# What is the relationship between the stability constant of a copper (II) electrolyte complex ion and the output voltage Ecell? [closed]

As a part of a school investigation I constructed a voltaic cell experiment where for each run I added a different ligands to the copper (II) sulfate half-cell electrolyte in a daniell cell to see if there was a change in the output voltage Ecell. When plotting a graph of this, it seemed that Ecell increased as the stability of the ligand increased (indicated by the stability constant Kc) however I am struggling to explain why. I presume its because the standard potential at the cathode Erhe increases with ligand stability due to greater oxidation.

• what are you talking about exactly? The electrolyte is already dissolved in water. Which ligand? are you talking about H2O? The Nernst Equation gives us the Open Circuit cell potential (are you measuring your cell potential at zero current?) It is expressed as $$Ecell = RT/(nF)*(ln(K/Q))$$ so it is a function of the concentrations and temperature – ChemEng Oct 6 '20 at 2:13

The redox potential of Copper does not depend directly on the stability of the complex Cu-ligand. It depends on the concentration of the residual $$\ce{Cu^{2+}}$$ ion in solution. And in any solution of $$\ce{CuSO4}$$, adding a ligand produces a decrease in concentration of the free ion $$\ce{Cu^{2+}}$$. So, according to Nernst's law, the potential $$\ce{Cu^{2+}/Cu}$$ becomes less positive.

For example, the standard redox potential of $$\ce{Cu^{2+}/Cu}$$ is + $$0.34$$ V. Let's consider the ligand chloride $$\ce{Cl-}$$. If enough chloride ions $$\ce{Cl-}$$ is added to a $$1$$ M $$\ce{Cu^{2+}}$$ solution so that at the end the excess of $$\ce{Cl}$$- ions is $$1$$ M, the complex $$\ce{CuCl4^{2-}}$$ will be formed, with a concentration $$1$$ M. The measured residual [$$\ce{Cu^{2+}}$$] concentration is $$2.5 ·10^{-6}$$ M. In this case, Nernst's law gives the following $$\ce{Cu^{2+}/Cu}$$ redox potential : $$\ce{E_{Cu^{2+}/Cu} = + 0.34 V + \frac{0.0592}{2}log(2.5·10^{-6}) = + 0.34 V - 0.0296·5.6 = +0.34 V -0.166 V = +0.174 V}$$ This +$$0.174$$ V is lower than the standard redox potential (+$$0.34$$ V).

The stability complex of the $$\ce{CuCl4^{2-}}$$ ion is : $$\ce{K = \frac{[CuCl4^{2-}]}{[Cu^{2+}][Cl^-]^4} = \frac{1}{2.5·10^{-6}·1^4} = 4·10^5 M^{-4}}$$

If another ligand is used, which makes a stronger complex with copper $$\ce{Cu^{2+}}$$, the stability constant is higher, and the redox potential will be still lower.

• Your first two sentences create false impression you may think standard redox potential of copper depends on Cu^2+ concentration. – Poutnik Oct 6 '20 at 14:57
• @ Poutnik. You are right. I will edit the coprrectkion. – Maurice Oct 6 '20 at 19:12

My speculation starts with a comment by ChemEng, who noted here that the Nernst Equation is essentially a function of concentration and temperature.

$$\ce{ Cu (s) + Cu(II)L <=> 2 Cu(I)L }$$

where the above reaction proceeding, given the general low solubility of Cuprous (absence a ligand) critically depends on the presence/stability of the highly soluble ligand L itself.

Also, a cited half-cell reaction in the Daniell cell is:

$$\ce{Cu(II)L + 2e− → Cu(s) }$$

Or, can it proceed, at least partially also, in steps dependent on solubility provided by the ligand:

$$\ce{Cu(II)L + e− → Cu(I)L }$$

$$\ce{Cu(I)L + e− → Cu }$$

So, again another possible example of the role of the select ligand in facilitating the cuprous (as an intermediate) concentration.

As such, my supposition is that the presence/stability of the soluble ligand impacts copper ion(s) concentration thereby impacting the Nernst equation.

[EDIT] Based on further research, my speculated explanation is likely too simple. Here is a more technical source on an advanced Daniell cell design. In recharging, note the mention of problematic Cu2O formation which is rectified, in part, by an increased Cu(NO3)2 electrolyte concentration.