Question

In the following redox reaction:
$Zn\left( s \right) + N{O_3}^ - \left( {aq} \right) + {H^ + }\left( {aq} \right) \to Z{n^{2 + }}\left( {aq} \right) + N{H_4}^ + \left( {aq} \right)$
$Zn\left( s \right)$ and $N{O_3}^ - \left( {aq} \right)$ respectively are,
A) Oxidant and reductant.
B) Reductant and oxidant.
C) Both oxidants.
D) Both reductant.

Answer 1

We can define Oxidation and Oxidizing agent as,
In the oxidation cycle the electrons are lost from a molecule. A compound that picks up electrons during oxidation is known as oxidizing agent.
Reducing and Reducing agent 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 answer:
Let us see what a redox reaction is.
Redox reactions:
All the redox reactions have two sections, they are decreased half and an oxidized half which happen simultaneously. 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 difference in the quantity of electrons in a redox reaction.
The reactions are,
$Zn\left( s \right) + N{O_3}^ - \left( {aq} \right) + {H^ + }\left( {aq} \right) \to Z{n^{2 + }}\left( {aq} \right) + N{H_4}^ + \left( {aq} \right)$
We can calculate the oxidation number of zinc as,
The oxidation number of zinc metal on the reactant side is zero and the oxidation number of zinc on the item side is two. Zinc loses electrons in the response and subsequently it goes about as a reducing agent.
The oxidation number of nitrogen in $N{O_3}^ -$ is,
$x - 6 = - 1$
$\Rightarrow x = 5$
We can calculate the oxidation number of nitrogen in $N{H_4}^ +$ is,
$x + 4 = + 1$
$\Rightarrow x = - 3$
The oxidation number of nitrogen in the product side is five and the oxidation number of nitrogen in the reactant side $- 3$. Nitrogen gains electrons in the reaction and hence it acts as an oxidizing agent.

So, the correct answer is Option B.

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.
The oxidation number of oxygen will be - 2 and in peroxides it will be - 1.