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