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)
- 7.64g
- 3.82g
-
1.91g
- 0.96g
The correct answer is: 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
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