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.

• You may find useful these links for text formatting (not to be applied to titles): Notation basics , Formatting of math/chem expressions and upright vs italic Feb 1, 2022 at 19:16
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• I have tried to watch videos and learn from the book, but I still dont understand how this should be solved. @Poutnik Feb 1, 2022 at 19:19
• In the book it was $NO^{3-}$, not $NO_3^-$, is it an error? @Poutnik Feb 1, 2022 at 19:29
• Note that oxidation numbers, in opposite to charges, are used to be written by Latin numbers with oreceeding sign, i.e. +IV, not 4+. Feb 1, 2022 at 19:29

The equation we need to balance is this:

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

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
• Formally correct, but more complicated than necessery. // Oxidized Cu changes ON by 2, reduced NO3- by 1, therefore Cu + 2 NO3- // Adding 4 H+ to maintain charge balance, together with H2O as product. Cu + 2 NO3- + 4H+ -> Cu^2+ + 2 NO2 + 2 H2O.// Do O and H inventory as independent check. // You are done. Feb 2, 2022 at 9:53
• For the record, that is not the proper reaction equation, there is a mix of NO2 and NO gases and the mechanics is under debate. Here is a citation from one of the latest articles: "Mechanistic Study of the Production of NOx Gases from the Reaction of Copper with Nitric Acid" Rebecca K. Carlson, Ping Yang*, Samuel M. Clegg, and Enrique R. Batista* Inorg. Chem. 2020, 59, 23, 16833–16842 doi.org/10.1021/acs.inorgchem.0c00607 Feb 2, 2022 at 12:57
• @AJKOER Factually it is correct, but for educative approach toward the OP it is unnecessery confusing detail. The reaction is about the outcome, not mechanism. At beginner level, it can be considered correct for concentrated HNO3. Mar 4, 2022 at 7:31