Calculate the mass of Ag deposited at cathode when a current of 2 amperes was passed through a solution of AgNO₃ for 15 minutes.
(Given : Molar mass of Ag = 108 g mol^–1 1F = 96500 C mol^–1)

Given,
Molar mass of Ag = 108 g mol^–1 
1F = 96500 C mol^–1

Q=It
Q= 2×15×60=1800 C
W(weight of substance deposited)= ZQ
Z= M/nF
n-factor here is 1
So Z = 108/1×96500

 =2.01 g
The mass of Ag deposited at the cathode is2.01 g