How many moles of copper would be deposited by passing 3F of electricity through a solution of copper (II) tetraoxosulpohate (IV)? (F = 96 500 C mol-)

The amount of substance deposited during electrolysis can be calculated using Faraday's laws of electrolysis, which state that the amount of substance deposited or liberated at an electrode is directly proportional to the amount of electricity passed through the solution. The formula for Faraday's law is: mass = (current × time × atomic mass) / (charge × valence) where: - current is the electric current (in Amperes) - time is the time the current flows (in seconds) - atomic mass is the molar mass of the substance (in grams per mole) - charge is the charge on one electron (in Coulombs) - valence is the number of electrons involved in the reaction In this case, we are trying to find the number of moles of copper deposited, so we can rearrange the formula to: moles = (current × time) / (charge × valence) We are given the charge on one electron (F = 96,500 C mol^-1), so we can use this value for charge. The valence of copper in copper (II) tetraoxosulphate (IV) is 2, so we can use this value for valence. We are also given the amount of electricity passed through the solution (3F). Therefore, we can calculate the number of moles of copper deposited as follows: moles = (current × time) / (3F × 2) We don't have information about the current or time, so we cannot calculate the exact number of moles deposited. However, we can see that the number of moles deposited is directly proportional to the current and time, and inversely proportional to the charge and valence. Since the amount of substance deposited is directly proportional to the amount of electricity passed through the solution, we can say that the option with the highest number of moles deposited is the correct answer. Therefore, the answer is: 1.5 moles.