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(iii) Another example of a reaction that does not agree with the theoretical prediction is the reaction between $\mathrm{Cu}^{2+}$ and $\mathrm{I}^{-}$ according to the equation below. $$ 2 \mathrm{Cu}^{2+}(\mathrm{aq})+4 \mathrm{I}^{-}(\mathrm{aq}) \rightleftharpoons 2 \mathrm{CuI}(\mathrm{s})+\mathrm{I}_{2}(\mathrm{aq}) \quad E_{\mathrm{cell}}^{\theta}=-0.39 \mathrm{~V} $$ White precipitate of $\ce{CuI}$ is formed during the reaction. Suggest a reason why the reaction proceeds despite the $E_{\text {cell }}^{\theta}$ being less than $0 \mathrm{~V}$.

The continuous precipitation of $\ce{CuI}$ from the solution will shift the equilibrium position to the right and drive the reaction forward.

Can I please ask what the words in italics mean? Because to my understanding solids are normally not considered in equilibrium calculations as they are assumed to have a constant concentration

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    $\begingroup$ It is answered here: rod.beavon.org.uk/feasibility.htm quite accurately. Read until the end. $\endgroup$
    – ACR
    Commented Aug 15, 2021 at 3:40
  • $\begingroup$ The problem is misstated, possibly by the book. The reference given in M. Farooq's comment (rod.beavon.org.uk/feasibility.htm, repeated here in case the other comment is lost) shows that in fact the negative cell potential applies to the reaction $\ce{2 Cu^{2+} + 2I^- -> 2 Cu^+ + I2}$ with all species in solution, not including the precipitation of the copper(I) iodide. $\endgroup$ Commented Aug 15, 2021 at 11:26
  • $\begingroup$ @M.Farooq so if I understand it right, it is due to the fact that the CuI is extremely insoluble and readily precipitated that leads to basically the Cu+ concentration always being extremely low, which is what shifts the eqm of the reaction between Cu2+ and I- to the right and this enables the reaction to proceed despite having a negative E knot? On a similar note, can I ask where can I read up on the relationship between Ksp and E cell of reactions? Or is it just that I have to infer it based on the question and the context. Thank you. $\endgroup$ Commented Aug 15, 2021 at 13:11
  • $\begingroup$ @CaptainTeemoOnDuty, You are right. Another key point is that all the standard values of electrode potential that you see in textbook tables are tabulated under standard conditions, i.e., all concentrations are 1 M. This is not the case, when copper (I) is precipitating as iodide. $\endgroup$
    – ACR
    Commented Aug 15, 2021 at 13:23

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