posted by Mango .
If 2.00 liters of 0.580 M CuSO4 solution is electrolyzed by passing 1.70 amps through the solution for 2.00 hr using inert electrodes.
What is the [Cu2+] in the solution at the end of the electrolysis?
Coulombs = A x sec = 1.70 x 2 x 60 x 60 = ??
One equivalent weight will be deposited by passing 96,495 coulombs. ONe equivalent weight = 63.54 g/2 = 31.8 but you need to do that more accurately.
31.8 x #coulombs/96,485 = g Cu deposited.
g Cu initially = 2L x 0.580M x 63.54 = ??
g Cu deposted from above = xx
g Cu remaining in solution = zz
M = moles/2L = zz
This assumes no electrolysis of the water (highly unlikely I think).