Question

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

The balanced half-reactions are as follows: (a) Oxidation of nitrous acid to nitrate ion in acidic solution: \[ \mathrm{HNO_{2}} + \mathrm{H_{2}O} \rightarrow \mathrm{NO_{3}^{-}} + 2e^{-} + 2\mathrm{H^{+}} \] (b) Oxidation of \(\mathrm{N}_{2}\) to \(\mathrm{N}_{2}\mathrm{O}\) in acidic solution: \[ \mathrm{N_{2}} + \mathrm{H_{2}O} \rightarrow \mathrm{N_{2}O} + 2e^{-} + 2\mathrm{H^{+}} \]

Step-by-step solution

Reaction (a) Nitrous acid to nitrate ion in acidic solution

Step 1: Write the unbalanced half-reaction Nitrous acid (\(\mathrm{HNO_{2}}\)) is converted into nitrate ion (\(\mathrm{NO_{3}^{-}}\)). \[ \mathrm{HNO_{2}} \rightarrow \mathrm{NO_{3}^{-}} \] Step 2: Find the change of oxidation state The oxidation state for nitrogen in nitrous acid is +3, while it is +5 in nitrate ion. The nitrogen gains 2 electrons in the process (it's being oxidized). \[ \mathrm{HNO_{2}} \rightarrow \mathrm{NO_{3}^{-}} + 2e^{-} \] Step 3: Balance the other atoms by adding water and protons \(\mathrm{H^{+}}\) To balance the oxygen atoms, we can add one water molecule to the left side. \[ \mathrm{HNO_{2}} + \mathrm{H_{2}O} \rightarrow \mathrm{NO_{3}^{-}} + 2e^{-} \] Now, we balance the hydrogen atoms by adding 2 protons \(\mathrm{H^{+}}\) to the right side. \[ \mathrm{HNO_{2}} + \mathrm{H_{2}O} \rightarrow \mathrm{NO_{3}^{-}} + 2e^{-} + 2\mathrm{H^{+}} \] The balanced half-reaction for the oxidation of nitrous acid is: \[ \mathrm{HNO_{2}} + \mathrm{H_{2}O} \rightarrow \mathrm{NO_{3}^{-}} + 2e^{-} + 2\mathrm{H^{+}} \]

Reaction (b) Oxidation of \(\mathrm{N}_{2}\) to \(\mathrm{N}_{2}$$\mathrm{O}\) in an acidic solution

Step 1: Write the unbalanced half-reaction Dinitrogen (\(\mathrm{N_{2}}\)) is converted into nitrous oxide (\(\mathrm{N_{2}O}\)). \[ \mathrm{N_{2}} \rightarrow \mathrm{N_{2}O} \] Step 2: Find the change of oxidation state The oxidation state of nitrogen in dinitrogen is 0, while it is +1 in nitrous oxide. Each nitrogen gains 1 electron in this process (they're being oxidized). \[ \mathrm{N_{2}} \rightarrow \mathrm{N_{2}O} + 2e^{-} \] Step 3: Balance the other atoms by adding water and protons \(\mathrm{H^{+}}\) To balance the oxygen atom, we can add one water molecule to the left side. \[ \mathrm{N_{2}} + \mathrm{H_{2}O} \rightarrow \mathrm{N_{2}O} + 2e^{-} \] Now, we balance the hydrogen atoms by adding 2 protons \(\mathrm{H^{+}}\) to the right side. \[ \mathrm{N_{2}} + \mathrm{H_{2}O} \rightarrow \mathrm{N_{2}O} + 2e^{-} + 2\mathrm{H^{+}} \] The balanced half-reaction for the oxidation of dinitrogen is: \[ \mathrm{N_{2}} + \mathrm{H_{2}O} \rightarrow \mathrm{N_{2}O} + 2e^{-} + 2\mathrm{H^{+}} \]