A current of 0.2ampere is passes through a solution of `CuSO_4` for 10 minutes calculate the man of Cu deposited on the cathode.

Explanation

Data: `l=1.5 amp, t=10 xx 60 =600s `
`Cu^(2+)+2e^(-) to Cu`
1 mole of `Cu^(2+)" requireds " 2 xx 96500C`
Formula: `I=Q/t`
`Q=It =1.5 xx 600 =900C`
`2 xx 96500" C deposits 63 g of Cu"`
900 C deposits.........? `=(63 xx 900)/(2 xx 96500) =0.2938g`