Using Reactions to Understand Chemistry: Examining CuI(s) Formation

[pivoxa15]

In my textbook, it had

2Cu2+(aq) + 4I-(aq) -> CuI(s) + I2(aq)

How does that work? First of all it's not balanced. It should be 2CuI(s). But how does CuI(s) form. It should be CuI+.

The book did say excess 4I- was usd to remove Ag+ since it was talking about back titrations. But the reaction as stated doesn't involve Ag. Has the book made mistakes?

 

[Gokul43201]

No, it's correct (except for the missing 2).

Cu(+2) is reduced to Cu(+1) and I(-1) is oxidized to I(0).

 

[pivoxa15]

I see. Thanks

 

[Borek]

Note: this reaction is interesting as its equilibrium is moved to the right thanks to the very low solubility of CuI.

 

[pivoxa15]

Note: this reaction is interesting as its equilibrium is moved to the right thanks to the very low solubility of CuI.

Borek
--
General Chemistry Software
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And the solubility is so low that CuI precipitates out as a solid.