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.
= 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.