JEE Advanced 2024 Paper 2 Q05 Chemistry Electrochemistry Galvanic Cells & EMF Medium

JEE Advanced 2024 Paper 2 · Q05 · Galvanic Cells & EMF

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

  1. A. $OH^-$ ions react with $N_2H_4$ at the anode to form $N_2(g)$ and water, releasing 4 electrons to the anode.
  2. 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.
  3. C. At the cathode, molecular oxygen gets converted to $OH^-$.
  4. D. Oxides of nitrogen are major by-products of the electrochemical process.
JEE Advanced multi-correct — pick every correct option, then check.
Reveal answer + step-by-step solution

Correct answer:A, C, D

Solution

At the anode, hydrazine is oxidized: $N_2H_4 + 4OH^- \to N_2 + 4H_2O + 4e^-$, so (A) is correct. At the cathode, $O_2 + 2H_2O + 4e^- \to 4OH^-$, so (C) is correct. The overall fuel-cell reaction is $N_2H_4 + O_2 \to N_2 + 2H_2O$, but at high $O_2$ : $N_2H_4$ ratios side reactions $N_2H_4 + 2O_2 \to 2NO + 2H_2O$ and $N_2H_4 + 3O_2 \to 2NO_2 + 2H_2O$ produce oxides of nitrogen as by-products, so (D) is correct. Hydrazine is oxidized at the anode (not split at the cathode), so (B) is wrong.

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