JEE Main 2020 — Electrochemistry Question with Solution
From: JEE Main 2020 (Online) 4th September Evening Slot
Question
250 mL of a waste solution obtained from the
workshop of a goldsmith contains 0.1 M AgNO3
and 0.1 M AuCl. The solution was electrolyzed
at 2V by passing a current of 1A for 15
minutes. The metal/metals electrodeposited will
be
[ = 0.80 V, = 1.69 V ]
[ = 0.80 V, = 1.69 V ]
Choose an option
Show full solutionCorrect option: C
Correct answer
COnly gold
Step-by-step explanation
Millimoles of Au+ = 0.1 × 250 = 25
Mole of Au+ = =
Charge passed = I × t = 1 × 15 × 60 = 900 C
moles of e– passed = =
Only gold will be deposited as quantity of charge passed is less than the amount of Au+ present.
Mole of Au+ = =
Charge passed = I × t = 1 × 15 × 60 = 900 C
moles of e– passed = =
Only gold will be deposited as quantity of charge passed is less than the amount of Au+ present.
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This is a previous-year question from JEE Main 2020, 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.