Question

When during electrolysis of a solution of,9650 coulmobs of charge pass through the electroplating bath, the mass of silver deposited on the cathode will be:

Answer 1

here, no. of moles of electrons = Q/F​=9650/96500​=0.1mole

Here, no. of moles of electrons = FQ​=965009650​=0.1mole

Now,

1 mole of AgNO3​ will produce one mole of monovalent silver ion on dissociation.

Therefore,

The number of electrons involved will also be one mole for one mole of AgNO3​.

Thus, 0.1 moles of silver will be produced by 0.1 mole of AgNO3​ .

Since,

Silver nitrate dissociates completely into Ag+ and NO3−​ ion.

Hence,

Mass of silver = no. of moles x molar mass= 0.1 x 108 = 10.8 g