Help in The Oxidation Number Method ?

[SC0]

Help in "The Oxidation Number Method"?

Homework Statement

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 ?
Someone explain please

Homework Equations

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

 

[Borek]

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.

 

[SC0]

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.

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.

 

[Borek]

As I signaled earlier - balance net ionic reaction first.

 

[DeeNos]

The Attempt at a Solution

The Oxidation Number Method is a technique used to balance redox reactions by assigning oxidation numbers to each element in the reaction and then using the changes in oxidation numbers to determine the number of electrons gained or lost. It is a useful tool for balancing reactions in which atoms change their oxidation state.

In the example given, Cu+HNO3 = Cu(NO3)2+NO2+H2O, the initial oxidation numbers are Cu=0, H=-1, N=+5, O=-2. After balancing the reaction, the oxidation numbers are Cu=+2, H=-1, N=+5, O=-2. This means that the copper atom has lost two electrons, while the nitrogen and oxygen atoms have remained unchanged.

The key to balancing redox reactions using the Oxidation Number Method is to ensure that the number of electrons lost is equal to the number of electrons gained. In this case, as you correctly pointed out, the number of electrons gained and lost is not the same in the final balanced equation. However, this is not a problem as long as the overall charge on both sides of the equation is balanced.

In this example, the overall charge on both sides of the equation is +2, so the number of electrons gained and lost does not need to be equal. This is because the overall charge is determined by the sum of the oxidation numbers, rather than the number of electrons gained or lost.

Therefore, the final balanced equation, Cu+4HNO3 = Cu(NO3)2+2NO2+2H2O, is correct and the Oxidation Number Method has been successfully used to balance the redox reaction. I hope this explanation helps clarify any confusion you may have had.