Chemistry, asked by TAZU14, 10 months ago

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?

Answers

Answered by qwsuccess
4

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

Similar questions