# Need help with electrochemistry lab

1. Nov 27, 2012

### qpham26

1. The problem statement, all variables and given/known data

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
2. Relevant equations
Cu2+ + 2e- ⇔ Cu(s) Eo = 0.34V

Ag+ + e- ⇔ Ag(s) Eo = 0.80V

Zn2+ + 2e- ⇔ Zn(s) Eo = -0.76V

Ecell = Eocell - $\frac{0.0592}{n}$logQ
3. The attempt at a solution
First question I have, so the value that we measure, is that the total Ecell ?
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)
Eo = 0.76 + 0.34 = 1.1 V

And now with [Zn2+] = 0.10M and the measured Ecell 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.

Last edited: Nov 28, 2012
2. Nov 29, 2012

### Staff: Mentor

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.

3. Nov 29, 2012

### qpham26

For Zn and Cu

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

I plugged it into wolfram alpha, so the answer cant 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$\frac{[Cu2+}{[Ag+]^2}$

4. Nov 29, 2012

### Staff: Mentor

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?

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

5. Nov 29, 2012

### qpham26

Hi Borek, what doesnt look Ok beside the Latex format?
was the chemical eq. correct?
and is the Nernst eq. set up right?

6. Nov 29, 2012

### Staff: Mentor

I was referring to formatting only.

7. Nov 29, 2012

### 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?