JEE Advanced · 2024 · 26 May 2024 · Paper 2 · Redox Reactions and Electrochemistry

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) $OH^{-}$ ions react with $N_{2} H_{4}$ at the anode to form $N_{2} (g)$ and water, releasing 4 electrons to the anode.✓ Correct
(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 $OH^{-}$ .✓ Correct
(D)Oxides of nitrogen are major by-products of the electrochemical process.

Answer: (A), (C)

Solution

In this hydrazine-oxygen fuel cell:

Anode reaction:

N_{2} H_{4} (aq) + 4 OH^{-} (aq) \to N_{2} (g) + 4 H_{2} O (l) + 4 e^{-}

Hydrazine is oxidized at the anode, releasing electrons. Statement (A) is correct.

Cathode reaction:

2 H_{2} O (l) + O_{2} (g) + 4 e^{-} \to 4 OH^{-} (aq)

Oxygen is reduced at the cathode to form hydroxide ions. Statement (C) is correct.

Statement (B) is incorrect because hydrazine reacts at the anode, not the cathode. Statement (D) is incorrect because nitrogen oxides are not major products; the main product is

N_{2}

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JEE Advanced · 2024 · 26 May 2024 · Paper 2 · Redox Reactions and Electrochemistry

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