Question

When during electrolysis of a solution of Ag No₃, 9650 coulombs of charge pass through the electroplating bath, the mass of silver deposite on the cathode will be:-

Options

• 10.8 g

• 21.6 g

• 108 g

• 1.08 g

Answer 1

10.8 g

Explanation:

\[\ce{Ag^{+} + e^{-} -> Ag}\]

Number of moles of Ag = `9650/96500 = 1/10` moles

\[\ce{Ag^{+} + e^{-} -> Ag}\]

∴ Mass of Ag produced = `1/10 xx 108` = 10.8 g