The remaining Cu2+ is 0.517 M .
Given ,
A volatile cell using Cu/Cu2+ and Sn/Sn2+ half cells is set up at standard conditions ,and each compartment has a volume of 345 ml .
Current = 0.17 A
Time = 48 hrs = 48 ( 3600) = 172800 s
we know , I = Q/t
thus , Q ,charge= It = 0.17 (17280) =29376 C
Then ,atomic mass of copper is 63 g
And valency of Cu is 2.
Thus , the equivalent mass of copper is 63/2 = 31.5
We know , 96500 coulombs of electricity produce copper = 31.5 g
29376 C of electricity produce copper =9.589 g
We know , 63.546 g of copper = 1 mol
9.589 g of copper = 0.1508 mol
We know ,
1 L of solution contains = 1 mole of Cu2+ ions
0.345 L of solution contains = o.345 moles of Cu2+ ions
the remaining Cu2+ ion in the solution is given by ,
total Cu2+ ions in the solution - copper deposited
= 0.345 mol - 0.152 mol =0.1942 mole of Cu2+
Thus , molarity ,M = 0.1942 mol/ 0.375L =0.517 M
Hence ,the remaining Cu2+ is 0.517 M .
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