3 faraday electricity was passed through the three electrolytic cells connected in series containing `Ag^(+)Ca^(2+) and Al^(+3)` ions respectively. The molar ration in which the three metl ions are liberated at the electrodes is :

Options

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

Answer

To solve the problem, we need to determine the molar ratio in which the metal ions \( \text{Ag}^+, \text{Ca}^{2+}, \text{Al}^{3+} \) are liberated at the electrodes when 3 Faraday of electricity is passed through the electrolytic cells connected in series.

### Step-by-Step Solution:

1. **Identify the number of moles produced per Faraday for each ion:**
- For silver (\( \text{Ag}^+ \)):
\[ \text{Ag}^+ + e^- \rightarrow \text{Ag} \]
1 Faraday produces 1 mole of Ag.

- For calcium (\( \text{Ca}^{2+} \)):
\[ \text{Ca}^{2+} + 2e^- \rightarrow \text{Ca} \]
2 Faraday produces 1 mole of Ca, so 1 Faraday produces \( \frac{1}{2} \) mole of Ca.

- For aluminum (\( \text{Al}^{3+} \)):
\[ \text{Al}^{3+} + 3e^- \rightarrow \text{Al} \]
3 Faraday produces 1 mole of Al, so 1 Faraday produces \( \frac{1}{3} \) mole of Al.

2. **Calculate the total moles produced with 3 Faraday:**
- For silver:
\[ 3 \text{ Faraday} \rightarrow 3 \text{ moles of Ag} \]
- For calcium:
\[ 3 \text{ Faraday} \rightarrow 3 \times \frac{1}{2} = 1.5 \text{ moles of Ca} \]
- For aluminum:
\[ 3 \text{ Faraday} \rightarrow 3 \times \frac{1}{3} = 1 \text{ mole of Al} \]

3. **Establish the molar ratio of the metals produced:**
- The moles of metals produced are:
- Ag: 3 moles
- Ca: 1.5 moles
- Al: 1 mole

- The molar ratio can be expressed as:
\[ \text{Ag} : \text{Ca} : \text{Al} = 3 : 1.5 : 1 \]

4. **Simplify the ratio:**
- To eliminate the decimal, we can multiply the entire ratio by 2:
\[ 3 \times 2 : 1.5 \times 2 : 1 \times 2 = 6 : 3 : 2 \]

5. **Final answer:**
- The molar ratio in which the three metal ions are liberated at the electrodes is:
\[ \text{Ag} : \text{Ca} : \text{Al} = 6 : 3 : 2 \]