An aqueous solution of hydrazine ($$N_2H_4$$) is electrochemically oxidized by $$O_2$$, thereby releasing chemical energy in the form of electrical energy. One of the products generated from the electrochemical reaction is $$N_2(g)$$.

Choose the correct statement(s) about the above process

The cell operates in alkaline medium, so all half-reactions are written with $$OH^-$$ or $$H_2O$$.

Case 1: Oxidation half-reaction (anode)
Hydrazine is oxidised to nitrogen gas. Balance it step-by-step:

1. Write the skeletal form: $$N_2H_4 \rightarrow N_2$$.

2. Balance nitrogen atoms (already equal).
3. To balance hydrogen, add 4 molecules of water on the product side: $$N_2H_4 \rightarrow N_2 + 4H_2O$$.
4. Balance oxygen by adding 4 $$OH^-$$ to the reactant side: $$N_2H_4 + 4OH^- \rightarrow N_2 + 4H_2O$$.
5. Finally, balance charge by adding 4 electrons on the product side: $$N_2H_4 + 4OH^- \rightarrow N_2 + 4H_2O + 4e^-$$.

This shows that $$OH^-$$ reacts with $$N_2H_4$$ at the anode to give $$N_2(g)$$ and water while releasing 4 electrons. Hence, Option A is correct.

Case 2: Reduction half-reaction (cathode)
Molecular oxygen gets reduced to hydroxide ions.

Write and balance in alkaline medium:
$$O_2 + 2H_2O + 4e^- \rightarrow 4OH^-$$.

Thus at the cathode $$O_2$$ is converted into $$OH^-$$. Hence, Option C is correct.

Why Options B and D are incorrect

B. The cathode reaction does not involve decomposition of $$N_2H_4$$; it involves the reduction of $$O_2$$. Therefore the statement about nascent hydrogen reacting with oxygen is wrong.

D. The overall cell reaction obtained by adding the two balanced half-reactions is
$$N_2H_4 + O_2 \rightarrow N_2 + 2H_2O$$.
No oxides of nitrogen (NO, $$NO_2$$, etc.) appear; hence they are not major by-products.

Therefore the correct statements are:
Option A and Option C.