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.

1 answer

1. 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.
   Then
   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).

   Answer ID
   524822

   Created
   April 3, 2011 9:31pm UTC

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   URL
   https://questions.llc/answers/524822