C)$3$
D)$2$

Answer 1

Hint: We can define oxidation and oxidizing agent as,
In the oxidation process the electrons are lost from an atom. A compound that gains electrons during oxidation is known as oxidizing agent.
Reducing and Reducing agents are Reduction is the gain of electrons by an atom and a compound that loses electrons during reduction is called a reducing agent.

Complete step by step solution:
Now we discuss about the concept of redox reaction as,
Redox reactions:
We have remembered that all the redox reactions have two parts, they are reduced half and an oxidized half which occur at the same time. The reduced half of the reaction accepts electrons and the oxidation number of the species gets decreased, while the oxidized half of the reaction loses electrons and the oxidation number of the species increases. There is no net variation in the number of electrons in a redox reaction.
We can write the given unbalanced redox reaction as follows.
$N{O_3}^ - + 4{H^ + } + {e^ - } \to 2{H_2}O + NO$
The oxidation numbers of nitrogen in the reactant side is find out as follows,
$N + \left( { - 6} \right) = - 1$
$ \Rightarrow N = - 1 + 6$
On simplifying we get,
$ \Rightarrow N = + 5$
The oxidation numbers of nitrogen in the product side is find out as follows,
$N + \left( { - 2} \right) = 0$
$ \Rightarrow N = + 2$
Now, we can balance the charges by adding the appropriate number of electrons.
$N{O_3}^ - + 4{H^ + } + 3{e^ - } \to 2{H_2}O + NO$
It is seen that the three electrons are needed to balance the equation.
Therefore,option C is correct.

Note:
Let us see few rules for oxidation numbers,
-A free element will be zero as its oxidation number.
-Monatomic ions will have an oxidation number equal to charge of the ion.
-In hydrogen, the oxidation number is ${\text{ + 1}}$, when combined with elements having less electronegativity; the oxidation number of hydrogen is -1.
-In compounds of oxygen, the oxidation number of oxygen will be -2 and in peroxides it will be -1.
-Group 1 elements will have +1 oxidation number.
-Group 2 elements will have +2 oxidation numbers.
-Group 17 elements will have -1 oxidation number.
-Sum of oxidation numbers of all atoms in neutral compounds is zero.