The weight ratio of Al and Ag deposited using the same quantity of current is:

Options

A. `9:108`
B. `2:12`
C. `108:9`
D. `3:8`

Explanation

To find the weight ratio of aluminum (Al) and silver (Ag) deposited using the same quantity of current, we can follow these steps based on Faraday's laws of electrolysis.

### Step 1: Understand Faraday's First Law of Electrolysis
According to Faraday's first law, the mass (M) of a substance deposited during electrolysis is directly proportional to the quantity of electricity (Q) passed through the electrolyte:
\[ M \propto Q \]

### Step 2: Introduce the Electrochemical Equivalent (Z)
We can express this relationship as:
\[ M = Z \cdot Q \]
where \( Z \) is the electrochemical equivalent of the substance.

### Step 3: Calculate the Electrochemical Equivalent (Z)
The electrochemical equivalent \( Z \) can be calculated using the formula:
\[ Z = \frac{M_a}{n} \]
where \( M_a \) is the atomic weight (or molar mass) of the element, and \( n \) is the valency (number of electrons transferred).

### Step 4: Determine the Values for Aluminum (Al)
- **Atomic Weight of Al**: 27 g/mol
- **Valency of Al**: 3 (since Al is in the form of \( Al^{3+} \) and accepts 3 electrons)

Using the formula:
\[ Z_{Al} = \frac{27}{3} = 9 \, \text{g/C} \]

### Step 5: Determine the Values for Silver (Ag)
- **Atomic Weight of Ag**: 108 g/mol
- **Valency of Ag**: 1 (since Ag is in the form of \( Ag^{+} \) and accepts 1 electron)

Using the formula:
\[ Z_{Ag} = \frac{108}{1} = 108 \, \text{g/C} \]

### Step 6: Calculate the Weight Ratio
Now, we can find the weight ratio of Al to Ag deposited using the same quantity of electricity (Q):
\[ \text{Weight Ratio} = \frac{M_{Al}}{M_{Ag}} = \frac{Z_{Al}}{Z_{Ag}} = \frac{9}{108} = \frac{1}{12} \]

### Final Answer
The weight ratio of Al to Ag deposited using the same quantity of current is \( 1:12 \).

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