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

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?

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.