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Chapter 22: Problem 57

Write complete balanced half-reactions for (a) oxidation of nitrous acid to nitrate ion in acidic solution, (b) oxidation of \(\mathrm{N}_{2}\) to \(\mathrm{N}_{2} \mathrm{O}\) in acidic solution.

Short Answer

Expert verified
The complete balanced half-reactions for (a) oxidation of nitrous acid to nitrate ion in acidic solution, and (b) oxidation of \(\mathrm{N}_{2}\) to \(\mathrm{N}_{2} \mathrm{O}\) in acidic solution are: a) HNO2 + 2H+ + 2e⟶ NO3- b) \(\mathrm{N}_{2}\) + H2O ⟶ \(\mathrm{N}_{2} \mathrm{O}\) + 2H+ + 2e-

Step by step solution

01

Identify the oxidation states of elements in nitrous acid and nitrate ion

Nitrous acid (HNO2) has oxygen with oxidation state -2, hydrogen +1, and nitrogen +3. In nitrate ion (NO3-), oxygen has -2, and nitrogen has +5.
02

Balance the half-reactions

The nitrogen in nitrous acid is oxidized from +3 to +5. The increase in oxidation state is by 2 units: HNO2 ⟶ NO3-
03

Balance the charges using protons (H+) and electrons

Add two H+ ions and two electrons to the reaction to balance the charge: HNO2 + 2H+ + 2e⟶ NO3- So, the complete balanced half-reaction is: HNO2 + 2H+ + 2e⟶ NO3- b) Oxidation of \(\mathrm{N}_{2}\) to \(\mathrm{N}_{2} \mathrm{O}\) in acidic solution:
04

Identify the oxidation states of elements in \(\mathrm{N}_{2}\) and \(\mathrm{N}_{2} \mathrm{O}\)

In \(\mathrm{N}_{2}\), the oxidation state of both nitrogens is 0. In \(\mathrm{N}_{2} \mathrm{O}\), the oxidation state of the nitrogen attached to oxygen is +2 and that of the other nitrogen is 0.
05

Balance the half-reactions

One of the nitrogens in \(\mathrm{N}_{2}\) is oxidized from 0 to +2. The increase in oxidation state is by 2 units: \(\mathrm{N}_{2}\) ⟶ \(\mathrm{N}_{2} \mathrm{O}\)
06

Balance the charges using protons (H+) and electrons

Add one water (H2O) on the reactants side to balance the O atoms and 2H+ and 2e- on the products side: \(\mathrm{N}_{2}\) + H2O ⟶ \(\mathrm{N}_{2} \mathrm{O}\) + 2H+ + 2e- So, the complete balanced half-reaction is: \(\mathrm{N}_{2}\) + H2O ⟶ \(\mathrm{N}_{2} \mathrm{O}\) + 2H+ + 2e-

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