Why does copper(II) sulfate react with potassium iodide in aqueous solution?

According to the list of standard reduction potentials [1, p. 5-79] $E^\circ(\ce{Cu^2+/Cu^+}) = \pu{+0.153 V},$ while $E^\circ(\ce{I2/I^-}) = \pu{+0.5355 V}.$ Doesn't it mean that iodine has more tendency to get electron and form $\ce{I-}$? If yes, then why does the reaction between $\ce{CuSO4}$ and $\ce{KI}$ occur?


### Reference


1. Haynes, W. M.; Lide, D. R.; Bruno, T. J. *CRC Handbook of Chemistry and Physics: A Ready-Reference Book of Chemical and Physical Data*, 97th ed.; Taylor & Francis Group (CRC Press): Boca Raton, FL, **2016**. ISBN 978-1-4987-5429-3.




This is literally a complex reaction! If I remember correctly when these are mixed there is an immediate color change that I attributed to a fast complex formation. Someone must check this out. I think I remember this.


$\ce{Cu^{2+} + I- <=> CuI+}$ ;  

$\ce{CuI+ + e- <=> CuI}$


Combine these two reactions to give:


$\ce{Cu^{2+} +I- + e- <=> CuI } ; ~~~~ \pu{ E^\circ = +0.877 V}$ [Langes Handbook]


We have our oxidant!


$\ce{I- <=> 1/2 I2 + e-} ; ~~~~ \pu{ E^\circ = -0.535 V}$


Combine these two reactions to give:


$\ce{Cu^{2+} +2 I- <=> CuI + 1/2 I2 } ; ~~~~ \pu{ E^\circ = = +0.342 V}$


This predicts a mixed product. I can't find a way to convert the $\ce{CuI}$ to $\ce{Cu}$ and $\ce{I2}$.


 2