The time required to deposit 4.5g of copper from CuSO\(_4\) solution by passing a current of 2.5 Amperes is (Cu = 64g ; 1F = 96500C/mol)
On ionization, Cu\(^{2+}\) + 2e\(^-\) → Cu
2 x 96,500 C liberated 64g Cu
( 2.5 x t) C will liberate 4.5g Cu (Recall that Q = I x t )
By simple proportion, we will have
\(\frac{193,000}{2.5t}\) = \(\frac{64}{4.5}\)
2.5t x 64 = 193,000 x 4.5
160 t = 868,500
t = \(\frac{868,500}{160}\)
t = 5,428.125 seconds
t = 5,428 sec
The correct answer is option C
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