Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: kensher on September 04, 2016, 03:19:40 PM
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Hello.
I've been struggelig with balancing redox reactions from my book. I try to balance them using the oxidation number method. The point were I struggle, is when I suppose to decide which coefficient numbers I should put where.
It would be extremely helpful if somebody could share which questions I should ask in order to figure out the right answer. I've attached some practice problems where I begin solving, but come up short.
Thanks in advance!
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I am not sure I follow what your problem is. Can you explain it step by step using second or third example? (The first one, with disproportionation, is rather specific, so it is not a good starting point for learning).
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I am not sure I follow what your problem is. Can you explain it step by step using second or third example? (The first one, with disproportionation, is rather specific, so it is not a good starting point for learning).
I gave it another shot. Same disappointing result. I just don't know how to set the correct coefficient number.
1) I first check if it's an equal amount of the oxidized or reduced atom on both sides of the equation, right?
2) I make sure that there is an equal number of electrons which have been oxidized and reduced. And then, I'm clueless.
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O2 as molecule needs 2x2 electrons.
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Concderning your first reaction:
NO2 + H2O = HNO3 + NO
This is a disproportionation reaction. Using oxidation numbers ypou may usde two methods:
1. write NO2 twice on the right side of equation - the firdt molecule is oxidized, the secon one - reduced.
or
2. Balance electrons from the right side of equation to the left one. Reaction becomes standard one.
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Concderning your first reaction:
NO2 + H2O = HNO3 + NO
This is a disproportionation reaction. Using oxidation numbers ypou may usde two methods:
1. write NO2 twice on the right side of equation - the firdt molecule is oxidized, the secon one - reduced.
or
2. Balance electrons from the right side of equation to the left one. Reaction becomes standard one.
Thanks for the reply. Think I actually managed to solve the other equation now after your tips.
I was just wondering if I should had written 4 at the first round circle or if the standard is to write it in the second red circle (see attachment)?
Can you please elaborate on the two methods you mentioned? With an super short example would be ideal.
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Just do normal balancing like to that for oxidation of ammonia
Method 1
NO2 + NO2 + H2O = NO + HNO3
Method 2
NO + HNO3 = NO2 + NO2 + H2O
...and try using editor of this forum at lest for the final result.
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See if this doesn't help:
http://www.chembuddy.com/?left=balancing-stoichiometry&right=oxidation-numbers-method
Note: of three methods mentioned on the ChemBuddy page (see the links in the left menu), ON method is probably the most lousy one.
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Thanks to both of you for answering.
I will check out the resources provided, and give it another go ☺
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Try this method on all redox reactions for balancing :
1.Write two half reactions separately .
2. Balance all atoms except H/O if they are not undergoing any change in ON for each half reaction.
3.Balance O by adding H2O
4.Balance H by adding H +
5.Balance charges by adding electrons
6 Equalize the number of electrons in both half reaction by multiplying it with a factor .
7.Now add both balanced half reactions cancelling out common things on both sides of the reaction.
8. If you want to make it in basic medium just add OH- = H+ in the final equation.
This is one of the simplest method taking care of charges and number of atoms .
Try it..