Posted by hriday on .
A current of 0.965 A is passed through 500 ml of 0.2 M solution of ZnSO4 for 10 min. The molarity of Zn2+ after deposition of Zn is
amperes x seconds = coulombs
coulombs/96,485 = Faraday's
1 Faraday will deposit 1 equivalent weight of Zn^+2 or 1/2 mole Zn^+2.
I get 0.965 x 10 min x (60 sec/min) = ?? coulombs.
That divided by 96,485 = 0.006 Faraday's and that will deposit 0.003 mole Zn^+2.
How many moles Zn^+2 in 500 mL 0.2 M ZnSO4. M x L = 0.2 x 0.50 = 0.1 mol
0.003 is deposited. How much is left. What is the molarity of that solution?