Is it possible for Cu to reduce Cu2+?

Question

I tried a slightly different set up of a Galvanic cell by inserting the $\ce{Zn}$ electrode into the $\ce{CuSO4}$ solution and the $\ce{Cu}$ electrode into the $\ce{ZnSO4}$ solution. This gave a voltage reading of 0.85 V. I also observed that the $\ce{Zn}$ electrode was getting coated with copper as the reaction progressed.

I had two hypotheses in mind -

1. The reaction is limited to only one half cell: the $\ce{Zn}$ electrode is oxidised and the $\ce{Cu^2+}$ in the solution is reduced. This explains the coating on the $\ce{Zn}$ electrode, but it does not explain the voltage. Doesn't the voltage mean that there is some interaction between the two cells?
2. This is my main question: Is it possible that the $\ce{Cu}$ atoms on the electrode are losing electrons that then travel through the circuit and reduce the $\ce{Cu^2+}$ ions in the second half cell? A test for $\ce{Cu^2+}$ ions in the $\ce{ZnSO4}$ solution after the reaction had progressed was inconclusive.

What exactly is happening in this set up?

Answer 1

Zinc is higher in the reactivity series than copper, so simply by placing a piece of zinc metal into a copper sulphate solution, you will get a displacement reaction where the zinc metal loses electrons and becomes $\ce{Zn^2+}$ ions, and $\ce{Cu^2+}$ ions gain those two electrons to become copper metal. This will happen until the zinc electrode is coated with copper and no more zinc ions can get into the solution.

The standard half-cell potentials are measured using standard 1M solutions, and deviations from this concentration cause the equilibria to shift according to Le Chatelier's principle:

$\ce{ Cu^{2+}_{(aq)} + 2e^{-}<=> Cu_{(s)}} \quad E^{o}=+0.337V$

Since the $\ce{[Cu^{2+}_{(aq)}]}$ is zero for the copper electrode, this system is far from equilibrium and the half-cell potential will be closer to zero. In the other half of the system, the $\ce{[Cu^{2+}_{(aq)}]}$ is less than 1M due to the reaction with the zinc electrode, but it is still significantly greater than zero. The result is a slight potential for copper metal to become ions and for the electrons to travel through the circuit to reduce some of of the copper(II) ions in the other half cell. Using the Nernst equation for this isn't valid since concentrations close to 1M are too high for accurate results on one hand (concentrations greater than $10^{-3}$M deviate from experimental values), and the use of the equation is on the assumption that the ion is in equilibrium. If we use it naively we do get something:

$E_{cell} = E^{o}_{cell} - (\frac{0.059}{2})log(Q) \approx 0-0.0295log(\frac{\ce{[0.0000000001]}}{\ce{[0.5]}}) = 0.37V$

As the approximation of the copper(II) ion concentration approaches zero the potential difference for the cell approaches infinity, which goes some way to demonstrate this is only valid within limits since you don't get a 100,000 Volts electric shock when you put a piece of metal into pure water. Nevertheless it describes a trend and supports the Le Chatelier interpretation.

However, if the circuit is connected using a voltmeter this has a sufficiently high resistance to prevent current from flowing, so although there is potential for the copper to reduce itself through the cell, this can only happen if electrons can flow. Even without the high resistance the reaction of the copper metal forming ions may be so slow, that very little current flows and few ions are formed which may explain the inconclusive test result.