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

Suneeta Sharma
24 Points
12 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
12 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
3 years 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