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

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

Grade:

3 Answers

Suneeta Sharma
24 Points
10 years ago

 

  • 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

    rohit kantheti
    20 Points
    10 years ago

    WIEGHT DEPOSITED=EQUIVALENT WIEGHT*CURRENT(i)*TIME TAKEN FOR THE DEPOSIT(t)/96500

                AND THUS 45=108*4.5*t/96500

    Himanshu gupta
    13 Points
    one year ago
    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
            
     
     
     

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