Need help with electrochemistry lab

[qpham26]

Homework Statement

Simplified lab procedure:

I have 3 wells (from well-plate)
one with Cu(NO₃)₂ and a piece of Cu(s)
one with AgNO₃ and a small piece of Ag(s)
last one with Zn(NO₃)₂ and a piece of Zn(s)
Salt bridges are soaked with KNO₃

All of the solutions above have molarity of 0.10M

A few drops of 6M NH₃ (aq) was added to the well with Cu(s)

And I used the volt meter to measure the voltage between Cu and the other two
Cu and Ag : 0.521 V
Cu and Zn : 0.619 V

And they want me to calculate the concentration of Cu²⁺
using Nernst equation

Homework Equations

Cu²⁺ + 2e^- ⇔ Cu(s) E^o = 0.34V

Ag⁺ + e^- ⇔ Ag(s) E^o = 0.80V

Zn²⁺ + 2e^- ⇔ Zn(s) E^o = -0.76V

E_cell = E^o_cell - \frac{0.0592}{n}logQ

The Attempt at a Solution

First question I have, so the value that we measure, is that the total E_cell ?
and let do the first case, since both are used for calculation [Cu²⁺]
For Zn and Cu

Zn(s) + Cu²⁺(aq) ⇔ Zn²⁺(aq) + Cu(s)
E^o = 0.76 + 0.34 = 1.1 V

And now with [Zn2+] = 0.10M and the measured E_cell I just need to plug this into the Nernst equation to get Q.

The same method would be applicable to Ag. Assuming I got the above correctly.
However, the answer I got for [Cu2+] is very small, 5 x 10^(-18)

and When I do the same thing for Ag
i got a different value 8.5 x 10^-5
So i think there is something wrong.

thanks for your time.

 

[Borek]

Hard to say if you are not doing any mistake without seeing exact calculations, but the logic behind things you wrote so far looks OK to me.

 

[qpham26]

For Zn and Cu
the equation I had was

0.619 = 1.10 - \frac{0.059}{2}log\frac{[Zn2+}{[Cu2+]}

I plugged it into wolfram alpha, so the answer can't have any mathematical errors.


as for Cu and Ag
Cu(s) + 2Ag+ → Cu²⁺(aq) + 2Ag(s)
the equation is

0.521 = 0.46 - \frac{0.059}{2}log\frac{[Cu2+}{[Ag+]^2}

 

[Borek]

0.619 = 1.10 - \frac{0.059}{2}log\frac{[Zn2+}{[Cu2+]}

Looks OK, unless I am wrong as well.

That is, it doesn't look OK - why don't you format it as a whole using LaTeX?

0.619 = 1.10 - \frac{0.059}{2}\log(\frac{[Zn^{2+}]}{[Cu^{2+}]})

 

[qpham26]

Hi Borek, what doesn't look Ok beside the Latex format?
was the chemical eq. correct?
and is the Nernst eq. set up right?

 

[Borek]

I was referring to formatting only.

 

[qpham26]

Borek, If the set up was right, how come i am getting 2 different answer?
does it have to do with the measurement then?