# 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, $\setminus \approx 0.000361 {\text{ mol Cu}}^{2 +}$) \color{red}{\tt{\frac{0.366" mol "Cu^(2+)}{1" L soln"}xx\frac{0.425" L soln"}{}}}?? (this one is $\setminus \approx 0.156 {\text{ mol Cu}}^{2 +}$) (Not sure if I need to do that one at all, but it's there.)

Aug 8, 2018

Based on the language of the question, it looks like $\left[{\text{Cu}}^{2 +}\right]$ 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 $C {u}^{2 +}$ to $C u \left(s\right)$.

The $\text{0.366 M}$ allows you to determine how many mols of ${\text{Cu}}^{2 +}$ was there to begin with. Conventional current is flowing out, so that electrons flow into ${\text{Cu}}^{2 +}$. Thus, $\left[{\text{Cu}}^{2 +}\right] \downarrow$.

${\text{0.366 mol"/cancel"L" xx 0.425 cancel"L" = "0.156 mols Cu}}^{2 +}$ initially

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

Then, $\text{2.68 C/s}$ flows out of copper(II) for $\text{262 s}$ so that $\left[{\text{Cu}}^{2 +}\right] \downarrow$. 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"^(-)))

$= {\text{0.00364 mols Cu}}^{2 +}$ would get consumed. (Not $\text{0.000364 mols}$.)

Thus, the mols left is:

${\text{0.156 mols Cu"^(2+) - "0.00364 mols Cu}}^{2 +}$

$= {\text{0.152 mols Cu}}^{2 +}$

And thus, the concentration leftover is:

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