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In an electrolytic cell, calculate the concentration of #\sf{Cu^(2+)}# remaining in 425 mL of solution that was originally 0.366 M #\sf{CuSO_4}#, after the passage of 2.68 A for 262 seconds?

I have most if not all of the dimensional analysis set up. However, I don't know where "originally 0.366 M..." fits in.

#\sf{\frac{262" sec"}{}xx\frac{2.68" C"}{1" sec"}xx\frac{1" mol "e^(-)}{96,485" C"}xx\frac{1" mol "Cu^(2+)}{2" mol "e^(-)}...?}# (which, by the way, #\approx0.000361" mol Cu"^(2+)#)


#\color{red}{\tt{\frac{0.366" mol "Cu^(2+)}{1" L soln"}xx\frac{0.425" L soln"}{}}}??# (this one is #\approx0.156" mol Cu"^(2+)#)

(Not sure if I need to do that one at all, but it's there.)

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ReportAnswer #1

Based on the language of the question, it looks like #["Cu"^(2+)]# will decrease until a smaller concentration is "remaining".

#["Cu"^(2+)]_f = "0.358 M"#


The point of an electrolytic cell is to force a nonspontaneous reaction to occur. That is accomplished by supplying a positive voltage at the cathode, to force electrons to flow in. In this case, we are doing so to reduce #Cu^(2+)# to #Cu(s)#.

The #"0.366 M"# allows you to determine how many mols of #"Cu"^(2+)# was there to begin with. Conventional current is flowing out, so that electrons flow into #"Cu"^(2+)#. Thus, #["Cu"^(2+)]darr#.

#"0.366 mol"/cancel"L" xx 0.425 cancel"L" = "0.156 mols Cu"^(2+)# initially

(How else would we know the final state? All we will get otherwise is the change...)

Then, #"2.68 C/s"# flows out of copper(II) for #"262 s"# so that #["Cu"^(2+)]darr#. So,

#262 cancel"s" xx (2.68 cancel"C")/cancel"s" xx (cancel("mol e"^(-)))/(96485 cancel"C") xx ("1 mol Cu"^(2+))/(2 cancel("mol e"^(-)))#

#= "0.00364 mols Cu"^(2+)# would get consumed. (Not #"0.000364 mols"#.)

Thus, the mols left is:

#"0.156 mols Cu"^(2+) - "0.00364 mols Cu"^(2+)#

#= "0.152 mols Cu"^(2+)#

And thus, the concentration leftover is:

#color(blue)(["Cu"^(2+)]_f) = ("0.152 mols Cu"^(2+))/("0.425 L") = color(blue)("0.358 M")#

answered by: Truong-Son N.
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In an electrolytic cell, calculate the concentration of #\sf{Cu^(2+)}# remaining in 425 mL of solution that was originally 0.366 M #\sf{CuSO_4}#, after the passage of 2.68 A for 262 seconds?
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