Animated Solution for Chemistry - Electrochemistry: Copper reduces ${\text{NO}}_3^{-}$ into $\text{NO}$ and ${\text{NO}}_2$ depending upon the concentration of ${\text{HNO}}_3$ in solution. (Assuming fixed $[{\text{Cu}}^{2+}]$ and $p_{\text{NO}}=p_{{\text{NO}}_2}$), the ${\text{HNO}}_3$ concentration at which the thermodynamic tendency for reduction of ${\text{NO}}_3^{-}$ into $\text{NO}$ and ${\text{NO}}_2$ by copper is same is $10^x$ M. The value of $2x$ is ……… . (Rounded off to the nearest integer)

The reaction of copper with nitric acid is a classic textbook phenomenon. We all remember the distinct observations: dilute nitric acid yields colorless nitric oxide ($\text{NO}$) gas, while concentrated nitric acid produces thick, toxic brown fumes of nitrogen dioxide (${\text{NO}}_2$). But have you ever wondered why the concentration dictates the product? This problem takes that qualitative observation and turns it into a rigorous thermodynamic battle!

In the language of electrochemistry, thermodynamic tendency is governed by the Gibbs free energy, which directly translates to the cell potential, $E_{\text{cell}}$.

When $E_1=E_2$, the system is at a tipping point. Any concentration higher than this will favor ${\text{NO}}_2$, and any concentration lower will favor $\text{NO}$.

Let's write down the Nernst equations for both half-cells.

Reaction 1: Production of $\text{NO}$ The Nernst equation for this 6-electron transfer is:

Reaction 2: Production of ${\text{NO}}_2$ The Nernst equation for this 2-electron transfer is:

Let's equate them and rearrange to group the standard potentials on one side:

Using the standard reduction potentials (typically provided in the exam data sheet), we know $E_{{\text{NO}}_3^{-}/\text{NO}}^{\circ }=0.96\text{ V}$ and $E_{{\text{NO}}_3^{-}/{\text{NO}}_2}^{\circ }=0.79\text{ V}$. The copper oxidation potential cancels out when we take the difference!

To combine the logarithms, we need a common denominator. Let's rewrite $\frac{1}{2}$ as $\frac{3}{6}$:

Using the power rule of logarithms, $3\log Q_2=\log Q_2^3$. Then, using the quotient rule:

Let's expand this massive fraction:

Look at the beauty of this expression! The $[{\text{Cu}}^{2+}{]}^3$ terms perfectly cancel out. Assuming standard pressure for the gases ($p_{\text{NO}}=p_{{\text{NO}}_2}=1\text{ bar}$), we are left with:

Since nitric acid is a strong acid, it dissociates completely. Therefore, $[{\text{H}}^{+}]=[{\text{NO}}_3^{-}]=[{\text{HNO}}_3]$.

Substitute this back into our simplified Nernst equation:

Solving for the logarithm:

The problem defines this concentration as $10^x\text{ M}$, which means $x=2.16$. We are asked to find the value of $2x$:

Rounding off to the nearest integer, we get our final answer: 4.

This problem is a masterpiece. It elegantly weaves together stoichiometry, the Nernst equation, logarithmic manipulation, and ionic equilibrium into a single, cohesive narrative!