Four faradays of electricity were passed through `AgNO_(3)(l) , CdSO_(4)(l), AlCl_(3) (l) and PbCl_(4) (l)` kept in four vessels using inert electrodes. The ratio of moles of Ag, Cd, Al and Pb deposited will be

Options

A. `12: 4 : 6: 3`
B. `1 : 2 : 3: 4`
C. `12 : 6 : 4 : 3`
D. `4 : 3 : 2 : 1`

Answer

To solve the problem, we need to determine the number of moles of silver (Ag), cadmium (Cd), aluminum (Al), and lead (Pb) deposited when four faradays of electricity are passed through their respective solutions.

### Step-by-Step Solution:

1. **Identify the reactions and the number of electrons required for deposition:**
- For Ag:
\[ \text{Ag}^+ + e^- \rightarrow \text{Ag} \quad (1 \text{ electron}) \]
- For Cd:
\[ \text{Cd}^{2+} + 2e^- \rightarrow \text{Cd} \quad (2 \text{ electrons}) \]
- For Al:
\[ \text{Al}^{3+} + 3e^- \rightarrow \text{Al} \quad (3 \text{ electrons}) \]
- For Pb:
\[ \text{Pb}^{4+} + 4e^- \rightarrow \text{Pb} \quad (4 \text{ electrons}) \]

2. **Calculate the moles of each metal deposited using the formula:**
\[ \text{Moles of metal} = \frac{Q}{nF} \]
where \( Q \) is the total charge (in coulombs), \( n \) is the number of electrons transferred per mole of metal, and \( F \) is Faraday's constant (approximately 96500 C/mol).

3. **Calculate the charge (Q) for 4 faradays:**
\[ Q = 4 \times F = 4 \times 96500 \text{ C} = 386000 \text{ C} \]

4. **Calculate moles of each metal deposited:**
- For Ag:
\[ \text{Moles of Ag} = \frac{4F}{1F} = 4 \text{ moles} \]
- For Cd:
\[ \text{Moles of Cd} = \frac{4F}{2F} = 2 \text{ moles} \]
- For Al:
\[ \text{Moles of Al} = \frac{4F}{3F} = \frac{4}{3} \text{ moles} \]
- For Pb:
\[ \text{Moles of Pb} = \frac{4F}{4F} = 1 \text{ mole} \]

5. **Express the ratio of moles of Ag, Cd, Al, and Pb:**
- The ratio of moles of Ag : Cd : Al : Pb is:
\[ 4 : 2 : \frac{4}{3} : 1 \]

6. **Convert the ratio to whole numbers:**
- To eliminate the fraction, multiply all terms by 3:
\[ 4 \times 3 : 2 \times 3 : \frac{4}{3} \times 3 : 1 \times 3 = 12 : 6 : 4 : 3 \]

7. **Final Ratio:**
- Thus, the final ratio of moles of Ag : Cd : Al : Pb is:
\[ 12 : 6 : 4 : 3 \]