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I was reading the Wikipedia page on nitric acid and saw the reaction of concentrated $\ce{HNO_3}$ and dilute $\ce{HNO_3}$ with $\ce{Cu}$.

It is mentioned that:

Copper reacts with dilute nitric acid at ambient temperatures with a 3:8 stoichiometry:

$$\quad\qquad\qquad\qquad\ce{3 Cu + 8 HNO_3 -> 3 Cu^{2+} + 2 NO + 4 H_2O + 6 NO_3^−}$$

With more concentrated nitric acid, nitrogen dioxide is produced directly in a reaction with 1:4 stoichiometry:

$$\ce{Cu + 4 H^+ + 2 NO_3^− -> Cu^{2+} + 2 NO_2 + 2 H_2O}$$

My question is: If concentrated nitric acid is a better oxidising agent, then why does it oxidises only 2 moles of $\ce{Cu}$ per 8 moles of nitric acid against the 3:8 ratio in dilute one. Why is there change of -1 in oxidation state of $\ce{N}$ in concentrated solution as opposed to -3 in dilute solution?

Does this indicate dilute $\ce{HNO3}$ is better oxidising agent than concentrated one?

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I think your assignments for the equations need to be inspected closely. Let's look at those equation with their redox potentials:

In dilute conditions (say, ratio of $\ce{HNO3}$ to $\ce{Cu}$ is 8:3): $$ \begin{align} \ce{NO3-(aq) + 4H3O+(aq) + 3e- &<=> NO(g) + 6H2O(l)} &E^\circ = \pu{0.957 V}\\ \ce{Cu(s) &<=> Cu^2+ (aq) + 2e-} &E^\circ = \pu{-0.342 V}\\ \ce{2NO3-(aq) + 8H3O+(aq) + 3Cu (s) &-> 3Cu^2+(aq) + 2NO(g) + 12H2O(l)}&E^\circ_{\text{rxn}} = \pu{0.615 V} \end{align}$$

In concentration conditions (say, ratio of $\ce{HNO3}$ to $\ce{Cu}$ is 4:1): $$ \begin{align} \ce{2NO3-(aq) + 4H3O+(aq) + 2e- &<=> N2O4(g) + 6H2O(l)} &&E^\circ = \pu{0.803 V}\\ \ce{Cu(s) &<=> Cu^2+ (aq) + 2e-} &&E^\circ = \pu{-0.342 V}\\ \ce{2NO3-(aq) + 4H3O+(aq) + Cu (s) &-> Cu^2+(aq) + N2O4(g) + 6H2O(l)}&&E^\circ_{\text{rxn}} = \pu{0.461 V} \end{align}$$

Because this concentrated reaction is exotermic, the colorless $\ce{N2O4}$, gas decomposes to reddish brown $\ce{NO2}$ gas according to $\ce{N2O4 (g) <=> 2NO2 (g)},~\Delta H^\circ = \pu{+57.2 kJ mol^{-1}}$

Based on $E^\circ_{\text{rxn}}$ values, one can assume that the concentrated reaction is less spontaneous. However, some dissolved $\ce{N2O4}$ can still undergo following redox reaction which is the most spontaneous reaction among three:

$$ \begin{align} \ce{N2O4(aq) + 4H3O+(aq) + 4e- &<=> 2NO(g) + 6H2O(l)} &E^\circ = \pu{1.035 V}\\ \ce{Cu(s) &<=> Cu^2+ (aq) + 2e-} &E^\circ = \pu{-0.342 V}\\ \ce{N2O4(aq) + 4H3O+(aq) + 2Cu(s) &-> 2Cu^2+(aq) + 2NO(g) + 6H2O(l)}&E^\circ_{\text{rxn}} = \pu{0.693 V} \end{align}$$

If we can consider the loss of $\ce{N2O4}$ to $\ce{NO2}$ is negligible, then total oxidation of $\ce{Cu}$ with both conc. $\ce{HNO3}$ and $\ce{N2O4}$ can be written as: $$ \begin{align} \ce{2NO3-(aq) + 8H3O+(aq) + 3Cu (s) &-> 3Cu^2+(aq) + 2NO(g) + 12H2O(l)}&E^\circ_{\text{rxn}} = \pu{1.154 V} \end{align}$$

Although it is identical reaction to that in dilute conditions, the reaction proceeds much more favorably (compare $E^\circ_{\text{rxn}}$ values in both conditions).

