Question

The amount of silver in mg deposited when \(9.65\) coulombs of electricity is passed through an aqueous solution of silver nitrate is \((Ag=108\,u,\ 1F=96500\,C\,mol^{-1})\)

• A. \(16.2\)
• B. \(21.2\)
• C. \(10.8\)
• D. \(6.4\)

Hint:

Use Faraday's law: mass deposited is directly proportional to charge passed.

Answer 1

The Correct Option is C

Step 1: For silver deposition: \[ Ag^+ + e^- \rightarrow Ag \]

Step 2: One mole of electrons deposits one mole of silver.

Step 3: Charge required to deposit \(108\) g of silver is: \[ 96500\,C \]

Step 4: Charge given is: \[ 9.65\,C \]

Step 5: Mass deposited: \[ m=\frac{108\times 9.65}{96500} \] \[ m=0.0108\,g \]

Step 6: Convert grams into mg: \[ 0.0108\,g=10.8\,mg \] \[ \boxed{10.8\,mg} \]