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

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Discussion Overview

The discussion revolves around the formation of CuI(s) from Cu2+(aq) and I-(aq) in the context of a chemical reaction presented in a textbook. Participants examine the balancing of the reaction and the implications of the species involved, particularly in relation to the solubility of CuI.

Discussion Character

  • Technical explanation, Debate/contested

Main Points Raised

  • One participant questions the balance of the reaction and suggests that CuI should be represented as CuI+ instead of CuI(s).
  • Another participant asserts that the textbook reaction is correct aside from the missing coefficient of 2 for CuI.
  • A later reply notes that the reaction's equilibrium shifts to the right due to the very low solubility of CuI, leading to its precipitation as a solid.
  • There is a mention of the role of excess I- in the context of back titrations, although its relevance to the reaction is questioned.

Areas of Agreement / Disagreement

Participants express differing views on the correctness of the textbook reaction and its balance. There is no consensus on whether the representation of CuI is accurate, and the discussion remains unresolved regarding the implications of the excess I- and the role of Ag+.

Contextual Notes

Some assumptions about the reaction conditions and the definitions of the species involved may be missing. The discussion does not resolve the mathematical steps regarding the balancing of the reaction.

pivoxa15
Messages
2,250
Reaction score
1
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?
 
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No, it's correct (except for the missing 2).

Cu(+2) is reduced to Cu(+1) and I(-1) is oxidized to I(0).
 
I see. Thanks
 
Note: this reaction is interesting as its equilibrium is moved to the right thanks to the very low solubility of CuI.
 

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