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
a.1 : 2 : 3
b.2 : 3 : 1
c.6 : 3 : 2
d.3 : 4 : 2

Options

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

Explanation

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

### Step 1: Understand the relationship between charge, moles, and Faraday's law
According to Faraday's law of electrolysis, the amount of substance liberated at an electrode is directly proportional to the quantity of electricity passed through the electrolyte. The formula we use is:
\[ n = \frac{Q}{nF} \]
where:
- \( n \) = number of moles of the substance deposited,
- \( Q \) = total electric charge (in coulombs),
- \( n \) = number of electrons required to deposit one mole of the substance,
- \( F \) = Faraday's constant (approximately \( 96500 \, C/mol \)).

### Step 2: Calculate the moles of silver (\( \text{Ag}^+ \))
For silver ions (\( \text{Ag}^+ \)):
- \( n = 1 \) (1 mole of electrons is required to deposit 1 mole of silver).
- Therefore, the moles of silver deposited:
\[ n_{\text{Ag}} = \frac{3F}{1F} = 3 \text{ moles} \]

### Step 3: Calculate the moles of calcium (\( \text{Ca}^{2+} \))
For calcium ions (\( \text{Ca}^{2+} \)):
- \( n = 2 \) (2 moles of electrons are required to deposit 1 mole of calcium).
- Therefore, the moles of calcium deposited:
\[ n_{\text{Ca}} = \frac{3F}{2F} = 1.5 \text{ moles} \]

### Step 4: Calculate the moles of aluminum (\( \text{Al}^{3+} \))
For aluminum ions (\( \text{Al}^{3+} \)):
- \( n = 3 \) (3 moles of electrons are required to deposit 1 mole of aluminum).
- Therefore, the moles of aluminum deposited:
\[ n_{\text{Al}} = \frac{3F}{3F} = 1 \text{ mole} \]

### Step 5: Determine the molar ratio
Now we have the number of moles of each metal:
- Moles of silver (\( \text{Ag} \)): 3
- Moles of calcium (\( \text{Ca} \)): 1.5
- Moles of aluminum (\( \text{Al} \)): 1

To express these in a simple ratio, we can multiply each by 2 to eliminate the fraction:
- Silver: \( 3 \times 2 = 6 \)
- Calcium: \( 1.5 \times 2 = 3 \)
- Aluminum: \( 1 \times 2 = 2 \)

Thus, the molar ratio of \( \text{Ag} : \text{Ca} : \text{Al} \) is:
\[ 6 : 3 : 2 \]

### Final Answer
The correct option is **c. 6 : 3 : 2**.

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