Determine the mass of copper deposited by 4.0 moles of electrons in the reaction represented by the equation below: Cu2(aq) + 2e- → Cu(s)

Determine the mass of copper deposited by 4.0 moles of electrons in the reaction represented by the equation below: Cu²_(aq) + 2e^- → Cu_(s)

Answer Details

To determine the mass of copper deposited, we need to use the concept of Faraday's law of electrolysis. This law states that the amount of substance produced at an electrode is directly proportional to the quantity of electricity passed through the electrolyte. The quantity of electricity is measured in coulombs (C) and is related to the number of moles of electrons (n) by the equation: Q = nF where Q is the quantity of electricity, n is the number of moles of electrons, and F is the Faraday constant, which is equal to 96,485 C/mol. In this case, we have 4.0 moles of electrons, so the quantity of electricity is: Q = nF = 4.0 mol x 96,485 C/mol = 385,940 C The reaction involves the reduction of Cu2+ ions to Cu metal, which requires two electrons per ion. So, for every two electrons that pass through the electrolyte, one Cu atom is produced. This means that the number of moles of Cu produced is half the number of moles of electrons: n(Cu) = 0.5 x n(e-) = 0.5 x 4.0 mol = 2.0 mol The molar mass of copper is 63.55 g/mol, so the mass of copper produced is: m = n x M = 2.0 mol x 63.55 g/mol = 127.1 g Therefore, the mass of copper deposited is approximately 127.1 g. The closest option to this value is which is 128 g.