Question

If 0.5 amp current is passed through acidified silver nitrate
solution for 100 minutes. The mass of silver deposited on
cathode, is (eq.wt.of silver nitrate = 108)
(a) 2.3523 g (b) 3.3575 g
(c) 5.3578 g (d) 6.3575 g

Answer 1

The mass of silver deposited on  cathode, is

(b) 3.3575 g

Current passed (I) = 0.5 A

Time (t) = 100 minutes = 100*60 = 6000 sec

Equivalent mass of Ag (E) = 108

We know, mass deposited (w) = EIt/96500

Replacing the values, we get,

w = (108*0.5*6000)/(96500) gm

= 3.3575gm

Option (b)  is correct.