Question

Passage pf current fpr 548 seconds through a silver coulometer results in the deposition of 0.746 g of silver. What is the current

Answer 1

Current deposited is 1.21 A.

The reaction which occurs is as follows:

Ag⁺ + e⁻ → Ag

Electron transfer(n) = 1

Molar mass of Ag = 108gm

So, 108gm of Ag is deposited by n Farads = 1F

By unitary method, 1gm of Ag would be deposited by 1/108 F

Similarly, 0.746gm of Ag would be deposited by [(1/108) × 0.746] F

We know, 1F = 96500C,

So, [(1/108) × 0.746] F = [(0.746/108) × 96500] C = 666.57 C

Charge (in C) = Current (in A) × Time (in sec)

⇒ Current = Charge/Time = 666.57/548 A = 1.21 A

Current produced is 1.21 A