When copper reacts with dilute nitric acid,

$$\ce{3Cu + 8HNO3 -> 3Cu(NO3)2 + 2NO +4H2O}$$

while with concentrated nitric acid,

$$\ce{Cu + 4HNO3 -> Cu(NO3)2 + 2NO2 + 2H2O}$$

As we see that with dilute acid oxidation state of nitrogen changes from $+5$ to $+2$ and with concentrated acid it changes from $+5$ to $+4$ but shouldn't it be vice- versa because since concentrated nitric acid is a good oxidizing agent and should show a huge change in oxidation number ?

Where am I going wrong?

  • $\begingroup$ Though not related, hot, alkaline potassium permanganate changes the O.N. of manganese from 7+ to 6+ to 4+ in MnO4. So, its O.N. change is not as drastic as in acidic medium i.e. from 7+ to 2+ to form Mn2+ ions. But it is a much stronger oxidising agent in basic medium than in acidic medium. $\endgroup$ – Aditya Roychowdhury Aug 23 '20 at 18:17
  • $\begingroup$ Do you know the reason $\endgroup$ – Anusha Aug 23 '20 at 18:39

You may reconcile both equations by stating that your second reaction occurs in all nitric acid solutions (concentrated or dilute). And in dilute solutions, you may admit that there is enough water to destroy $\ce{NO2}$ according to the following equation : $$\ce{3NO2 + H2O⟶ NO + 2HNO3}$$ And multiplying this new equation by $2$, and adding to your second equation (after multiplication by $3$) gives your first equation


Because nitric acid is such a good oxidizing agent, when sufficiently concentrated it could oxidize any nitric oxide that tries to form.

There are various ways to view this effect. See answers to this question for some ideas.


Not the answer you're looking for? Browse other questions tagged or ask your own question.