What mass of copper will be deposited by the liberation of Cu2+ when 0.1F of electricity flows through an aqueous solution of a copper (ll) salt? [Cu = 64]

What mass of copper will be deposited by the liberation of Cu²⁺ when 0.1F of electricity flows through an aqueous solution of a copper (ll) salt? [Cu = 64]

Answer Details

The amount of substance that is deposited during an electrolysis reaction can be calculated using Faraday's laws of electrolysis. According to Faraday's laws, the amount of substance deposited is directly proportional to the amount of electricity that flows through the solution. The formula that relates the amount of substance to the amount of electricity is: Amount of substance = (Electricity in Coulombs) / (Faraday constant x Valency) where the Faraday constant is the amount of electric charge per mole of electrons, and the valency is the number of electrons involved in the redox reaction. For the given problem, the copper (II) ion has a valency of 2, and we are given that 0.1 F of electricity flows through the solution. The Faraday constant is 96,485 Coulombs per mole of electrons. Therefore, the amount of copper that will be deposited can be calculated as: Amount of substance = (0.1 F) / (2 x 96,485 C/mol) = 5.18 x 10^-5 mol The molar mass of copper is 64 g/mol, so the mass of copper that will be deposited is: Mass of copper = Amount of substance x Molar mass = 5.18 x 10^-5 mol x 64 g/mol = 0.00331 g ≈ 3.2 g (rounded to one decimal place) Therefore, the answer is 3.2g.