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?


  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.
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    $\begingroup$ The normal potentials are valid if and only if the concentrations of the ions are $1$ M. It is not the case here. For example, the $\ce{Cu^+}$ concentration is never as high as $1$ M. $\endgroup$
    – Maurice
    Dec 4, 2022 at 15:51
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    $\begingroup$ chemistry.stackexchange.com/questions/53768 $\endgroup$
    – andselisk
    Dec 4, 2022 at 16:30
  • $\begingroup$ When potassium iodide, is added to a copper sulphate solution, a redox reaction occurs in which the iodide ion converts cupric ion, to cuprous ion, (blue to white precipitate) before being oxidised to iodine gas. $\endgroup$ Dec 4, 2022 at 17:45
  • $\begingroup$ See this link: chemistry.stackexchange.com/questions/138362/… $\endgroup$
    – AChem
    Dec 5, 2022 at 3:15

1 Answer 1


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 $\ce{I2}$.

  • $\begingroup$ Those are the standard potentials. The potentials are for one mole of electrons regardless of how the equation is written. DeltaG[0] = E[0]nF accounts for the energy change with different equations. $\endgroup$
    – jimchmst
    Dec 6, 2022 at 3:42
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    $\begingroup$ Well, whatever the reason for downvoting the post was, its formatting, or rather lack of thereof, would be most obvious one. Another issue is grumbling about downvotes is pointless, or even detrimental for the site. $\endgroup$
    – Mithoron
    Dec 7, 2022 at 16:23
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    $\begingroup$ It would be great if you put some effort into writing chemical equations in a more readable format, especially since the site does make the necessary tool available and it's not difficult to learn. $\endgroup$
    – Buck Thorn
    Dec 7, 2022 at 17:31

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