Write a balanced half-reaction for the oxidation of gaseous nitric oxide (NO) to aqueous nitrous acid (HNO2) in acidic aqueous solution.

NO + H₂O → HNO₂ + 1 e- + 1 H⁺

Step-by-step explanation:

NO ⇒ oxidation number of N = +2

HNO₂ ⇒ oxidation number of N= +3

Therefore, NO has to lose 1 electron to be oxidized to HNO₂. We write the half-reaction with 1 electron (1 e-) in the products side.

NO → HNO₂ + 1 e-

Now, we have 0 electrical charges in the reactants side, and a total of -1 electrical charge in the products side. As the reaction is in acidic aqueous solution, we have to add H⁺ ions to balance the charges. We perform the balance by adding 1 H⁺ (positive charge) to neutralize the negative charge in the side of the products:

NO → HNO₂ + 1 e- + 1 H⁺

Now, we perform the mass balance. We have:

N: 1 atom in both sides

O: 1 atom in reactants side and 2 atoms in products side

H: 0 atoms in reactants side, 2 atoms in products side.

Thus, we have to add 1 H₂O molecule to the reactants side to equal the masses:

NO + H₂O → HNO₂ + 1 e- + 1 H⁺

Finally, the oxidation half-reaction is:

NO + H₂O → HNO₂ + 1 e- + 1 H⁺

every chemical equation adheres to the law of conservation of mass, which states that matter cannot be created or destroyed. therefore, there must be the same number of atoms of each element on each side of a chemical equation.