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JEE Advanced 2024Paper 2
Exam
JEE Advanced
Year
2024
Paper
Paper 2
Subject
Chemistry
Chapter
Electrochemistry

JEE Advanced 2024 · Paper 2 · Chemistry · Electrochemistry

An aqueous solution of hydrazine (N_2 H_4) is electrochemically...

An aqueous solution of hydrazine (N2H4)\left(\mathrm{N}_2 \mathrm{H}_4\right) is electrochemically oxidized by O2\mathrm{O}_2, thereby releasing chemical energy in the form of electrical energy. One of the products generated from the electrochemical reaction is N2( g)\mathrm{N}_2(\mathrm{~g}). Choose the correct statement(s) about the above process :

Options

  1. AOH−\mathrm{OH}^{-}ions react with N2H4\mathrm{N}_2 \mathrm{H}_4 at the anode to form N2( g)\mathrm{N}_2(\mathrm{~g}) and water, releasing 4 electrons to the anode.
  2. BAt the cathode, N2H4\mathrm{N}_2 \mathrm{H}_4 breaks to N2( g)\mathrm{N}_2(\mathrm{~g}) and nascent hydrogen released at the electrode reacts with oxygen to form water.
  3. CAt the cathode, molecular oxygen gets converted to OH−\mathrm{OH}^{-}.
  4. DOxides of nitrogen are major by-products of the electrochemical process.

Correct answer

A. OH−\mathrm{OH}^{-}ions react with N2H4\mathrm{N}_2 \mathrm{H}_4 at the anode to form N2( g)\mathrm{N}_2(\mathrm{~g}) and water, releasing 4 electrons to the anode.

Step-by-step 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.
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