JEE Advanced 2024 Paper 2, Chemistry Section 2 Q1: Electrochemical Cell And Nernst Equation

JEE Advanced2024Paper 2Chemistry Section 2

An aqueous solution of hydrazine ( $N_{2} H_{4}$ ) 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
$O H^{-}$ ions react with $N_{2} H_{4}$ at the anode to form $N_{2} (g)$ and water, releasing 4 electrons to the anode.
B
At the cathode, $N_{2} H_{4}$ 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 $O H^{-}$ .
D
Oxides of nitrogen are major by-products of the electrochemical process.

Solution

This is the hydrazine–oxygen fuel cell. Hydrazine ( $N$ in $- 2$ state) is the fuel; molecular oxygen is the oxidant. Oxidation (electron loss) occurs at the anode, reduction at the cathode.

Anode (oxidation of hydrazine):

$N_{2} H_{4} + 4 O H^{-} ⟶ N_{2} + 4 H_{2} O + 4 e^{-} .$

Hydrazine, not $O_{2}$ , is consumed here. This matches option (A).

Cathode (reduction of oxygen):

$O_{2} + 2 H_{2} O + 4 e^{-} ⟶ 4 O H^{-} .$

$O_{2}$ is reduced to hydroxide ion. This matches option (C).

Option (B) is incorrect because $N_{2} H_{4}$ is oxidised at the anode, not the cathode, and the mechanism does not proceed via nascent hydrogen. Option (D) is also incorrect: under normal cell operation, nitrogen oxides are not produced, only $N_{2}$ and water.

Net reaction: $N_{2} H_{4} + O_{2} ⟶ N_{2} + 2 H_{2} O .$

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JEE Advanced 2024 Paper 2, Chemistry Section 2 Q1: Electrochemical Cell And Nernst Equation

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