An electrolytic cell contains a solution of `Ag_(2)SO_(4)` and have platinum electrodes. A current is passed until 1.6gm of `O_(2)` has been liberated at anode. The amount of silver deposited at cathode would be

Options

A. 107.88gm
B. 1.6gm
C. 0.8gm
D. 21.60gm.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Determine the amount of charge passed using the amount of oxygen liberated at the anode.

Given that 1.6 g of O₂ is liberated at the anode, we can calculate the moles of O₂ produced:

\[ \text{Molar mass of } O_2 = 32 \text{ g/mol} \]

\[ \text{Moles of } O_2 = \frac{1.6 \text{ g}}{32 \text{ g/mol}} = 0.05 \text{ mol} \]

### Step 2: Calculate the number of electrons involved in the reaction at the anode.

The reaction at the anode for the liberation of oxygen can be represented as:

\[ 2 \text{H}_2O \rightarrow O_2 + 4 \text{H}^+ + 4e^- \]

From this reaction, we see that 1 mole of O₂ corresponds to 4 moles of electrons. Therefore, the moles of electrons (n) involved in liberating 0.05 moles of O₂ is:

\[ n = 0.05 \text{ mol O}_2 \times 4 \text{ mol e}^- / \text{mol O}_2 = 0.20 \text{ mol e}^- \]

### Step 3: Calculate the total charge (Q) using Faraday's constant.

Faraday's constant (F) is approximately 96500 C/mol. The total charge can be calculated as:

\[ Q = n \times F = 0.20 \text{ mol} \times 96500 \text{ C/mol} = 19300 \text{ C} \]

### Step 4: Determine the equivalent weight of silver (Ag).

The reduction reaction at the cathode is:

\[ \text{Ag}^+ + e^- \rightarrow \text{Ag} \]

The equivalent weight of silver (E₂) is given by:

\[ E_2 = \frac{\text{Molar mass of Ag}}{\text{Change in oxidation state}} = \frac{108 \text{ g/mol}}{1} = 108 \text{ g} \]

### Step 5: Use Faraday's second law of electrolysis to find the amount of silver deposited.

According to Faraday's second law:

\[ \frac{W_1}{W_2} = \frac{E_1}{E_2} \]

Where:
- \( W_1 \) is the mass of O₂ liberated (1.6 g)
- \( W_2 \) is the mass of silver deposited
- \( E_1 \) is the equivalent weight of O₂ (8 g)
- \( E_2 \) is the equivalent weight of Ag (108 g)

Rearranging gives:

\[ W_2 = W_1 \times \frac{E_2}{E_1} \]

Substituting the known values:

\[ W_2 = 1.6 \text{ g} \times \frac{108 \text{ g}}{8 \text{ g}} = 1.6 \text{ g} \times 13.5 = 21.6 \text{ g} \]

### Conclusion

The amount of silver deposited at the cathode is **21.6 g**.

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