# Why does HNO3 not give off H2 when reacting with Cu?

A dilute solution of which acid is most likely to produce a reduction product other than $$\ce{H_{2}}$$ when it reacts with a metal?

(A) $$\ce{HF}$$

(B) $$\ce{HCl}$$

(C) $$\ce{HNO3}$$

(D) $$\ce{H2SO4}$$

$$\tiny{\text{Question from 2012 local chemistry olympiad}}$$

The given answer is (C), $$\ce{HNO3}$$ but I don't understand why.

First off, what does it mean by reduction product -- is it the product that is formed when the acid is reduced in the redox reaction between the metal and the acid?

Second, why is the answer $$\ce{HNO3}$$?

Whenever any metal is added to an acid, the metal is oxidised to one of its positive ions, at the cost of either the acid which is reduced or water which is reduced to hydrogen. The product formed by reduction is what is referred to as the reduction product.

The reduction potential of $\ce{NO3-}$ is $0.8$ which makes the reduction of nitrate instead of hydrogen ion $\ce{H+}$ ($E^\circ=0.0$) more favourable. Thus it is the nitrate in nitric acid which is reduced instead of $\ce{H+}$ or $\ce{H2O}$. But at very low concentration of nitrate ion and low temperature, as explained by Nernst equation, the reduction potential decreases significantly and combined with overpotential effects, it is $\ce{H2}$ which is formed and not the reduction product of nitrate which is $\ce{NO2}$.

In the case of halic acids, the halogens cannot be further reduced. In the case of the sulfate anion, although it has a reduction potential of $0.17$, owing to overpotential effects, sulfate is not reduced, instead, water is reduced yielding $\ce{H2}$. At high concentration, the reduction potential rises sufficiently to overcome the overpotential effects, allowing the reduction of sulfate to form $\ce{SO2}$ at higher concentrations of sulfuric acid. Hence under normal conditions, only nitric acid gives a product other than hydrogen gas.

Reduction potential data

$$\ce{SO4^{2-} + 4H+ + 2e^- -> SO2 + 2H2O}(E^\circ=0.17)$$ $$\ce{NO3- + 2H+ + e- -> NO2 + H2O} (E^\circ=0.8)$$ $$\text{Nernst equation } E=E^\circ- \frac{RT}{nF}\ln Q$$ $Q$= reaction quotient, $E$=redox potential, $E^\circ$=Standard redox potential

EDIT
The oxidation of metal at the cost of reduction of another substance occurs whenever it is thermodynamically favourable. Usually, (except silver, copper) metals have negative reduction potentials, which makes their oxidation at the cost of hydrogen thermodynamically favourable as the $\Delta G<0$. This leads to a spontaneous redox reaction. This is not true if there is a kinetic or physical barrier to reaction. For example, Iron is rendered passive by nitric acid due to the formation of an oxide layer at the surface which prevents any further contact and hence stops the reaction.

• I found this answer to a similar question on Yahoo Answers: "If the metal is below hydrogen in the reactivity series, then the oxidizing agent is the nitrate ion, not hydrogen." I don't understand how the metal being below H in the reactivity series determines the oxidizing agent. Could you please explain how this is? Mar 13 '14 at 4:53
• You should really mention standard conditions. Dilute Nitric acid is close to 1M which is one standard condition. Mar 14 '14 at 7:11
• @user2617804 The $\circ$ in the superscript of the redox potential symbol $E$, already indicates the standard state as per the IUPAC recommendation. Mar 14 '14 at 11:23
• @user2612743 Activity series is another way to state the redox potentials. The elements above hydrogen have negative reduction potential and can hence liberate hydrogen by reducing it in water. The elements below have positive electrode potentials and hence cannot reduce hydrogen, therefore they must reduce the nitrate anion which has a more positive reduction potential. Mar 14 '14 at 11:25