Рассчитайте массу выделившегося на катоде серебра, если при электролизе с помощью инертного электрода раствора нитрата серебра на аноде Выделяется 12 г кислорода.

Molar mass of oxygen (O₂) = 32 g/mol Molar mass of silver (Ag) = 108 g/mol

The balanced equation for the electrolysis of silver nitrate (AgNO₃) is: 2 AgNO₃ (aq) -> 2 Ag (s) + O₂ (g) + 2 NO₂ (g)

According to the equation, for every 4 moles of AgNO₃ electrolyzed, 2 moles of Ag are deposited on the cathode, and 1 mole of O₂ is produced on the anode.

Given that 12 g of O₂ is generated, we can calculate the amount of Ag deposited as follows:

1 mole of O₂ = 32 g 2 moles of Ag = 2 × 108 g

Therefore, the amount of Ag deposited is (12 g O₂) × (2 moles Ag / 32 g O₂) = 0.75 moles Ag.

To calculate the mass of Ag, we use the equation:

Mass of Ag = Moles of Ag × Molar mass of Ag = 0.75 moles × 108 g/mol = 81 g.

Thus, the mass of silver deposited on the cathode is 81 g.