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⁺

the answer is 75.0%.