Question

Calculate the [Cu2') remaining in 425 mL of a solution that was originally 0.366 M CuSO4 after passage of 2.68 A for 282 s and the deposition of Cu at the cathode.

Answer 1

Answer:

$[Cu]^{remaining}=0.357 M$

Explanation:

Hello,

In this case, we may use the following equation in order to compute the moles of copper that are processed, considering it goes from Cu⁰ to Cu²⁺, so two electrons are transferred:

$n_{Cu}=\frac{2.68C/s*282s}{96500C/mol*2} =3.9x10^{-3}mol$

After that, we compute the initial moles of copper in the solution, considering that the concentration of copper (II) equals the concentration of copper:

$n_{Cu,0}=0.366mol/L*0.425L=0.156mol$

In such a way, we can subtract the process moles to the initial moles to compute the remaining moles of copper:

$n_{Cu}^{remaining}=0.156mol-0.0039mol=0.152mol$

Finally, the concentration is:

$[Cu]^{remaining}=0.152mol/0.425L$

$[Cu]^{remaining}=0.357 M$

Regards.