What mass of Cu would be produced by the cathodic reduction of Cu\(^{2+}\) when 1.60A of current passes through a solution of CuSO\(_4\) for 1 hour. (F=96500Cmol\(^{-1}\) , Cu=64)

A.

7.64g

B.

3.82g

C.

1.91g

D.

0.96g

Correct answer is C

M = \(\frac{MmIT}{96500n}\)

Where

M=mass

Mm = Molar mass = 64g/mol

I = current = 1.6A

T= Time =1h r=3600s

N= No of Charge = +2

M = \(\frac{64x \times 1.6 \times 3600}{96500 \times 2}\)

M=1.91g

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