Chemical Forums

Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: renots on February 13, 2006, 11:44:50 AM

Title: balanced chemical equation for the reaction of Cu(NO3)2 (aq) with KI (aq)
Post by: renots on February 13, 2006, 11:44:50 AM

sry i cant figure the answer to this problem out and its due b4 my lab in like 2 hrs

Write the balanced chemical equation for the reaction of Cu(NO₃)₂ (aq) with KI (aq) including states. Note: Two of the products are CuI and I₂.  Which of the above reactants change oxidation states?  What are the changes?

if anyone could help i would greatly appreciate it.

Title: Re:easy question
Post by: Borek on February 13, 2006, 11:55:04 AM

Cu²⁺ + I^- -> CuI + I₂

This is not balanced, but that's relatively easy (note: charge must be balanced as well as atoms). Write oxidation numbers near all ions/atoms - which ones are changing?

Title: Re:easy question
Post by: renots on February 13, 2006, 12:05:31 PM

what happens to the K and the NO? do they get used up or something?

Title: Re:easy question
Post by: Mitch on February 13, 2006, 12:44:10 PM

K+ is just a spectator ion, but if you must you can think of KNO3 being formed, but its not accurate way to think.

Title: Re:easy question
Post by: renots on February 13, 2006, 12:52:59 PM

Cu(NO3)2 + KI -----> CuI + I₂ + X

would the X be kno3, and how would i balance that?  could i take the kno3 completely out of the equation?  when im balancing i just end up going in a never ending circle of increasing coefficiencts

Title: Re:easy question
Post by: AWK on February 14, 2006, 02:52:16 AM

This is a quite easy redox reaction, which can be split into two steps. The fiirst one is a double exchange reaction
Cu(NO3)2 + 2KI = CuI2 + 2KNO3
The next step is a redox one
2CuI2 = 2CuI +  I2

After joining both reaction together (with proper coefficients):
2Cu(NO3)2 + 4KI =

Title: Re:easy question
Post by: Borek on February 14, 2006, 04:50:54 AM

Quote from: AWK on February 14, 2006, 02:52:16 AM

2CuI2 = 2CuI +  I2

Interesting thing is that this reaction equilibrium is driven by low CuI solubility. Comparison of potentials of both half-reactions (Cu²⁺/Cu⁺, I₂/I^-) suggests that no redox reaction should take place.