Balancing a chemical reaction with oxidation numbers

We have been teached how to balance chemical reactions with oxidation numbers in school, but somehow I can't understand it.


This is the equation to balance
$$\ce{Cu(s) + NO3−(aq) + H+(aq) -> Cu^2+(aq) + NO2(g) + H2O(l)}$$


At first I looked at the oxidation numbers. The only oxidation numbers that change are $\ce{Cu}:\ 0\ \rightarrow \ 2+$ and $\ce{N}:\ 1-\rightarrow \ 4+$, both are oxidized.


Then I'm a bit lost. In a typical redox reaction I would create partial reactions for both oxidation and reduction, but nothing is reduced in this reaction. I made two partial reactions with $\ce{Cu}$ and $\ce{N}$ being oxidized, but I think that it's not how it should be done. I came to a very different result than my book.


Is there anyone who knows how I should balance this reaction? I'm a bit lost.




The equation we need to balance is this:


Cu (s) + NO3− (aq) + H+ (aq) ⟶ Cu2+(aq) + NO2 (g) + H2O (l)


Using [simple oxidation state rules](https://en.wikipedia.org/wiki/Oxidation_state#Simple_approach_without_bonding_considerations)


The oxidation states on the left hand side:


* Cu is 0 - Elements are always 0
* O in NO3 is -2 (per atom), adding to -6 for 3 atoms of O
* Since the oxidation states of all the atoms in a molecule must add to the charge of the molecule, N is equal to:
-1 (charge of the molecule) - (-6) (charge of all the oxygen atoms). Hence, N is -1 - (-6) = +5
* Finally, H+ is +1, since the oxidation numbers of all monoatomic ions as the charge of the ions


The oxidation states on the right hand side:


* Cu is +2 (monoatomic ion)
* O is -2 per atom, which adds to -4 for 2 atoms of O
* N is +4, since the overall molecule is neutral
* O (in H2O) is -2
* H is +1 (H is +1 in all its compounds) for each atom, adding up to +2, which also accounts for the overall molecule charge of 0


Only the following oxidation states are changing:


* Cu from 0 to +2
* N from +5 to +4


Now we can split the equation into half-reactions:


* Cu ⟶ Cu2+ + 2e-
* NO3- + e- ⟶ NO2


Balancing the equations:


* Cu ⟶ Cu2+ + 2e-
* NO3- + e- + 2H+ ⟶ NO2 + H2O


Multiplying the equations:


* Cu ⟶ Cu2+ + 2e-
* 2NO3- + 2e- + 4H+ ⟶ 2NO2 + 2H2O


Balancing the equations together:


* Cu + 2NO3- + 2e- + 4H+ ⟶ Cu2+ + 2e- + 2NO2 + 2H2O
* Cu + 2NO3- + 4H+ ⟶ Cu2+ + 2NO2 + 2H2O


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