JEE Main 2014ChemistryElectrochemistryMediumMCQ

JEE Main 2014 — Electrochemistry Question with Solution

JEE Main 2014 (09 Apr Online)

Question

A current of 10.0 A flows for 2.00 h through an electrolytic cell containing a molten salt of metal X. This results in the decomposition of 0.250 mol of metal X at the cathode. The oxidation state of X in the molten salt is:

F=96,500 C

Choose an option

Show full solutionCorrect option: C
Correct answer
C3+

Step-by-step explanation

Quantity of Electricity  = 1 0 × 3 6 0 0 × 2 9 6 5 0 0
                                   =0.746≈0.75F.
If oxidation state is n+ for 1 mol deposition of a metal:
Electricity=nF (Faradays)
For 0.250 mol=0.250 nF
∴ 0.250 nF=0.75 F
n=3

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About this question

This is a previous-year question from JEE Main 2014, 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.