How many coulombs of electric charge mst be passed through a solution of Silver Nitrate to coat a copper sheet of area 100 cm sq on both sides with 0.005 mm thickness . density of Ag is 10.5 g/cm3 . (Ag=108)

Mass of Ag = Density X volume
  = 10.5 x [(100x2) x 0.0005]  ------0.005 mm converted to cm
  = 1.05 gms
Moles of Ag deposited = 1.05 / 108 = 9.72 x 10-3 moles of Ag.

The reaction occurring during deposition of Ag :

Ag+ (aq) + e- ----------> Ag(s)

1 electron is needed to deposit 1 atom of Ag+
Hence, 
96500 Coulombs = 1mole of e- = deposit 1 mol of Ag+ i.e. 108 gms

Thus to deposit

amount of charge

= 9.72 x 10-3 moles of Ag x 96500 /108

= 8.68 C.
 

  • -8

Mass deposited = Density of the metal × Surface area × thickness.

Mass deposited = Density of the metal × Surface area × thickness.

  • -5
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