When the electrolysis of silver sulphate was carried out by Pt electrodes, 1.6 g oxygen was liberated at the anode, the amount of silver deposited at cathode will be :

Question

Accepted Answer

To determine the amount of silver deposited at the cathode during the electrolysis of silver sulphate, we need to consider the stoichiometry of the reactions occurring at both the anode and the cathode.

### Explanation for Option 1 (21.6 g):

1. **Electrolysis Reactions**:
   - At the anode, oxygen is liberated from the decomposition of water:
     $$
     2H_2O \rightarrow O_2 + 4H^+ + 4e^-
     $$
   - At the cathode, silver ions are reduced to form silver metal:
     $$
     Ag^+ + e^- \rightarrow Ag
     $$

2. **Moles of Oxygen Liberated**:
   - Given that 1.6 g of oxygen is liberated, we calculate the moles of oxygen:
     $$
     \text{Moles of } O_2 = \frac{1.6 \, \text{g}}{32 \, \text{g/mol}} = 0.05 \, \text{mol}
     $$

3. **Faraday's Law of Electrolysis**:
   - According to Faraday's laws, the amount of substance deposited or liberated at an electrode is directly proportional to the quantity of electricity passed.
   - For every mole of $ O_2 $ produced, 4 moles of electrons are required.

4. **Calculating Moles of Electrons**:
   - Moles of electrons for 0.05 mol of $ O_2 $:
     $$
     0.05 \, \text{mol} \times 4 = 0.2 \, \text{mol of electrons}
     $$

5. **Silver Deposition**:
   - Each mole of electrons deposits one mole of silver:
     $$
     \text{Moles of } Ag = 0.2 \, \text{mol}
     $$
   - Mass of silver deposited:
     $$
     \text{Mass of } Ag = 0.2 \, \text{mol} \times 108 \, \text{g/mol} = 21.6 \, \text{g}
     $$

Thus, the correct amount of silver deposited at the cathode is 21.6 g, which corresponds to Option 1.

### Explanation for Why Other Options are Incorrect:

- **Option 2 (108 g)**:
  - This would imply that 1 mole of silver is deposited, which would require 1 mole of electrons. However, only 0.2 moles of electrons are involved, making this option incorrect.

- **Option 3 (1.6 g)**:
  - This suggests that only 0.0148 moles of silver are deposited, which is inconsistent with the 0.2 moles of electrons calculated. Hence, this option is incorrect.

- **Option 4 (0.8 g)**:
  - This would mean only 0.0074 moles of silver are deposited, which is far less than the 0.2 moles of electrons available. Therefore, this option is incorrect.

In conclusion, the correct answer is Option 1, with 21.6 g of silver being deposited at the cathode.