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Topic: Cu(NO3)2 + 2NaI help, please.  (Read 14138 times)

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Offline magician13134

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Cu(NO3)2 + 2NaI help, please.
« on: November 02, 2009, 11:15:23 PM »
In chem class we did this reaction:
Cu(NO3)2 + 2NaI     ->     2NaNO3 + CuI2
And were told our net ionic equation was correct:
2Cu2+ + 4I-     ->    2CuI + I2

Then we were asked:
"I2 is a better oxidizing agent than Cu2+, does this fit your observations? " (We also verified with hexane that there was I2 in the filtrate...)
It looks to me that Iodine is losing electrons, not gaining them. Someone mentioned that we had to "flip the equation", but I have NO idea why, it doesn't make sense. Can anyone please help me understand this? And... Sorry for the late notice, but I would love to know this by about 9AM tomorrow... Thanks :)

Offline AWK

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Re: Cu(NO3)2 + 2NaI help, please.
« Reply #1 on: November 03, 2009, 01:30:10 AM »
THis is a redox reaction
Cu2+ => Cu+
I- => I2
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Offline Borek

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Re: Cu(NO3)2 + 2NaI help, please.
« Reply #2 on: November 03, 2009, 03:45:30 AM »
What is solubility of CuI? Do you know LeChatelier's principle?
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Offline magician13134

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Re: Cu(NO3)2 + 2NaI help, please.
« Reply #3 on: November 03, 2009, 09:02:52 AM »
Right... That's what I thought the Redox reaction was... But according to the question, Iodine should be gaining electron(s), right?

And I am not familiar with LeChatelier's principle, no, I'll look into that though.

Thanks for both responses!

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