Investigate the loss in mass of the anode when a copper

... toms decomposed. The current remains the same all around the circuit.

Current is the measure of the number of charges that pass a point per second (C/s). Therefore if you increase the current, there will be more electrons available at the cathode to make copper atoms. So with greater current the mass at the cathode will increase. This in turn means that at the anode there will be a greater loss in mass as current increases, because the copper will need to decompose more and match the number coulombs being pumped out of the power pack.

I expect that the loss in mass at the anode will be directly proportional to the current. To explain why, I will first write out the ionic equation at the anode: Cu (s) Cu2+(aq) +2 e Every copper atom loses two electrons and a copper ion at the anode, so one mole of copper atoms goes to two moles of electrons plus one mole of copper ions. The current is a measure of how many electrons flow through the circuit in a given time, so if the time is kept constant, the current only affects the number of electrons passing. Therefore, the current would only affect the number of copper atoms decomposed, and their combined mass. In fact, if the mass of copper required is doubled, the current must be doubled to allow double the number of electrons to pass through the circuit in the same amount of time, thus losing twice the overall mass. If 63.5g (~1 mole) of copper atoms must be decomposed at the anode in one second, a current of 193,000 amps would be needed. This is given by using Faraday“s law and Faraday“s constant. Faraday“s Law: Current (I) = Charge (Q) / Time (T) Faraday“s Constant: 1 mole of electrons have about 96,500 coulombs of charge. Cu2+(aq) + 2e- Gives Cu(s) 1 mole + 2 moles Gives 1 mole 1 mole + 2 x 96,500 Gives 63.5g I = Q / T Q = 2 x 96000 Coulombs T = 1 Second I = 2 x 96,500 / 1 I = 193,000 Amps So for every 193,000A, one mole of copper atoms will be decomposed. Therefore, if I know that the current will be directly proportional to the loss in mass at the anode, I can predict the shape of the graph:

Knowing these equations and the relative atomic masses from the periodic table, I can calculate the exact losses in mass at the anode. I will show the method I use for one calculation and then just calculate the rest. Thi ...