Wierd reaction?

  • Thread starter pivoxa15
  • Start date
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Main Question or Discussion Point

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 dosen't involve Ag. Has the book made mistakes?
 

Answers and Replies

Gokul43201
Staff Emeritus
Science Advisor
Gold Member
6,987
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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).
 
2,234
1
I see. Thanks
 
Borek
Mentor
28,129
2,636
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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