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JEE Main 20199 AprilEvening Shift

Exam: JEE Main
Year: 2019
Date: 9 April
Shift: Evening Shift
Subject: Chemistry
Chapter: Electrochemistry

JEE Main 2019 · 9 April · Evening Shift · Chemistry · Electrochemistry

A solution of Ni (N O _ 3 ) _ 2 is electrolyzed between platinum...

A solution of

Ni{\left(N{O}_{3}\right)}_{2}

is electrolyzed between platinum electrode

0.1

Faraday electricity. How many mole of

Ni

will be deposited at the cathode?

Options

A $0.10$
B $0.15$
C $0.05$
D $0.20$

Correct answer

C. $0.05$

Step-by-step solution

N{i}^{+ 2} + 2{e}^{-} \to Ni

2F

charge gives = 1 mole of

Ni

0.1F

charge gives

= \dfrac{1}{2} \times 0.1 = 0.05moleNi

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