$\text{An aqueous solution of hydrazine}\left(N_2{H}_4\right)\text{is electrochemically oxidized by}{O}_2,\text{thereby releasing}$

$\text{chemical energy in the form of electrical energy.}\text{One of the products generated from the electrochemical}$

$\text{reaction is}{\mathrm{ N}}_2\left(g\right).$

Correct Answer:
The correct statements describing the electrochemical process are:
1. OH⁻ ions react with N₂H₄ at the anode to form N₂(g) and water, releasing 4 electrons to the anode.
2. At the cathode, molecular oxygen gets converted to OH⁻.
3. Oxides of nitrogen are major by-products of the electrochemical process.

Step-by-Step Explanation:

1. Understanding the Hydrazine-Oxygen Cell Process:
In a hydrazine-oxygen fuel cell, an aqueous solution of hydrazine ($N_2{H}_4$) acts as the fuel at the anode (where oxidation occurs), and molecular oxygen ($O_2$) acts as the oxidant at the cathode (where reduction occurs) in an alkaline medium.

2. Anode Reaction (Oxidation Half-Cell):
At the anode, hydrazine reacts with hydroxide ions (${\mathrm{OH}}^{-}$) present in the aqueous alkaline electrolyte. Hydrazine is oxidized to nitrogen gas ($N_2$) and water, releasing 4 electrons to the electrode.

The balanced oxidation half-reaction at the anode is:

$N_2{H}_4\left(aq\right)+4{\mathrm{OH}}^{-}\left(aq\right)\to N_2\left(g\right)+4H_{2}O\left(l\right)+4e^{-}$

This confirms that hydroxide ions react with hydrazine at the anode to produce nitrogen gas and water while releasing 4 electrons.

3. Cathode Reaction (Reduction Half-Cell):
At the cathode, oxygen gas accepts the electrons coming from the anode through an external circuit and combines with water molecules to form hydroxide ions (${\mathrm{OH}}^{-}$).

The balanced reduction half-reaction at the cathode is:

$O_2\left(g\right)+2H_2\mathrm{O}\left(l\right)+4e^{-}\to 4{\mathrm{OH}}^{-}\left(aq\right)$

This confirms that molecular oxygen gets converted to ${\mathrm{OH}}^{-}$ at the cathode.

4. Overall Cell Reaction:
Combining both half-reactions gives the overall cell reaction:

$N_2{H}_4\left(aq\right)+O_2\left(g\right)\to N_2\left(g\right)+2H_{2}O\left(l\right)$

Thus, the reaction generates harmless nitrogen gas and water as products.