Evaluate Q in Nernst Equation?

Question

The question goes as follows:
"Standard cell potentials are determined with 1.0 M solutions of ions. In this experiment, we will use 0.10 M solutions of Zn2+, Cu2+, Pb2+, and Ag1+ to minimize the amount of hazardous waste generated. In this series of questions, you will figure out what effect this will have on the voltages you observe.

Consider a cell consisting of a Cu2+/Cu couple, and a Ni2+/Ni couple.

a. Starting with 1.0 M solutions of ions (standard conditions), evaluate Q in the Nernst Equation."

OK, so, Q=[products]^coeff/[reactancts]^coeff. I'm figured out th equation to be 2Cu^2+ + Ni<---> Ni^2+ + 2Cu^1+ from the Standard Reduction Potential Table. The Q equation would be= [Ni]/[Cu^2+]^2, right? but I have no idea how to get the Molarities!! Is Cu^2+ 0.1M??? What's Ni?? PLease help!

Answer 1

Not really.
First of all, let me write the releted standard reduction potentials here:
Cu(2+)(aq) + e− → Cu+(aq) Eo = +0.16 V
Cu(2+)(aq) + 2e− → Cu(s) Eo = +0.34 V
Cu+(aq) + e− → Cu(s) Eo = +0.52 V
It is very clear the potential to reduce Cu(2+) to Cu+ is much smaller than to reduce Cu+ to Cu, so the overall effect is to directly reduce Cu2+ to Cu. I would not deny that there may be a very small cncentration of Cu+ formed in the solution, but that is covered already by the overall reaction:
Cu(2+)(aq) + 2e− → Cu(s) Eo = +0.34 V
You only use Cu(2+) --> Cu+ when you can avoid Cu+ to be further reduced to Cu.

Now, even if the reaction would be:
2Cu(2+) + Ni<---> Ni(2+) + 2Cu+
Q should NOT be: [Ni]/[Cu(2+)]^2, rather Q should be: [Ni(2+)]*[Cu+]^2/[Cu(2+)]^2. Please do not forget [Cu+]^2, and do not put solid metal "Ni" into the equation.
The correct reaction is:
Cu(2+) + Ni(s) <==> Ni(2+) + Cu(s)
and thus the correct Q:
Q = [Ni(2+)] / [Cu(2+)]
The question says: "Starting with 1.0 M solutions of ions ", that means for both Cu(2+) and Ni(2+), so Q = 1.

You for sure need to study hard.