While I agree generally with curiousbrain’s answer, I don’t think that the charge density alone is the culprit.
Rather, $\ce{Cu+}$ is a $\mathrm{d^{10}}$ ion which therefore has no real preference for any ligand shell — much like zinc(II). All 10 d-electrons will always populate antibonding orbitals with respect to the $\ce{M-OH2}$ coordinate bond, weakening them and creating a highly labile ligand sphere.
On the contrary, $\ce{Cu^2+}$ has a rather well defined strongly Jahn-Teller distorted octahedral ligand sphere. While the lower $\mathrm{d}_{xy}$, $\mathrm{d}_{xz}$, $\mathrm{d}_{yz}$ and $\mathrm{d}_{z^2}$ orbitals are antibonding[1] and fully populated, the very strongly antibonding $\mathrm{d}_{x^2 - y^2}$ orbital is only populated by a single electron, strengthening the copper-water interactions and making the ligand sphere less labile. This may well be a reason why the displacemen of an electron is favourable in aquaeous media.
Notes:
[1]: If one looks at the orbital scheme of a typical octahedral complex, additionally includes π interactions between ligands and metal, and finally also considers the Jahn-Teller distortion, it becomes evident that all metal-centred orbitals are antibonding to a certain extent. However, $\mathrm{d}_{x^2-y^2}$ is much more strongly antibonding than all other d-orbitals.