Question

Then during electrolysis of a solution of agno3 , 9650 coulombs of charge pass through the electroplating bath, the mass of silver deposited in the cathode will be

Answer 1

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

Now,

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

Therefore,

.number of electrons involved will also be one mole for one mole of AgNO .
 
 Thus, 0.1 mole of silver will be produced by 0.1 mole of AgNO .

Since,

silver nitrate dissociates completely into Ag  and NO ion.

Hence,

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

Answer 2

The mass of silver deposited on cathode will be 10.8 g.

Number of moles of electrons = $\frac{Q}{F}$ = $\frac{9650}{96500}$ = 0.1 mole

When silver nitrate is dissociated, we know that one mole of monovalent silver ion is produced.

As a result, one mole of electrons will be produced.

Thus, 0.1 moles of silver nitrate will generate 0.1 moles of silver.

Therefore, the silver mass = (number of moles) (molar mass)

                                             = (0.1) (108) g

                                             = 10.8g

Hence, the silver deposited is 10.8 g.

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