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: 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
There is an explanation video available below.