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

Answer

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

### Step 1: Understand the Electrochemical Reactions
- For each metal ion, we need to understand the reduction reaction that occurs at the cathode:
- **Silver (Ag⁺)**:
\[ \text{Ag}^+ + e^- \rightarrow \text{Ag} \]
- **Calcium (Ca²⁺)**:
\[ \text{Ca}^{2+} + 2e^- \rightarrow \text{Ca} \]
- **Aluminum (Al³⁺)**:
\[ \text{Al}^{3+} + 3e^- \rightarrow \text{Al} \]

### Step 2: Calculate Moles of Each Metal Deposited
- Using Faraday's laws of electrolysis, we can determine how many moles of each metal will be deposited for 3 Faraday of electricity.

#### For Silver (Ag⁺):
- 1 Faraday deposits 1 mole of Ag.
- Therefore, 3 Faraday will deposit:
\[ 3 \text{ Faraday} \times 1 \text{ mole/Faraday} = 3 \text{ moles of Ag} \]

#### For Calcium (Ca²⁺):
- 2 Faraday deposits 1 mole of Ca.
- Therefore, 3 Faraday will deposit:
\[ 3 \text{ Faraday} \times \frac{1 \text{ mole}}{2 \text{ Faraday}} = \frac{3}{2} \text{ moles of Ca} \]

#### For Aluminum (Al³⁺):
- 3 Faraday deposits 1 mole of Al.
- Therefore, 3 Faraday will deposit:
\[ 3 \text{ Faraday} \times \frac{1 \text{ mole}}{3 \text{ Faraday}} = 1 \text{ mole of Al} \]

### Step 3: Write the Molar Ratio
- Now we have the moles of each metal:
- Ag: 3 moles
- Ca: \( \frac{3}{2} \) moles
- Al: 1 mole

- To express this as a ratio, we can convert all quantities to a common denominator. The least common multiple of the denominators (1, 2, and 1) is 2. Thus, we can express the moles as:
- Ag: \( 3 = \frac{6}{2} \)
- Ca: \( \frac{3}{2} = \frac{3}{2} \)
- Al: \( 1 = \frac{2}{2} \)

- Therefore, the molar ratio of Ag : Ca : Al is:
\[ 6 : 3 : 2 \]

### Final Answer
The molar ratio in which the three metal ions are liberated at the electrodes is **6 : 3 : 2**.

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