Question

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:

• A. \(OH^-\) ions react with \(N_2H_4\) at the anode to form \(N_2(g)\) and water, releasing 4 electrons to the anode.
• B. At the cathode, \(N_2H_4\) breaks to \(N_2(g)\) and nascent hydrogen released at the electrode reacts with oxygen to form water.
• C. At the cathode, molecular oxygen gets converted to \(OH^-\).
• D. Oxides of nitrogen are major by-products of the electrochemical process.

Hint:

Electrochemical processes involve oxidation at the anode and reduction at the cathode. Analyze the half-reactions to determine the products.

Answer 1

The Correct Option is A

The overall reaction for the electrochemical oxidation of hydrazine is: \[ N_2H_4(aq) + O_2 \longrightarrow N_2 + 2H_2O. \] At the anode (oxidation): \[ N_2H_4 + 4OH^- \longrightarrow N_2 + 4H_2O + 4e^-. \] At the cathode (reduction): \[ O_2 + 2H_2O + 4e^- \longrightarrow 4OH^-. \] Net reaction: \[ N_2H_4 + O_2 \longrightarrow N_2 + 2H_2O. \] Hence, \(OH^-\) ions participate in the reaction at the anode, and molecular oxygen gets converted to \(OH^-\) at the cathode. No oxides of nitrogen are formed as by-products.