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)
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Correct Answer: Option C
Explanation:
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
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