Help in The Oxidation Number Method ?

1. Mar 2, 2013

SC0

Help in "The Oxidation Number Method"?

1. The problem statement, all variables and given/known data

I just learned how to balance the redox reactions with this method, but there is something I don't understand ...
Look at this example :
Cu+HNO3 = Cu(NO3)2+NO2+H2O
after making the number of electrons gained and lost equal we get this equation :
Cu+2HNO3 = Cu(NO3)2+2NO2+H2O
Now the number of electrons gained and lost is the same but we must continue balancing the equation and so we get at last :
Cu+4HNO3 = Cu(NO3)2+2NO2+2H2O
Didn't we just ruin the equation again ? ( the number of electrons lost and gained isn't the same ) Or am I wrong ?

2. Relevant equations

Cu+HNO3 = Cu(NO3)2+NO2+H2O

2. Mar 2, 2013

Staff: Mentor

Re: Help in "The Oxidation Number Method"?

Please elaborate on what you are doing, step by step.

My guess is that you should try to balance net ionic reaction first, and it will become obvious what the problem is (or rather - why there is no problem). But I can be wrong, hard to say without details.

3. Mar 3, 2013

SC0

Ok here is the original equation :
Cu+HNO3 = Cu(NO3)2+NO2+H2O

Now according to "The Oxidation Number Method", I must balance the charges first
and if you look at the equation you will find that :
Cu -> lost 2electrons
N -> gained 1electron

Then balance the charges by multiplying N by 2, so I multiplied HNO3 on the reactant side and multiplied NO2 on the product side
The equation now should be like this :
Cu+2HNO3 = Cu(NO3)2 + 2NO2 + H2O

Now charges are balanced, but the equation isn't, and to do this we should make a little change, it will become like this :
Cu+4HNO3 = Cu(NO3)2 + 2NO2 + 2H2O

Now the equation is fully balanced but we multiplied 2HNO3 again and it became 4HNO3 ...
My question is didn't we - by doing that - unbalanced the charges again ?
Hope I made it clear enough
Thanks.

4. Mar 3, 2013

Staff: Mentor

As I signaled earlier - balance net ionic reaction first.