I have learnt in the qualitative analysis of cupric ions that upon treating its aqueous solution with excess $\ce{KI}$, triiodide ions are formed which can be further tested using starch.
The explanation provided is that $\ce{Cu^2+}$ ions oxidise iodide ions to iodine which combines with additional $\ce{KI}$ to form $\ce{KI3}$.
$$\ce{2Cu^2+ + 2I- -> 2Cu+ + I2} \tag1$$ $$\ce{I2 + KI -> KI3 } \tag2$$
I looked up the reduction potentials of cupric to cuprous and iodine to iodide, and noticed that the latter is greater than the former.
I figured that maybe the reaction is reversible and lies to the left but the little amount of iodine formed is sufficient to turn starch blue black as it is a sensitive test.
But my doubt stems from the fact that it is generally regarded that cupric ions can oxidise iodide to iodine and also that cupric iodide does not exist (at least not in aqueous solutions). These statements would imply that the reaction (1) would lie to the right hand side which according to me should not be true.
Is it true that the reaction (1) actually lies to the right possibly due to formation of cuprous iodide which is a precipitate and the triiodide complex.
If not, then wouldn't it be incorrect for us to say that cupric ions oxidise iodide ions?
