Answer to Question #144784 in General Chemistry for Joshua

Question

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

Answer 1

 mass of Ag deposited = 56g 

RAM of Ag = 107.9 g mol^-1 (from Periodic Table) 

Moles= (Ag) 56 ÷ 107.9 = 0.519 mole = 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 ÷ 3600 = 3.1 hours