Introduction:
In an electrolyte cell, a chemical reaction is driven by the passage of an electric current. The cell consists of two electrodes, an anode and a cathode, immersed in an electrolyte solution. In this case, the electrolyte solution contains Ag2SO4 and the electrodes are made of platinum (Pt).

Given information:
- Amount of O2 liberated at the anode = 1.6 gm
- Electrolyte solution contains Ag2SO4
- Electrodes are made of Pt

Theory:
In an electrolyte cell, the anode is the electrode where oxidation occurs, and the cathode is the electrode where reduction occurs. During the electrolysis process, positive ions migrate towards the cathode and are reduced, while negative ions migrate towards the anode and are oxidized.

Solution:
To determine the amount of Ag liberated at the cathode, we need to understand the reactions occurring at both electrodes and the stoichiometry involved.

Anode reaction:
At the anode, oxidation takes place. In this case, O2 is liberated, indicating the oxidation of a species containing oxygen. The reaction can be represented as follows:
2 Ag2SO4 → 2 Ag+ + SO4^2- + O2↑

From the given information, we know that 1.6 gm of O2 is liberated at the anode. To determine the amount of Ag liberated at the cathode, we need to calculate the moles of O2 using its molar mass (32 g/mol) and then use the stoichiometry of the reaction.

Moles of O2:
Moles = Mass / Molar mass = 1.6 gm / 32 g/mol = 0.05 mol

Stoichiometry:
From the balanced equation, we can see that 2 moles of O2 are produced for every 2 moles of Ag2SO4 oxidized. This means that 2 moles of Ag are also produced for every 2 moles of O2 liberated.

Since 1 mole of Ag has a molar mass of 107.87 g/mol, we can calculate the amount of Ag liberated at the cathode as follows:

Moles of Ag = 0.05 mol × (2 mol Ag / 2 mol O2) = 0.05 mol

Mass of Ag = Moles × Molar mass = 0.05 mol × 107.87 g/mol = 5.39 g

Therefore, the amount of Ag liberated at the cathode would be 5.39 grams.

Conclusion:
In the electrolyte cell containing Ag2SO4 and Pt electrodes, when 1.6 gm of O2 is liberated at the anode, the amount of Ag liberated at the cathode would be 5.39 grams. This is determined by calculating the moles of O2 using its molar mass and then using the stoichiometry of the reaction to determine the amount of Ag.

21.6 gm by Faraday's second law