Question

How long (in minutes) must a \(2\ \text{A}\) current be passed through \(\text{AgNO}_{3}\) solution to deposit \(5.4\ \text{g Ag}\)? [Molar mass of Ag \(=108\ \text{g mol}^{-1}\)]

• A. \(40.2\ \text{min}\).
• B. \(54.5\ \text{min}\).
• C. \(30.1\ \text{min}\).
• D. \(60.6\ \text{min}\).

Hint:

Find the charge required for 0.05 mol of silver, then divide by the current.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
\(\text{Ag}^{+}+e^{-}\rightarrow\text{Ag}\): one faraday (\(96500\ \text{C}\)) deposits one mole (\(108\ \text{g}\)) of silver.

Step 2: Charge required
Moles of Ag \(= \dfrac{5.4}{108} = 0.05\). Charge \(Q = 0.05\times 96500 = 4825\ \text{C}\).

Step 3: Time
\(t = \dfrac{Q}{I} = \dfrac{4825}{2} = 2412.5\ \text{s}\), which is \(\dfrac{2412.5}{60}\approx 40.2\ \text{min}\).

Final Answer:
The current must flow for about 40.2 minutes. \[ \boxed{\text{(A) }40.2\ \text{min}} \]