Balancing a chemical reaction with oxidation numbers

Question

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.

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