JEE Main 2019 — Electrochemistry Question with Solution
From: JEE Main 2019 (Online) 9th April Evening Slot
Question
A solution of Ni(NO3)2 is electrolysed between
platinum electrodes using 0.1 Faraday
electricity. How many mole of Ni will be
deposited at the cathode?
Choose an option
Show full solutionCorrect option: D
Correct answer
D0.05
Step-by-step explanation
Cathode reaction :
Ni+2 + 2e- Ni(s)
From 2 mole of electrons 1 mole of Ni is deposited at the cathode.
So from 0.1 F or 0.1 mole of electrons = 0.05 mole of Ni is deposited at the cathode.
Ni+2 + 2e- Ni(s)
From 2 mole of electrons 1 mole of Ni is deposited at the cathode.
So from 0.1 F or 0.1 mole of electrons = 0.05 mole of Ni is deposited at the cathode.
Practice this on the real CBT interface
Solve this JEE Main question (and the rest of the Electrochemistry chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.
Solve interactively →About this question
This is a previous-year question from JEE Main 2019, 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.