Why is CuI2 not stable in solution while other copper(II) halide salts are?

Question

Why is $\ce{CuI2}$ (copper (II) iodide) not stable in solution while other copper(II) halide salts are? And how can you tell this from the redox potentials?

Answer 1

It is common knowledge that copper (II) iodide ($\ce{CuI2}$) does not exist. Thus, your question "why is $\ce{CuI2}$ not stable in solution?" is clear on that, but unlike this previous question, you wasn't clear about what happens (or what you have observed). I have no knowledge of what would happen when some one add $\ce{KI}$ solution to $\ce{CuSO4}$ solution. Yet, I can speculate what would happen by considering few redox half reactions:

$$\ce{I2 + 2e- <=> 2I-} \qquad E^\circ = \pu{0.536 V} \tag{1}$$ $$\ce{Cu^2+ + 2e- <=> Cu} \qquad E^\circ = \pu{0.342 V} \tag{2}$$ $$\ce{Cu^2+ + e- <=> Cu+} \qquad E^\circ = \pu{0.153 V} \tag{3}$$

For convenience, let;s rewrite equation $(1)$ in oxidation format:

$$\ce{2I- <=> I2 + 2e-} \qquad E^\circ = \pu{-0.536 V} \tag{4}$$

Sum of $(2)$ and $(4)$ would gives the appropriate redox equation:

$$\ce{Cu^2+ + 2I- <=> I2 + Cu_{(s)}} \qquad E_\mathrm{EMF}^\circ = \pu{-0.194 V} \tag{5}$$

Similarly, sum of $(3)\times 2$ and $(4)$ would gives the appropriate redox equation:

$$\ce{2Cu^2+ + 2I- <=> I2 + 2Cu+} \qquad E_\mathrm{EMF}^\circ = \pu{-0.383 V} \tag{6}$$

The excess $\ce{I-}$ would react with $\ce{Cu+}$ and $\ce{CuI}$ would precipitate as a result:

$$\ce{Cu+ + I- -> CuI_{(s)}} \qquad K_\mathrm{sp} = 1 \times 10^{-12} \tag{7}$$

Therefore, albeit forward reactions of equations $(5)$ and $(6)$ are not thermodynamically favorable (e.g., for equation $(5)$, $K_\mathrm{eq} \approx 2.7 \times 10^{-7}$), both forward reactions persist and they would be driven to completion because of one product separate as a solid (Le chatelier principle).

Let's now consider oxidation half reactions of other three halides:

$$\ce{2F- <=> F2 + 2e-} \qquad E^\circ = \pu{-2.866 V} \tag{8}$$ $$\ce{2Cl- <=> Cl2 + 2e-} \qquad E^\circ = \pu{-1.358 V} \tag{9}$$ $$\ce{2Br- <=> Br2 + 2e-} \qquad E^\circ = \pu{-1.087 V} \tag{10}$$

If $\ce{Cu^2+ -> Cu+}$ reduction happens, the respective three redox reactions are:

$$\ce{2Cu^2+ + 2F- <=> F2 + 2Cu+} \qquad E_\mathrm{EMF}^\circ = \pu{-2.713 V} \tag{11}$$ $$\ce{2Cu^2+ + 2Cl- <=> Cl2 + 2Cu+} \qquad E_\mathrm{EMF}^\circ = \pu{-1.205 V} \tag{12}$$ $$\ce{2Cu^2+ + 2Br- <=> Br2 + 2Cu+} \qquad E_\mathrm{EMF}^\circ = \pu{-0.934 V} \tag{13}$$

And if $\ce{Cu^2+ -> Cu}$ reduction happens, the respective three redox reactions are:

$$\ce{Cu^2+ + 2F- <=> F2 + Cu} \qquad E_\mathrm{EMF}^\circ = \pu{-2.524 V} \tag{14}$$ $$\ce{Cu^2+ + 2Cl- <=> Cl2 + Cu} \qquad E_\mathrm{EMF}^\circ = \pu{-1.016 V} \tag{15}$$ $$\ce{Cu^2+ + 2Br- <=> Br2 + Cu} \qquad E_\mathrm{EMF}^\circ = \pu{-0.745 V} \tag{16}$$

Since values of $E_\mathrm{EMF}^\circ$s are so high negatives, these reactions are practically not happen. For example, $K_\mathrm{eq}$s are in the range of $6.14 \times 10^{-26}$ (for eq. $(16)$) and $1.55 \times 10^{-92}$ (for eq. $(11)$). Also keep in mind that, contrary to $\ce{CuI}$, $\ce{CuBr}$ and $\ce{CuCl}$ are also water soluble and $\ce{CuCl}$ is water soluble as well as hygroscopic.