Redox Reactions And Electrochemistry Ques 325

Question: How many atoms of calcium will be deposited from a solution of $ CaCl_2 $ by a current 0.25 mA following for 60 seconds [BHU 2004]

Options:

A) $ 4.68\times 10^{18} $

B) $ 4.68\times 10^{15} $

C) $ 4.68\times 10^{12} $

D) $ 4.68\times 10^{9} $

Show Answer

Answer:

Correct Answer: A

Solution:

Given, Current (i) = 25 mA = 0.025 A Time (t) = 60 sec Q = i t $ =60\times 0.025=1.5 $ coulombs No. of electrons $ =\frac{1.5\times 6.023\times 10^{23}}{96500} $ $ {e^{-}}=9.36\times 10^{18} $ $ Ca\to C{a^{2+}}+2{e^{-}} $ $ 2{e^{-}} $ are required to deposite one Ca atom $ 9.36\times 10^{18}\ {e^{-}} $ will be used to deposite $ =\frac{9.36\times 10^{18}}{2} $ $ =4.68\times 10^{18} $ .



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