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Topic: Redox Reaction Balancing  (Read 3274 times)

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Offline ashishK

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Redox Reaction Balancing
« on: April 21, 2015, 11:50:48 AM »
The equation is: H2O2 + I2 = HIO3 + H2O

I know that the I2 molecule in LHS undergoes oxidises and the total change in oxidation number is 10. But the confusion is that the peroxide atom in H2O2 undergoes reduces and form oxygen atoms in two compounds i.e. HIO3 and H2O. So which compound containing the oxygen atom should I choose?
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Online Borek

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Re: Redox Reaction Balancing
« Reply #1 on: April 21, 2015, 02:04:21 PM »
Why don't you balance just by inspection?
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Offline ashishK

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Re: Redox Reaction Balancing
« Reply #2 on: April 22, 2015, 12:12:21 PM »
Actually we got a chapter on "Redox Reaction", so this problem is meant to be solved by Oxidation number method, that's the reason.
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Offline Hunter2

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Re: Redox Reaction Balancing
« Reply #3 on: April 23, 2015, 01:42:26 AM »
There is a receipe to solve this kind of questions.

Oxidation if oxygen is involved in acidic envirement:

Add water on educt side and get hydrogenions on product side.

Reduction works vise versa.

So try now your example. How you go from iodine to iodine acid and from Peroxide to Water.

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