Redox Reactions And Electrochemistry Question 171
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} $ .
BETA
