Why does the reaction between copper(II) ion and iodide ion proceed even though the E_cell of the set up is negative?


> 
> (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


