Question

Passage of current for 548 seconds through a silver coulometer results in the deposition
of 0.746g of silver. What is the current in Ampere?
​

Answer 1

Passage of 1.22 A current for 548 seconds through a silver coulometer results in the deposition

of 0.746g of silver.

• If 1F gives 108g of Ag

Or 96500 Columb gives 108g Ag

1 g of Ag requires $\frac{96500}{108}C$

According to question,

                          0.746g will require  $\frac{0.746x96500}{108}C$

                         =666.56C

• As Q=It

Therefore required current will be I=Q/t

                                 $\frac{666.56}{548}=1.22 Amps$