Answer:
NO + H₂O → HNO₂ + 1 e- + 1 H⁺
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⁺