Question

4.
The amount of Cu deposited when 10 A of current
is passed for 10 minute through Cuso,(aq) is
(1) 1.97 g
(2) 2.89 g
(3) 4 g
(4) 3.5 g​

Answer 1

I=10A

t=10 min=600 sec

no of e- exchanged= 2

molwt=63.54g

F=96500

according to the formula,

w/mol.wt×n = i×t/F

w=i×t×molwt/n×F

w=10×600×63.54/2×96500

w=381240/193000

w=1.97g

I hope this helps ( ╹▽╹ )