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.

  • 1
    $\begingroup$ 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 $\endgroup$
    – Poutnik
    Feb 1, 2022 at 19:16
  • $\begingroup$ Welcome to chemistry SE! Before asking, read available textbooks, google relevant keywords and study search results. Check for duplicates or related Q/As by insertion the search term site:stackexchange.com. Search other valuable sites with the term site:libretexts.org or site:hyperphysics.phy-astr.gsu.edu or site:en.wikipedia.org . This prevents cases of asking for answers that have been already written in many ways on many places. // Review also Chemistry SE: resources-for-learning-chemistry $\endgroup$
    – Poutnik
    Feb 1, 2022 at 19:18
  • $\begingroup$ I have tried to watch videos and learn from the book, but I still dont understand how this should be solved. @Poutnik $\endgroup$
    – James
    Feb 1, 2022 at 19:19
  • $\begingroup$ In the book it was $NO^{3-}$, not $NO_3^-$, is it an error? @Poutnik $\endgroup$
    – James
    Feb 1, 2022 at 19:29
  • $\begingroup$ Note that oxidation numbers, in opposite to charges, are used to be written by Latin numbers with oreceeding sign, i.e. +IV, not 4+. $\endgroup$
    – Poutnik
    Feb 1, 2022 at 19:29

1 Answer 1


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

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
  • 2
    $\begingroup$ 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. $\endgroup$
    – Poutnik
    Feb 2, 2022 at 9:53
  • 2
    $\begingroup$ 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 $\endgroup$
    – AJKOER
    Feb 2, 2022 at 12:57
  • 2
    $\begingroup$ @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. $\endgroup$
    – Poutnik
    Mar 4, 2022 at 7:31
  • $\begingroup$ Agreed. Learning the language a bit will help in learning advanced principles. Much easier to balance Cu + 4 HNO3 = Cu[NO3]2 +2NO2 +2 H2O when one knows what nitric acid and nitrate ion are. If the actual reaction is a consecutive series and one is concerned about mechanism each reaction deserves its own equation. [The abstract of the above link offers nothing] $\endgroup$
    – jimchmst
    2 days ago

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