JEE Main 2024 — Electrochemistry Question with Solution
From: JEE Main 2024 (Online) 31st January Morning Shift
Question
One Faraday of electricity liberates gram atom of copper from copper sulphate. is ________.
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Step-by-step explanation
To find the value of x when one Faraday of electricity liberates x times gram atom of copper from copper sulphate, we need to understand how electricity interacts with copper ions in solution. The key reaction is:
This shows that copper ions (Cu2+) gain two electrons (e-) to become copper metal (Cu). From electrochemistry, we know that:
- 2 Faradays of electricity are required to deposit 1 mole (or 1 gram atom) of copper.
- Therefore, 1 Faraday will deposit half of that amount, which is 0.5 mole, or in other terms, 0.5 gram atom of copper.
- Expressing 0.5 in the form of x times gives us .
This means x equals 5.
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This is a previous-year question from JEE Main 2024, covering the Electrochemistry chapter of Chemistry. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.