An aqueous solution of hydrazine $\left(\mathrm{N}_2 \mathrm{H}_4\right)$ is electrochemically oxidized by…

An aqueous solution of hydrazine $\left(\mathrm{N}_2 \mathrm{H}_4\right)$ is electrochemically oxidized by $\mathrm{O}_2$, thereby releasing chemical energy in the form of electrical energy. One of the products generated from the electrochemical reaction is $\mathrm{N}_2(\mathrm{~g})$. Choose the correct statement(s) about the above process :
  1. $\mathrm{OH}^{-}$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.
  2. 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.
  3. At the cathode, molecular oxygen gets converted to $\mathrm{OH}^{-}$.
  4. Oxides of nitrogen are major by-products of the electrochemical process.

Solution

\(\underset{\substack{\text { Oxidation } \\ \text { (Anode) }}}{\stackrel{-2\qquad}{N_2H_4}}+\underset{\substack{\text { Reduction } \\ \text { (Cathode) }}}{\stackrel{0}{O_2}} \longrightarrow \stackrel{0}{N_2}+\mathrm{H}_2 \mathrm{O}^{-2}\) At anode: \(\mathrm{N}_2 \mathrm{H}_4+4 \mathrm{OH}^{-} \longrightarrow \mathrm{N}_2+4 \mathrm{H}_2 \mathrm{O}+4 e^{-}\) At cathode: \(\mathrm{O}_2+2 \mathrm{H}_2 \mathrm{O}+4 e^{-} \longrightarrow 4 \mathrm{OH}^{-}\) Complete reaction: \(\mathrm{N}_2 \mathrm{H}_4+\mathrm{O}_2 \longrightarrow \mathrm{N}_2+2 \mathrm{H}_2 \mathrm{O}\) Statements (A) and (C) are correct.

Asked in: JEE Advanced 2024 (Paper 2)

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