Question

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+)`)

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`\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.)

Answer 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"`

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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")`