JEE Advanced 2024 · Paper 2 · Chemistry Q5

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ChemistryMultiple correct+4 / -2Question 39 of 51

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

Choose the correct statement(s) about the above process

\ce{OH-}

ions react with

\ce{N2H4}

at the anode to form

\ce{N2(g)}

and water, releasing 4 electrons to the anode.

At the cathode,

\ce{N2H4}

breaks to

\ce{N2(g)}

and nascent hydrogen released at the electrode reacts with oxygen to form water.

At the cathode, molecular oxygen gets converted to

\ce{OH-}

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

Answer and worked solution

Official answer: A, C

The cell. This is a hydrazine-oxygen fuel cell in alkaline medium.

Anode (oxidation): $\ce{N2H4 + 4OH- -> N2 + 4H2O + 4e-}$
Cathode (reduction): $\ce{O2 + 2H2O + 4e- -> 4OH-}$
Overall: $\ce{N2H4 + O2 -> N2 + 2H2O}$

Options.

(A) True: this is exactly the anode reaction.
(B) False: hydrazine is oxidised at the anode, not the cathode.
(C) True: $\ce{O2}$ is reduced to $\ce{OH-}$ at the cathode.
(D) False: the product is $\ce{N2}$ , and nitrogen oxides are not formed.

Topics:Batteries and corrosion Balancing redox reactions Galvanic cells and EMF

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JEE Advanced 2024 · Paper 2, Chemistry question 5