Baring in mind that 1 mole of Copper = 63.5 gHow much current is necessary to liberate/ produce 0.5 kg of Copper in one hour?Working also if possible, thanks.
The reduction of copper ions into copper metal is described by the chemical equation Cu2+ + 2e- -----> Cu Current in amperes (I) equals charge in coulombs (q) divided by time in seconds (t) t = 1 hour = 60 x 60 = 3600 seconds Now we need to determine how much charge is necessary in one hour. This can be done using stoichiometry. We know that for every 2 moles of electrons, we produce 1 mole of copper metal. (500 g Cu) x (1 mol Cu / 63.5 g Cu) x (2 mol e / 1 mol Cu) x (96500 C / 1 mol e) = 1519685 C Divide charge by time, and we get around 422 AMPS of current, which is a tremendous amount of electricity.
