Step 1: Understanding the Concept:
This problem involves Faraday's First Law of Electrolysis, which states that the mass of a substance deposited or liberated at any electrode is directly proportional to the quantity of electricity passed through the electrolyte.
Step 2: Key Formula or Approach:
1. Write the half-reaction for the deposition of silver.
2. Determine the number of moles of electrons required to deposit one mole of the substance.
3. Calculate the total charge (in Coulombs) required to deposit one mole of the substance. This is equal to (n-factor $\times$ F), where n is the number of moles of electrons in the half-reaction and F is the Faraday constant.
4. Use a ratio to find the mass deposited by the given charge. The formula is:
\[ \text{Mass deposited (w)} = Z \times Q = \left(\frac{E}{F}\right) \times Q \]
where E is the equivalent weight (Molar Mass / n-factor), F is the Faraday constant, and Q is the charge passed.
Step 3: Detailed Explanation:
1. Half-reaction: Silver is deposited from Ag$^+$ ions.
\[ Ag^+ + 1e^- \rightarrow Ag(s) \]
This shows that 1 mole of electrons is required to deposit 1 mole of silver. So, the n-factor is 1.
2. Molar mass and charge:
- Molar mass of Ag = 108 g/mol.
- Charge required to deposit 1 mole of Ag = $1 \times F = 1 \times 96500$ C.
3. Calculation:
We can set up a proportion:
If 96500 C of charge deposits 108 g of Ag,
Then 9.65 C of charge will deposit 'w' g of Ag.
\[ \frac{w}{9.65 \text{ C}} = \frac{108 \text{ g}}{96500 \text{ C}} \]
\[ w = \frac{108 \times 9.65}{96500} \text{ g} \]
\[ w = \frac{108}{10000} \text{ g} = 0.0108 \text{ g} \]
The question asks for the amount in milligrams (mg).
\[ w (\text{in mg}) = 0.0108 \text{ g} \times 1000 \frac{\text{mg}}{\text{g}} = 10.8 \text{ mg} \]
Step 4: Final Answer:
The amount of silver deposited is 10.8 mg. Therefore, option (C) is correct.