Question

What mass of silver salt is deposited when a current of 2.6 A is passed though a solution of a silver salt for 40 minutes? (ag=108 F=96500C)

Answer 1

We know that

Time (t)=$40\times60=2400sec$

$I=2.6A$

Charge $(Q)=it$

$Q=2.6\times40\times60=6240C$

96500C charge deposit=1 mol of Ag

1C charge deposit =$\frac{108}{96500}$

6240C charge deposit =

$$\frac{108}{96500}\times6240=6.98gm$$