Why does Cu2+ react with I- to give CuI?

[phantomvommand]

Why does Cu2+ react with I- to give CuI?
According to my teacher,
2Cu2+ + 2I- <--> 2Cu+ + I2-- reaction 1, Ecell < 0
2Cu+ + 2I- --> 2CuI (s) -- reaction 2
Fall in [Cu2+] due to reaction 2 leads to forward reaction in reaction 1 being favoured (by LCP), so Ecell becomes more positive, and thus reaction 1 is now spontaneous and the reaction proceeds fully to form CuI(s).
But doesn't [I-] in reaction1 also fall due to reaction 2? Why does LCP apply, and why does it shift equilibrium rightwards?

 

[Borek]

CuI (s)

That's already an answer, but additionally: check the Ksp.

 

[phantomvommand]

That's already an answer, but additionally: check the Ksp.

I understand that CuI will form. But I was thinking that the reaction would stop after running out of (the very little) Cu+ to precipitate away, since no more Cu+ is formed due to the non-spontaneous reaction 1. I am thinking no more Cu+ forms because LCP doesn't favour either side of reaction one, since both Cu+ and I- are used up in reaction 2.

 

[mjc123]

Fall in [Cu2+] due to reaction 2 leads to forward reaction in reaction 1 being favoured (by LCP),

That should be "fall in [Cu⁺] due to reaction 2" - not [Cu²⁺].
Because CuI has very low solubility, the concentration of Cu⁺ is very small. Essentially, as soon as you make any Cu⁺, it is removed by precipitation, so reaction 1 continues to go rightwards.

 

[phantomvommand]

That should be "fall in [Cu⁺] due to reaction 2" - not [Cu²⁺].
Because CuI has very low solubility, the concentration of Cu⁺ is very small. Essentially, as soon as you make any Cu⁺, it is removed by precipitation, so reaction 1 continues to go rightwards.

yes, apologies for the typo. So reaction 1 goes rightwards because the effect of a fall in [Cu+] is much greater than for [I-], given the extremely low [Cu+], so LCP still favours the rightward direction? Or does it go rightwards 'by chance', ie the reaction continues when some Cu+ is formed again by chance

 

[mjc123]

No "chance". It's an equilibrium; leftward and rightward reactions are both occurring all the time. The very low [Cu⁺] means the rate of the leftward reaction is low.

 

[phantomvommand]

No "chance". It's an equilibrium; leftward and rightward reactions are both occurring all the time. The very low [Cu⁺] means the rate of the leftward reaction is low.

I think you are saying that [Cu+] forms due to the rightward reaction (which exists, but just happens to a very small extent). And then the [Cu+] is precipitated away. I would agree with this.
But does the LCP argument still stand though? That for the same fall in amount of Cu+ and I- due to reaction 2, the effect of the fall is much greater for [Cu+] due to the extremely low quantities, and so the rightward reaction is favoured, in addition to the already existing (but very slow) rightward reaction. I ask this because my teacher explicitly said the Ecell 'turns positive' for reaction 1 due to the forward reaction now being favoured by LCP.

 

[mjc123]

Yes. You can quantify this using the K_sp and the Nernst equation.