Answer:
[tex]2H_2O(l)+NO(g)\rightarrow NO_3^-(aq) +4H^+(aq)+3e^-[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out possible to set up the initial version of the half-reaction as follows:
[tex]NO\rightarrow NO_3^-[/tex]
Now, we assign the oxidation numbers to both nitrogen atoms:
[tex]N^{2+}O\rightarrow N^{5+}O_3^-[/tex]
It means three electrons are carried, two water molecules are needed on the left side in order to balance the oxygen atoms and consequently four hydronium ions to balance hydrogen atoms in acidic media:
[tex]2H_2O(l)+NO(g)\rightarrow NO_3^-(aq) +4H^+(aq)+3e^-[/tex]
Regards!
