In passing `3F` of electricity through three electrolytic cells connect in series containing `Ag^(o+),Ca^(2+),` and `Al^(3+)` ions, respectively. The molar ratio in which the three metal ions are liberated at the electrodes is

Options

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

Explanation

To solve the problem of determining the molar ratio in which the metal ions \( \text{Ag}^+, \text{Ca}^{2+}, \) and \( \text{Al}^{3+} \) are liberated at the electrodes when passing 3 Faraday of electricity through three electrolytic cells connected in series, we can follow these steps:

### Step 1: Determine the number of moles of each metal ion liberated

1. **Silver Ion (\( \text{Ag}^+ \))**:
- The half-reaction for silver ions is:
\[ \text{Ag}^+ + e^- \rightarrow \text{Ag} \]
- This reaction shows that 1 mole of \( \text{Ag}^+ \) requires 1 Faraday of electricity.
- Therefore, for 3 Faraday:
\[ \text{Moles of } \text{Ag} = 3 \text{ Faraday} \times 1 \text{ mole/Faraday} = 3 \text{ moles} \]

2. **Calcium Ion (\( \text{Ca}^{2+} \))**:
- The half-reaction for calcium ions is:
\[ \text{Ca}^{2+} + 2e^- \rightarrow \text{Ca} \]
- This reaction shows that 1 mole of \( \text{Ca}^{2+} \) requires 2 Faraday of electricity.
- Therefore, for 3 Faraday:
\[ \text{Moles of } \text{Ca} = \frac{3 \text{ Faraday}}{2 \text{ Faraday/mole}} = \frac{3}{2} \text{ moles} \]

3. **Aluminum Ion (\( \text{Al}^{3+} \))**:
- The half-reaction for aluminum ions is:
\[ \text{Al}^{3+} + 3e^- \rightarrow \text{Al} \]
- This reaction shows that 1 mole of \( \text{Al}^{3+} \) requires 3 Faraday of electricity.
- Therefore, for 3 Faraday:
\[ \text{Moles of } \text{Al} = \frac{3 \text{ Faraday}}{3 \text{ Faraday/mole}} = 1 \text{ mole} \]

### Step 2: Calculate the molar ratio

Now we have the moles of each metal ion:
- Moles of \( \text{Ag} = 3 \)
- Moles of \( \text{Ca} = \frac{3}{2} \)
- Moles of \( \text{Al} = 1 \)

To find the molar ratio, we express these in a simplified form:
- The ratio is \( 3 : \frac{3}{2} : 1 \).

To eliminate the fraction, we can multiply each term by 2:
- \( 3 \times 2 : \frac{3}{2} \times 2 : 1 \times 2 \)
- This gives us \( 6 : 3 : 2 \).

### Final Molar Ratio
Thus, the molar ratio in which the three metal ions are liberated at the electrodes is:
\[ \text{Ag} : \text{Ca} : \text{Al} = 6 : 3 : 2 \]