Question

Identify the substance oxidized, the substance reduced, the oxidizing agent, and the reducing agent?

Answer 1

When oxidation takes place then oxidation state of central atom increases while in$2HN{O_3}{\text{ }} + {\text{ }}3{H_3}As{O_3}{\text{ }} \to {\text{ }}2NO{\text{ }} + {\text{ }}3{H_3}As{O_4}{\text{ }} + {\text{ }}{H_2}O$ case of reduction the oxidation state of central atom decreases. Thus we will find the oxidation of each compound in the given reaction. The substance which gets oxidized will behave as a reducing agent while the substance which gets reduced will behave as an oxidizing agent.

Complete Answer:
The reaction is given as:
$2HN{O_3}{\text{ }} + {\text{ }}3{H_3}As{O_3}{\text{ }} \to {\text{ }}2NO{\text{ }} + {\text{ }}3{H_3}As{O_4}{\text{ }} + {\text{ }}{H_2}O$
Since the reaction must be a redox reaction because we know that oxidation and reduction takes place simultaneously only in redox reaction. Since we know that when the oxidation state of an atom increases then it is said to be oxidized while if its oxidation state decreases then it is said to be reduced.
Now we will find the oxidation state of various elements from the above reaction as:
$(i){\text{ }}HN{O_3}$:
Let the oxidation state of nitrogen be x then it can be found as:
$\Rightarrow {\text{ }} + 1{\text{ }} + {\text{ }}x{\text{ }} + {\text{ }}\left( { - 2 \times 3} \right){\text{ }} = {\text{ }}0$
$\Rightarrow {\text{ }} + 1{\text{ }} + {\text{ }}x{\text{ - 6 }} = {\text{ }}0$
$\Rightarrow {\text{ }}x{\text{ }} = {\text{ + 5}}$
The oxidation state of nitrogen here is ${\text{ + 5}}$.
$(ii){\text{ NO}}$:
Let the oxidation state of nitrogen be x then it can be found as:
$\Rightarrow {\text{ }}x{\text{ }} + {\text{ }}\left( { - 2} \right){\text{ }} = {\text{ }}0$
$\Rightarrow {\text{ }}x{\text{ - 2 }} = {\text{ }}0$
$\Rightarrow {\text{ }}x{\text{ }} = {\text{ + 2}}$
The oxidation state of nitrogen here is${\text{ + 2}}$.
Hence the oxidation state of nitrogen decreases from ${\text{ + 5}}$ to ${\text{ + 2}}$. Hence we can say that reduction of nitrogen takes place. Thus $HN{O_3}$ is reduced to ${\text{NO}}$.
$(iii){\text{ }}{H_3}As{O_3}$:
Let the oxidation state of arsenic be x then it can be found as:
$\Rightarrow {\text{ }}\left( { + 1 \times 3} \right){\text{ + }}x{\text{ }} + {\text{ }}\left( { - 2 \times 3} \right){\text{ }} = {\text{ }}0$
$\Rightarrow {\text{ + 3 + }}x{\text{ - 6 }} = {\text{ }}0$
$\Rightarrow {\text{ }}x{\text{ }} = {\text{ + 3}}$
The oxidation state of arsenic here is${\text{ + 3}}$.
$(iv){\text{ }}{H_3}As{O_4}$:
Let the oxidation state of arsenic be x then it can be found as:
$\Rightarrow {\text{ }}\left( { + 1 \times 3} \right){\text{ + }}x{\text{ }} + {\text{ }}\left( { - 2 \times 4} \right){\text{ }} = {\text{ }}0$
$\Rightarrow {\text{ + 3 + }}x{\text{ - 8 }} = {\text{ }}0$
$\Rightarrow {\text{ }}x{\text{ }} = {\text{ + 5}}$
The oxidation state of arsenic here is ${\text{ + 5}}$. Hence the oxidation of arsenic increases from ${\text{ + 3}}$ to ${\text{ + 5}}$ , therefore we can say that ${H_3}As{O_3}$ is being oxidized to ${H_3}As{O_4}$.
We know that if a substance gets oxidized then it will act as a reducing agent for other substances. Therefore ${H_3}As{O_3}$ will behave as a reducing agent and $HN{O_3}$ will behave as an oxidizing agent.

Note:
Oxidation is also defined as the addition of oxygen to a compound which can be seen in conversion of ${H_3}As{O_3}$ to ${H_3}As{O_4}$. Reduction can also be defined as removal of oxygen which can be clearly seen in conversion from $HN{O_3}$ to ${\text{NO}}$.