Calculate the amount in moles of silver deposited when 9650C of electricty is passed through a solution of silver salt [= 96500 C$mol^{-1}$]

1. Analyze the question

We are asked to calculate the number of moles of silver ($Ag$) deposited when a specific quantity of electricity (charge) is passed through a silver salt solution.

Given:

• Quantity of electricity ($Q$) = $9650\text{ C}$
• Faraday’s constant ($F$) = $96500\text{ C mol}^{-1}$

2. Explain the concept

To solve this, we use Faraday’s First Law of Electrolysis, which states that the mass of a substance deposited is proportional to the quantity of electricity passed. The relationship is:

Where:

• $Q$ is the quantity of electricity in Coulombs ($C$).
• $n$ is the number of moles of the substance.
• $z$ is the valency (or number of electrons involved in the reaction). For silver ($Ag^+$), the reaction is $Ag^+ + e^- \rightarrow Ag(s)$, so $z = 1$.
• $F$ is Faraday’s constant ($96500\text{ C mol}^{-1}$).

3. Show the solution

Step 1: Identify the formula Since we need to find the amount in moles ($n$), we rearrange the formula:

Step 2: Substitute the values

• $Q = 9650\text{ C}$
• $z = 1$
• $F = 96500\text{ C mol}^{-1}$

Step 3: Calculate

4. Identify the answer

Comparing our result to the given options: A. 0.05 B. 10.80 C. 10.00 D. 0.10

The calculated value matches option D.

5. Tip

When solving electrolysis problems, always check the charge of the metal ion. For Group 1 metals (like $Ag^+$, $Na^+$, $K^+$), $z=1$. For Group 2 metals (like $Cu^{2+}$, $Mg^{2+}$, $Zn^{2+}$), $z=2$. Remembering this saves you time during the exam!