Calculate the time to deposit 1.27 g of copper at cathode when a current of 2 A was passed through the solution of CuSO4. (Molar mass of Cu = 63.5 g mol−1, 1 F = 96500 C mol−1)
Given, mass of Cu deposited (z) = 1.27 g
Current, I = 2A
Molar mass of Cu = 63.5 g mol−1, 1 F = 96500 C mol−1
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 =
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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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