Calculate the time required to deposit 56g of silver from a silver nitrate solution using a current of 4.5A.

							
 
mass Ag deposited = 56g 
MM = 107.9 g mol-1 (from Periodic Table) 
n (Ag) 56 ÷ 107.9 = 0.519 mol = n(e)


n(e) = 0.519 mol 
F = 96,500 C mol-1

Q = 0.519 x 96,500 = 50,083.5 C

Q = 50,083.5 C   I = 4.5 A 
 
t = Q/I
t = 50,083.5 ÷ 4.5 = 11,129.67 seconds 
t = 11,129.67 ÷ 60 = 185.5 minutes 
t = 185.5 ÷ 60 = 3.1 hours


						

							
WIEGHT DEPOSITED=EQUIVALENT WIEGHT*CURRENT(i)*TIME TAKEN FOR THE DEPOSIT(t)/96500
            AND THUS 45=108*4.5*t/96500
						

							
Mass of Ag= 56 gm
Molar mass of Ag= 107.9 gm/mol
           Number of moles of Ag= mass÷molar mass
                                                     = 56÷107.9
                                                     = 0.518999 
                        Ag+ + e-= Ag 
      To deposite 1 mole Ag, 1 mole e- is passed
         So,  to deposite 0.518999 mole, 
   Number of e- passed=0.518999×1=0.518999 moles
                •1 mole consist of 96500C
            So, 0.518999 mole charge contain is, 
     Amount of charge passed= 96500×0.518999
                                                   =50083.4035C
          Time taken = charge÷current
                               = 50083.4035÷4.5
                     time = 11129.645 second
                      time= 1129.645÷60=185.494 minutes
                      time = 185.494÷60=3.092 hour