The user wants me to identify the correct statements about the electrochemical oxidation of hydrazine (N₂H₄) by oxygen (O₂).

This process describes a hydrazine-oxygen fuel cell. In any electrochemical cell, oxidation occurs at the anode and reduction occurs at the cathode.

• Fuel (reductant): Hydrazine, $\mathrm{N}_2 \mathrm{H}_4$, will be oxidized at the anode.
• Oxidant: Oxygen, $\mathrm{O}_2$, will be reduced at the cathode.

The problem states the solution is aqueous and the options mention $\mathrm{OH}^-$, which implies the reaction occurs in a basic (alkaline) medium. Let's determine the half-reactions.

Step 1: Anode (Oxidation) Reaction

Hydrazine is oxidized to nitrogen gas. The oxidation state of Nitrogen in $\mathrm{N}_2 \mathrm{H}_4$ is -2, and in $\mathrm{N}_2$ it is 0. This is an oxidation process.

1. Write the basic species: $\mathrm{N}_2 \mathrm{H}_4 \rightarrow \mathrm{N}_2$
2. Balance atoms other than O and H: Nitrogen is already balanced (2 atoms on each side).
3. Balance O atoms by adding H₂O: There are no oxygen atoms in the initial species.
4. Balance H atoms by adding H₂O and OH⁻: There are 4 H atoms on the left. In a basic medium, we balance H by adding water to the side that needs H and hydroxide ions to the opposite side. To balance 4 H atoms, we need to produce 4 $\mathrm{H}_2\mathrm{O}$ on the right and add 4 $\mathrm{OH}^-$ on the left. $\mathrm{N}_2 \mathrm{H}_4 + 4\mathrm{OH}^- \rightarrow \mathrm{N}_2 + 4\mathrm{H}_2\mathrm{O}$
5. Balance the charge by adding electrons (e⁻): The left side has a total charge of -4. The right side has a charge of 0. We add 4 electrons to the right side to balance the charge. $\mathrm{N}_2 \mathrm{H}_4(\text{aq}) + 4\mathrm{OH}^-(\text{aq}) \rightarrow \mathrm{N}_2(\text{g}) + 4\mathrm{H}_2\mathrm{O}(\text{l}) + 4\mathrm{e}^-$ This is the balanced half-reaction at the anode.

Step 2: Cathode (Reduction) Reaction

Oxygen is reduced. The oxidation state of Oxygen in $\mathrm{O}_2$ is 0. In a basic solution, it is reduced to hydroxide ions, $\mathrm{OH}^-$, where its oxidation state is -2.

1. Write the basic species: $\mathrm{O}_2 \rightarrow \mathrm{OH}^-$
2. Balance atoms other than O and H: Not applicable.
3. Balance O atoms: There are 2 O atoms on the left and 1 on the right. We place a coefficient of 2 in front of $\mathrm{OH}^-$. $\mathrm{O}_2 \rightarrow 2\mathrm{OH}^-$
4. Balance H atoms: There are 2 H atoms on the right and none on the left. We add 2 $\mathrm{H}_2\mathrm{O}$ to the left to provide H atoms. Now we have 4 H on the left and 2 on the right, which is not balanced. Let's try balancing H with $\mathrm{H}_2\mathrm{O}$. To get 2 H on the left, we need one $\mathrm{H}_2\mathrm{O}$. Let's reconsider. Correct balancing involves adding $\mathrm{H}_2\mathrm{O}$ to balance O, then $\mathrm{H}^+$, then converting to basic. Alternative method: $\mathrm{O}_2 + 2\mathrm{H}_2\mathrm{O} \rightarrow 4\mathrm{OH}^-$. O is balanced (4 on both sides), H is balanced (4 on both sides).
5. Balance the charge: The left side has a charge of 0. The right side has a charge of -4. We add 4 electrons to the left side. $\mathrm{O}_2(\text{g}) + 2\mathrm{H}_2\mathrm{O}(\text{l}) + 4\mathrm{e}^- \rightarrow 4\mathrm{OH}^-(\text{aq})$ This is the balanced half-reaction at the cathode.

Step 3: Evaluate the options

• A: $\mathrm{OH}^{-}$ ions react with $\mathrm{N}_2 \mathrm{H}_4$ at the anode to form $\mathrm{N}_2(\mathrm{~g})$ and water, releasing 4 electrons to the anode. Our derived anode reaction is $\mathrm{N}_2 \mathrm{H}_4 + 4\mathrm{OH}^- \rightarrow \mathrm{N}_2 + 4\mathrm{H}_2\mathrm{O} + 4\mathrm{e}^-$. This statement accurately describes the anode half-reaction. Thus, statement A is correct.

• B: At the cathode, $\mathrm{N}_2 \mathrm{H}_4$ breaks to $\mathrm{N}_2(\mathrm{~g})$ and nascent hydrogen released at the electrode reacts with oxygen to form water. Hydrazine ($\"{N}_2 \mathrm{H}_4$) is the fuel and is oxidized at the anode, not the cathode. Oxygen is reduced at the cathode. Thus, statement B is incorrect.

• C: At the cathode, molecular oxygen gets converted to $\mathrm{OH}^{-}$. Our derived cathode reaction is $\mathrm{O}_2 + 2\mathrm{H}_2\mathrm{O} + 4\mathrm{e}^- \rightarrow 4\mathrm{OH}^-$. In this process, molecular oxygen ($\"{O}_2$) is consumed, and hydroxide ions ($"{OH}^-}) are produced. Thus, statement C is correct.

• D: Oxides of nitrogen are major by-products of the electrochemical process. The overall reaction is obtained by adding the anode and cathode half-reactions: $\mathrm{N}_2 \mathrm{H}_4 + 4\mathrm{OH}^- + \mathrm{O}_2 + 2\mathrm{H}_2\mathrm{O} + 4\mathrm{e}^- \rightarrow \mathrm{N}_2 + 4\mathrm{H}_2\mathrm{O} + 4\mathrm{e}^- + 4\mathrm{OH}^-$ Canceling common terms, we get: $\mathrm{N}_2 \mathrm{H}_4(\text{aq}) + \mathrm{O}_2(\text{g}) \rightarrow \mathrm{N}_2(\text{g}) + 2\mathrm{H}_2\mathrm{O}(\text{l})$ The products are nitrogen gas and water, which are environmentally benign. The formation of nitrogen oxides ($\"{NO}_x$) is not a major pathway in a controlled electrochemical process like a fuel cell. Thus, statement D is incorrect.

Conclusion: The correct statements are A and C.