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 )

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

Tardigrade Admin · Accepted Answer

Given current (i)=0.5 amp Time (t)=100 minutes × 60=6000 sec Equivalent weight of silver nitrate (E)=108. According to Faraday's first law of electrolysis W=(E i t/96500)=(108 × 0.5 × 6000/96500)=3.3575 g .

Q. 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 $= 108$ )

$2.3523 g$

$3.3575 g$

$5.3578 g$

$6.3575 g$

Given current $(i) = 0.5 am p$
Time $(t) = 100$ minutes $\times 60 = 6000 sec$
Equivalent weight of silver nitrate $(E) = 108$ .
According to Faraday's first law of electrolysis
$W = \frac{E i t}{96500} = \frac{108 \times 0.5 \times 6000}{96500} = 3.3575 g .$

Q. 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 )

2.3523g

3.3575g

5.3578g

6.3575g