JEE Advanced · Chemistry · 17. Electrochemistry
An aqueous solution of hydrazine \(\left(\mathrm{N}_2 \mathrm{H}_4\right)\) is electrochemically oxidized by \(\mathrm{O}_2\), thereby releasing chemical energy in the form of electrical energy. One of the products generated from the electrochemical reaction is \(\mathrm{N}_2(\mathrm{~g})\).
Choose the correct statement(s) about the above process :
- A \(\mathrm{OH}^{-}\)ions react with \(\mathrm{N}_2 \mathrm{H}_4\) at the anode to form \(\mathrm{N}_2(\mathrm{~g})\) and water, releasing 4 electrons to the anode.
- B At the cathode, \(\mathrm{N}_2 \mathrm{H}_4\) breaks to \(\mathrm{N}_2(\mathrm{~g})\) and nascent hydrogen released at the electrode reacts with oxygen to form water.
- C At the cathode, molecular oxygen gets converted to \(\mathrm{OH}^{-}\).
- D Oxides of nitrogen are major by-products of the electrochemical process.
Answer & Solution
Correct Answer
(C) At the cathode, molecular oxygen gets converted to \(\mathrm{OH}^{-}\).
Step-by-step Solution
Detailed explanation
\(\underset{\substack{\text { Oxidation } \\ \text { (Anode) }}}{\stackrel{-2\qquad}{N_2H_4}}+\underset{\substack{\text { Reduction } \\ \text { (Cathode) }}}{\stackrel{0}{O_2}} \longrightarrow \stackrel{0}{N_2}+\mathrm{H}_2 \mathrm{O}^{-2}\)
At anode:
\(\mathrm{N}_2 \mathrm{H}_4+4 \mathrm{OH}^{-} \longrightarrow \mathrm{N}_2+4 \mathrm{H}_2 \mathrm{O}+4 e^{-}\)
At cathode:
\(\mathrm{O}_2+2 \mathrm{H}_2 \mathrm{O}+4 e^{-} \longrightarrow 4 \mathrm{OH}^{-}\)
Complete reaction:
\(\mathrm{N}_2 \mathrm{H}_4+\mathrm{O}_2 \longrightarrow \mathrm{N}_2+2 \mathrm{H}_2 \mathrm{O}\)
Statements (A) and (C) are correct.
At anode:
\(\mathrm{N}_2 \mathrm{H}_4+4 \mathrm{OH}^{-} \longrightarrow \mathrm{N}_2+4 \mathrm{H}_2 \mathrm{O}+4 e^{-}\)
At cathode:
\(\mathrm{O}_2+2 \mathrm{H}_2 \mathrm{O}+4 e^{-} \longrightarrow 4 \mathrm{OH}^{-}\)
Complete reaction:
\(\mathrm{N}_2 \mathrm{H}_4+\mathrm{O}_2 \longrightarrow \mathrm{N}_2+2 \mathrm{H}_2 \mathrm{O}\)
Statements (A) and (C) are correct.
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