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Topic: Changing Oxidation States and their products  (Read 8653 times)

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Offline CopperSmurf

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Changing Oxidation States and their products
« on: October 20, 2007, 10:29:15 PM »
Hi. I'm new around here and just starting out in inorganic chemistry. Just wondering if someone could help me with the products of the following reactions, which involve oxidation states (I'm guessing the products so plz correct me if I'm wrong): 
HCl + KNO2 -->                 NO (g) + KCl
KI + KNO2 -->                   KI + KNO2      (no reaction)
KI + KNO2 + H2SO4 -->      NO (g) + HI + K2SO4
CrCl3 + NaClO + HCl -->      NaCl + H2O + Cr + Cl2 (g)
KI + CuSO4 -->                 I2 (g) + CuI (s) + K2SO4
(g) = gas , (s) = solid
I don't need the equations to be balanced because the problem I'm having is how do I really know if those are the products that formed and how would I know for sure? (I can't make these)

Offline FeLiXe

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Re: Changing Oxidation States and their products
« Reply #1 on: October 21, 2007, 06:17:11 AM »
balancing is always sensible. in the first case you would notice that your reaction is not possible. there will be no reaction

2nd is probably no reaction

in the 3rd one you need to oxidise the I- to I2 and then water will form as you will notice when you balance the equation. by the way concentrated sulfuric acid may as well oxidise KI. the question is not very well chosen

the Cr one does not work with stabilities. i am not quite sure what they are looking for

last one: maybe
Math and alcohol don't mix, so... please, don't drink and derive!

Offline CopperSmurf

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Re: Changing Oxidation States and their products
« Reply #2 on: October 21, 2007, 07:32:51 AM »
I'm not sure what you mean by "the Cr one does not work with stabilities"  ??? is the equation making it more unstable?

And about the last one, heehee I was also thinking of "maybe" since it doesn't make sense to me when I use the Frost Diagram because there should be enough energy (I think, if I'm reading the diagram right) for the reaction to proceed but confused on what forms as a product...

By the way, thanks for such a speedy reply!

Offline FeLiXe

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Re: Changing Oxidation States and their products
« Reply #3 on: October 21, 2007, 08:30:56 AM »
sure, but actually I just happened to look at this forum again after a long time

I am saying that Cr(III) could never oxidise Cl-. I guess that CrCl3 is oxidised to Cr2O52- and hypochlorite is reduced to Cl-

the trick with Cu(I) is that CuI is a very stable compound. that's why this is formed in many reactions
Math and alcohol don't mix, so... please, don't drink and derive!

Offline CopperSmurf

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Re: Changing Oxidation States and their products
« Reply #4 on: October 21, 2007, 10:00:00 AM »
Thanks!  :D

Offline CopperSmurf

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Re: Changing Oxidation States and their products
« Reply #5 on: October 21, 2007, 08:56:44 PM »
CrCl3 + NaClO + HCl -->
this part still stumps me though... pretty sure there should be chlorine gas (Cl2) but what happens to the Cr?

Offline Borek

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Re: Changing Oxidation States and their products
« Reply #6 on: October 22, 2007, 02:45:25 AM »
As FeLiXe already wrote - Cl- is the only stable chlorine form produced here. Chromium is most likely oxidized to chromate, or even dichromate. That will depend on the solution pH.
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