Question

Give a balanced equation for reaction of nitrogen dioxide with water.

Answer 1

Hint: To answer this question, you must recall the reactions of nitrogen and its oxides. Nitrogen dioxide when reacted with water undergoes a disproportionation reaction.

Complete step by step solution:
- We know that nitrogen has the atomic number 7. It has 5 valence electrons. It can show a variety of oxidation numbers varying from $ - 3$ in compounds like ammonia to $ + 5$ in compounds like nitric acid.
- During a reaction, such elements having a great range of oxidation numbers possibly undergo changes in their oxidation number to form new compounds. We know that in nitrogen dioxide, nitrogen exists in an oxidation state of $ + 4$. Its oxidation number can either increase or decrease. Oxygen being a highly electronegative element does not show any changes in the oxidation state.
- Thus, in the reaction between nitrogen dioxide and water, the only species whose oxidation number changes is nitrogen. Thus, it undergoes a disproportionation reaction, i.e., it undergoes both oxidation as well as reduction. The products formed are nitric acid (oxidation number = $ + 5$) and nitrous acid (oxidation number = $ + 3$).
The reaction is given as: $2N{O_2} + {H_2}O \to HN{O_2} + HN{O_3}$

Additional information:
This reaction of nitrogen dioxide with water is known as the Ostwald’s process which is widely used for the industrial preparation of nitric acid.

Note: If there is an increase in the oxidation number of an element, the element is said to be oxidized and if there is a decrease in the oxidation number, the element is said to be reduced. The sum of the oxidation states of all the atoms in a compound must be zero and equal to the charge in case of ions.