Calculate the time to deposit 1.5 g of silver at cathode when a current of 1.5 A is passed through the solution of `AgNO_(3)`.

Explanation

The reaction at the cathode is :
`{:(Ag^(+)(aq)+e^(-)rarr Ag(s)),(108 g " "96500 C):}`
108 g of silver is deposited by passing charge =96500 C
1.5 g of silver is deposited by passing charge `=((96500 C)xx(1.5 g))/((108 g))=1340.27 C =1340.27 `As
Time to deposited 1.5 g of silver `=("charge"(Q))/("current"(I))=((1340.27 As))/((1.5 A))`
=893.51 s or 14.85 min