posted by bekah on .
How many grams of copper will be deposited at the cathode of an electrolytic cell if a current of 680.0 mA is run through a 2.5 M solution of copper sulfate for 20.0 minutes?
Q = coulombs = amperes x seconds
Q = 0.680A x 20m x (60s/m) = ? = about 816
96,485 coulombs will deposit (63.54/2)g Cu. You have 816 coulombs; therefore,
63.54/2 x (816/96,485) = about 0.3g if I punched the right buttons.