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. 108 g
B. 1.6 g
C. 0.8 g
D. 21.60 g

Answer

To solve the problem step by step, we need to determine the amount of silver deposited at the cathode when 1.6 grams of oxygen is liberated at the anode in an electrolytic cell containing a solution of Ag₂SO₄.

### Step 1: Calculate the moles of O₂ liberated
First, we need to calculate the number of moles of O₂ that corresponds to the mass given.

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

Using the formula:

\[ \text{Number of moles of } O_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{1.6 \text{ g}}{32 \text{ g/mol}} = 0.05 \text{ mol} \]

### Step 2: Determine the number of equivalents of O₂
Next, we need to calculate the number of equivalents of O₂. The reaction for the liberation of O₂ at the anode can be represented as:

\[ 2H_2O \rightarrow O_2 + 4H^+ + 4e^- \]

From this reaction, we see that 1 mole of O₂ corresponds to 4 moles of electrons (4 equivalents).

Thus, the number of equivalents of O₂ is:

\[ \text{Number of equivalents of } O_2 = \text{moles of } O_2 \times 4 = 0.05 \text{ mol} \times 4 = 0.2 \text{ equivalents} \]

### Step 3: Calculate the amount of silver deposited
The reduction of silver ions at the cathode can be represented as:

\[ Ag^+ + e^- \rightarrow Ag \]

Here, 1 equivalent of silver corresponds to 1 mole of silver deposited. Therefore, the amount of silver deposited can be calculated using the number of equivalents of electrons transferred, which is equal to the equivalents of O₂ liberated.

Since we have 0.2 equivalents of electrons, we can calculate the mass of silver deposited:

\[ \text{Mass of } Ag = \text{equivalents of } Ag \times \text{molar mass of } Ag \]

The molar mass of silver (Ag) is approximately 108 g/mol. Therefore:

\[ \text{Mass of } Ag = 0.2 \text{ equivalents} \times 108 \text{ g/mol} = 21.6 \text{ g} \]

### Conclusion
The amount of silver deposited at the cathode is **21.6 grams**.