Question

$\mathbf{8}\mathit{N}{\mathit{H}}_{\mathbf{3}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\mathbf{6}\mathit{N}{\mathit{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\to }\mathbf{7}{\mathit{N}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\mathbf{12}{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{\left(}\mathbf{l}\mathbf{\right)}$Identify the oxidizing agent and the reducing agent in the following reaction, and explain your answer:

Short answer

Answer: You need to identify the oxidizing agent and reducing agent of the given reaction.

Step-by-step solution

Oxidation state of nitrogen in NH3 

Oxidation state of Hydrogen attached to non-metal is +1.

Let, oxidation state of nitrogen in NH₃ is x.

No you can write,

$$\begin{gathered} x+3\left(+1\right)=0 \\ x=-3 \end{gathered}$$

Oxidation state of nitrogen in NO2 

Oxidation state of Oxygen attached to non-metal is -2.

Let, oxidation state of nitrogen in NO₂ is y.

No you can write,

$$\begin{gathered} y+2\left(-2\right)=0 \\ y=+4 \end{gathered}$$

Oxidation state of nitrogen in N2 

Oxidation state of nitrogen in N₂ is 0.

Conclusion

In the given reaction

$8N{H}_3\left(g\right)+6N{O}_2\left(g\right)\to 7N_2\left(g\right)+12H_{2}O\left(l\right)$

The oxidation state of nitrogen in NH₃ is -3 which increases to 0 in N₂. Therefore, NH₃ undergoes oxidation in this given reaction. Hence, it acts as a reducing agent.

The oxidation state of nitrogen in NO₂ is +4 which reduces to 0 in N₂. Therefore, NO₂ undergoes reduction in this given reaction. Hence, it acts as an oxidizing agent.