An Investigation of Faraday’s Laws

... node, I will be able to calculate the direct proportionality between the number of Coulombs passed through the solution and the amount of metal removed from the anode.

I will take some readings from my graph in order to calculate the direct proportionality:

When number of coulombs = 800

Then number of grams of copper removed from anode = 0.245

When number of coulombs = 1000

Then number of grams of copper removed from anode = 0.305

When number of coulombs = 1200

Then number of grams of copper removed from anode = 0.367

When number of coulombs = 1400

Then number of grams of copper removed from anode = 0.429

When number of coulombs = 1600

Then number of grams of copper removed from anode = 0.490

When number of coulombs = 1800

Then number of grams of copper removed from anode = 0.550

In order now to investigate Faraday's second law, using number of grams in one mole of copper. Cu = 64, therefore, number of moles removed from the anode during electrolysis = x/60

0.245/60 = 0.00408 and 800/0.00408 = 196078

0.305/60 = 0.00508 and 1000/0.00508 = 196850

0.367/60 = 0.00612 and 1200/0.00612 = 196078

0.429/60 = 0.00715 and 1400/0.00715 = 195804

0.490/60 = 0.00817 and 1600/0.00817 = 195838

0.550/60 = 0.00916 and 1800/0.00916 = 196507

The correct value for the above is two Faradays, which is equal to 193000. My average value of 196193 seems to be a little too high. I will attempt to explain this problem in my evaluation. However, ...