Q. 10

Calculate the tim

Answer :

Given, mass of Cu deposited (z) = 1.27 g


Current, I = 2A


Molar mass of Cu = 63.5 g mol1, 1 F = 96500 C mol1


Reaction taking place: Cu2+ + 2e- Cu


By Faraday’s first law,


W= zQ (since, Q = It)


W = zIt …(1)


Where, W, amount of substance deposited = 1.27g


I, current = 2A and t = time


Z = =


Substituting all values in equation (1),


1.27 =


t = 1930 s


Therefore, time to deposit 1.27 g of copper at cathode is 1930s.


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