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Topic: Electrochemistry balancing using half reaction method  (Read 3624 times)

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Offline 12312521521512

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Electrochemistry balancing using half reaction method
« on: January 16, 2015, 10:39:16 PM »
the question says to balance the following using the half reaction method

TeO32-+N2O4 -> TeNO3-

OK so for the TeO32-
I know that the charge on O is -2 so Te has to be 4
since 4-6=-2

For N2O4 N is 8

On the product side I said that NO3 has a charge of -1 so Te has a charge of +1. Then did 1+(-2x3)+x=-1 which gave me a value of 5 for N

I have no idea how to write the half reactions as there is only one term on the product side.
I tried doing N2O4 --> 2TeNO3
TeO3 --> TeNO3 but there are missing terms when set up that way so it doesn't balance? How do you go about solving this question

Offline Borek

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Re: Electrochemistry balancing using half reaction method
« Reply #1 on: January 17, 2015, 03:25:07 AM »


So, which one it is?

For N2O4 N is 8

No, oxidation number of N is not 8. Are you sure you have not forgot oxidation number is a property of an atom, and there are two N atoms in the molecule?

Can't say the chemistry behind this reaction (nor behind the Te/NO3- compound) makes sense to me.

I would try to treat TeNO3 (or whatever its expected formula is) as a salt - that is, separately Te (or Te+) and separately NO3-. After you have balanced half reactions just combine them, and then combine as many Te/NO3 pairs as possible.
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Offline mjc123

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Re: Electrochemistry balancing using half reaction method
« Reply #2 on: January 17, 2015, 04:23:05 PM »
The equation as written is impossible to balance. The RHS has exactly one negative charge and three O atoms per Te atom; the LHS has two negative charges and more than 3 O atoms per Te. There is no way you can balance Te, O and charge simultaneously. Something must be missing.

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