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 of determining the ratio of moles of Ag, Cd, Al, and Pb deposited when 4 Faradays of electricity are passed through their respective solutions, we can follow these steps:

### Step 1: Understand the concept of Faraday's laws of electrolysis
Faraday's laws state that the amount of substance deposited during electrolysis is directly proportional to the quantity of electricity passed through the electrolyte. The relationship can be expressed as:
\[ n = \frac{Q}{F} \]
where \( n \) is the number of moles of the substance deposited, \( Q \) is the total charge passed, and \( F \) is Faraday's constant (approximately 96500 C/mol).

### Step 2: Determine the charge passed
In this case, we are given that 4 Faradays of electricity are passed. Therefore, the total charge \( Q \) can be calculated as:
\[ Q = 4 \times F \]

### Step 3: Calculate the moles of each metal deposited
1. **For Silver (Ag)**:
- The half-reaction for silver is:
\[ \text{Ag}^+ + e^- \rightarrow \text{Ag} \]
- This indicates that 1 mole of Ag is deposited per 1 Faraday.
- Thus, for 4 Faradays:
\[ n_{\text{Ag}} = \frac{4}{1} = 4 \text{ moles of Ag} \]

2. **For Cadmium (Cd)**:
- The half-reaction for cadmium is:
\[ \text{Cd}^{2+} + 2e^- \rightarrow \text{Cd} \]
- This indicates that 1 mole of Cd is deposited per 2 Faradays.
- Thus, for 4 Faradays:
\[ n_{\text{Cd}} = \frac{4}{2} = 2 \text{ moles of Cd} \]

3. **For Aluminum (Al)**:
- The half-reaction for aluminum is:
\[ \text{Al}^{3+} + 3e^- \rightarrow \text{Al} \]
- This indicates that 1 mole of Al is deposited per 3 Faradays.
- Thus, for 4 Faradays:
\[ n_{\text{Al}} = \frac{4}{3} \text{ moles of Al} \]

4. **For Lead (Pb)**:
- The half-reaction for lead is:
\[ \text{Pb}^{4+} + 4e^- \rightarrow \text{Pb} \]
- This indicates that 1 mole of Pb is deposited per 4 Faradays.
- Thus, for 4 Faradays:
\[ n_{\text{Pb}} = \frac{4}{4} = 1 \text{ mole of Pb} \]

### Step 4: Write the ratio of moles of metals deposited
Now we have:
- Moles of Ag = 4
- Moles of Cd = 2
- Moles of Al = \( \frac{4}{3} \)
- Moles of Pb = 1

To express these in a common ratio, we can multiply each term by 12 (the least common multiple of the denominators) to eliminate fractions:
- Moles of Ag = \( 4 \times 12 = 48 \)
- Moles of Cd = \( 2 \times 12 = 24 \)
- Moles of Al = \( \frac{4}{3} \times 12 = 16 \)
- Moles of Pb = \( 1 \times 12 = 12 \)

Thus, the ratio of moles of Ag : Cd : Al : Pb is:
\[ 48 : 24 : 16 : 12 \]
This simplifies to:
\[ 12 : 6 : 4 : 3 \]

### Final Answer
The ratio of moles of Ag, Cd, Al, and Pb deposited is:
\[ 12 : 6 : 4 : 3 \]