How many moles of copper would be deposited by passing 1 Faraday of electricity through a \(CuCl_2\) solution?
1 Faraday of electricity is equal to the charge of 1 mole of electrons, which is 96,485 coulombs. The reduction of copper(II) ions in solution to copper metal requires 2 electrons per ion, so 1 Faraday of electricity will deposit 0.5 moles of copper.
1 Faraday * (2 electrons/Cu ion) / (1 mole electrons/96,485 coulombs) * (1 mole Cu/2 electrons) = 0.5 moles Cu
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