Late edition:

I have tried to avoid criticizing literature values on this subject, but my answer so far arose questions from concerned people. Thus, I'd try to do my best to handle it as delicate as possible.

To my knowledge, there shouldn't be a question about whether conc. $\ce{HNO3}$ is better oxidizing agent than dilute $\ce{HNO3}$. The oxidizing reagent here is the $\ce{NO3-}$ ion, as my equations described above. For example, any $\ce{NO3-}$ ion (say $\ce{KNO3}$) will oxidize $\ce{Cu}$ in an acid medium (say, $\ce{HCl}$). In using dilute or concentrated $\ce{HNO3}$, we conveniently provide both $\ce{NO3-}$ ions and $\ce{H3O+}$ for our redox reaction. This reaction is highly based on reaction conditions. I think nobody knows exact process, but everybody speculate by the results. That's the reason you are seeing too many different equations in different sources. I'm using the comprehensive table in: http://sites.chem.colostate.edu/diverdi/all_courses/CRC%20reference%20data/electrochemical%20series.pdf, which I believe more realistic. Now, consider following 2 questions:

The question 1: If concentrated nitric acid is a better oxidizing agent, then why does it oxidizes only 2 moles of Cu per 8 moles of nitric acid against the 3:8 ratio in dilute one?

Answer: Conc. $\ce{HNO3}$ has more $\ce{NO3-}$ available to react. When available, two moles of it react with 1 mole of $\ce{Cu}$ to give more stable product, $\ce{N2O4}$ ($\Delta H^\circ_f = \pu{+9.16 kJ/mol}$) compared to $\ce{NO}$ ($\Delta H^\circ_f = \pu{+90.25 kJ/mol}$).

The question 2: Why is there change of -1 in oxidation state of $\ce{N}$ in concentrated solution as opposed to -3 in dilute solution?

Answer: Look at the last series of reactions involving $\ce{N2O4}$ given in my original answer. Eventual change of oxidation state of $\ce{N}$ is same in both cases. This reaction is not just a matter of power. I don't think anybody studied this process in details yet. I'd like to make my point by giving some known redox processes of of $\ce{NO3-}$.

$$ \begin{align} \ce{NO3-(aq) + 3H3O+(aq) + 2e- &<=> HNO2(aq) + 4H2O(l)} &E^\circ = \pu{0.934 V}\\ \ce{NO3-(aq) + 4H3O+(aq) + 3e- &<=> NO(g) + 6H2O(l)}&E^\circ = \pu{0.957 V} \end{align}$$

Both processes are under relatively similar conditions (say, $\pu{2M}$ vs $\pu{3M}~\ce{HNO3}$). In process (1), the oxidation state of $\ce{N}$ changes from +5 to +3 while in (2), it was +5 to +2. Which process dominate when they were subjected to oxidize appropriate amount of $\ce{Cu}$?

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  • $\begingroup$ Sorry, but I am unable to grasp what is your final answer to the two questions: "(1) If concentrated nitric acid is a better oxidising agent, then why does it oxidises only 2 moles of Cu per 8 moles of nitric acid against the 3:8 ratio in dilute one. (2) Why is there change of -1 in oxidation state of N in concentrated solution as opposed to -3 in dilute solution?" $\endgroup$ – Gaurang Tandon May 18 '18 at 1:24
  • $\begingroup$ Thanks for your answer..So, do you mean to say that the reaction provided in wiki is incomplete and doesn't show the true picture?.. $\endgroup$ – user35508 May 18 '18 at 10:08

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