JEE Challenger
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ChemMedAdv 2024 • Paper 2

Electrochemical Oxidation of Hydrazine in Fuel Cell

An aqueous solution of hydrazine (N2H4\text{N}_2\text{H}_4) is electrochemically oxidized by O2\text{O}_2, thereby releasing chemical energy in the form of electrical energy. One of the products generated from the electrochemical reaction is N2(g)\text{N}_2(\text{g}).

Choose the correct statement(s) about the above process

Options

A

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

Correct
B

At the cathode, N2H4\text{N}_2\text{H}_4 breaks to N2(g)\text{N}_2(\text{g}) and nascent hydrogen released at the electrode reacts with oxygen to form water.

C

At the cathode, molecular oxygen gets converted to OH−\text{OH}^-.

Correct
D

Oxides of nitrogen are major by-products of the electrochemical process.

Topics & Concepts

ElectrochemistryElectrochemical Cell

Step-by-Step Solution

To determine the correct statement(s) regarding the electrochemical oxidation of hydrazine (N2H4\text{N}_2\text{H}_4) in a hydrazine-oxygen (N2H4−O2\text{N}_2\text{H}_4-\text{O}_2) fuel cell, let us analyze the half-cell reactions taking place at the electrodes in an alkaline aqueous medium.

1. Reactions in a Hydrazine-Oxygen Fuel Cell

  • At the Anode (Oxidation):
    Hydrazine (N2H4\text{N}_2\text{H}_4) is oxidized at the anode in the presence of hydroxide ions (OH−\text{OH}^-), producing nitrogen gas (N2\text{N}_2), water (H2O\text{H}_2\text{O}), and releasing electrons to the anode. N2H4(aq)+4OH−(aq)⟶N2(g)+4H2O(l)+4e−\text{N}_2\text{H}_4(\text{aq}) + 4\text{OH}^-(\text{aq}) \longrightarrow \text{N}_2(\text{g}) + 4\text{H}_2\text{O}(\text{l}) + 4e^-

  • At the Cathode (Reduction):
    Molecular oxygen (O2\text{O}_2) is reduced at the cathode by taking up electrons in the presence of water to form hydroxide ions (OH−\text{OH}^-). O2(g)+2H2O(l)+4e−⟶4OH−(aq)\text{O}_2(\text{g}) + 2\text{H}_2\text{O}(\text{l}) + 4e^- \longrightarrow 4\text{OH}^-(\text{aq})

  • Overall Cell Reaction:
    Adding the two half-reactions gives: N2H4(aq)+O2(g)⟶N2(g)+2H2O(l)\text{N}_2\text{H}_4(\text{aq}) + \text{O}_2(\text{g}) \longrightarrow \text{N}_2(\text{g}) + 2\text{H}_2\text{O}(\text{l})

2. Evaluation of Options

  • Option (A):
    As established by the anodic oxidation reaction, OH−\text{OH}^- ions react with N2H4\text{N}_2\text{H}_4 at the anode to produce N2(g)\text{N}_2(\text{g}) and H2O\text{H}_2\text{O}, transferring 4e−4e^- per molecule of hydrazine to the anode.
    Option (A) is CORRECT.\text{Option (A) is CORRECT.}

  • Option (B):
    Hydrazine is oxidized at the anode, not at the cathode. The cathode reaction involves the reduction of oxygen. No nascent hydrogen is generated as part of the cathode reaction.
    Option (B) is INCORRECT.\text{Option (B) is INCORRECT.}

  • Option (C):
    At the cathode, molecular oxygen (O2\text{O}_2) accepts electrons and is reduced to hydroxide ions (OH−\text{OH}^-).
    Option (C) is CORRECT.\text{Option (C) is CORRECT.}

  • Option (D):
    The only products of the reaction are nitrogen gas (N2\text{N}_2) and water (H2O\text{H}_2\text{O}). Oxides of nitrogen (NOx\text{NO}_x) are not formed during this electrochemical process, making it an environmentally clean energy source.
    Option (D) is INCORRECT.\text{Option (D) is INCORRECT.}

Conclusion

The correct options are A and C.

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Electrochemical Oxidation of Hydrazine in Fuel Cell | Chemistry PYQ Solution - JEE Challenger