How many seconds would it take to deposit 17.3 g of Ag (atomic mass = 107.87) from a solution of AgNO3 using a current of 10.00 amp? a) 3.09 x 103 s b) 7.74 x 102 s c) 4.64 x 103 s d) 1.55 x 103 s e) 5.16 x 102 s
Mass of Ag = 17.3 gm
Moles of Ag = 17.3 g / 107.86 g/mol = 0.16039 Moles
Let us write half-reaction at cathode
Ag+ + e- ====> Ag(s)
number of moles of electrons required = 0.16039 Mole e -
Let us calcualte moles of electrons to coulombs of charge with Faraday's constant.
0.16039 Mole e - x 96485 = 15.475.528 coulombs
Hence 15.475.528 coulombs of charge is required
Let us calculate the time using coulombs of charge and current
10 A x t = 15.475.528
t = 15475 Seconds or 1.55 x 103 seconds
Hence Option D is correct answer
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