Copper reduces ${\mathrm{NO}}_3^{-}$ into $\mathrm{NO}$ and ${\mathrm{NO}}_2$ depending upon the concentration of ${\mathrm{HNO}}_3$ in solution. (Assuming fixed $\left[{\mathrm{Cu}}^{2+}\right]$ and ${\mathrm{P}}_{\mathrm{NO}}={\mathrm{P}}_{{\mathrm{NO}}_2}$), the ${\mathrm{HNO}}_3$ concentration at which the thermodynamic tendency for reduction of ${\mathrm{NO}}_3^{-}$ into $\mathrm{NO}$ and ${\mathrm{NO}}_2$ by copper is same is ${10}^x \mathrm{M}$. The value of $2x$ is ___. (Rounded-off to the nearest integer)
[Given, ${\mathrm{E}}_{{\mathrm{Cu}}^{2+}/\mathrm{Cu}}^{\mathrm{o}}=0.34 \mathrm{V},{\mathrm{E}}_{{\mathrm{NO}}_3^{-}/\mathrm{NO}}^{\mathrm{o}}=0.96 \mathrm{V},{\mathrm{E}}_{{\mathrm{NO}}_3^{-}/{\mathrm{NO}}_2}^{\mathrm{o}}=0.79 \mathrm{V}$ and at $298 \mathrm{K}$, $\frac{\mathrm{RT}}{\mathrm{F}}\left(2.303\right)=0.059$]

If the partial pressure of $\mathrm{NO}$ and ${\mathrm{NO}}_2$ gas is taken as $1$ bar, then Answer is $4$, else the question is bonus.

${\mathrm{NO}}_3^{-}+4{\mathrm{H}}^{+}+3{\mathrm{e}}^{-}⟶\mathrm{NO}+2{\mathrm{H}}_2\mathrm{O}$   ${\mathrm{E}}_{{\mathrm{NO}}_3^{-}/\mathrm{NO}}^{\mathrm{o}}=0.96 \mathrm{V}$

${\mathrm{NO}}_3^{-}+2{\mathrm{H}}^{+}+{\mathrm{e}}^{-}⟶{\mathrm{NO}}_2+{\mathrm{H}}_2\mathrm{O}$   ${\mathrm{E}}_{{\mathrm{NO}}_3^{-}/{\mathrm{NO}}_2}^{\mathrm{o}}=0.79$

Let $\left[{\mathrm{HNO}}_3\right]=\mathrm{y}\Rightarrow \left[{\mathrm{H}}^{+}\right]=\mathrm{y}$ and $\left[{\mathrm{NO}}_3^{-}\right]=\mathrm{y}$ 

for same thermodynamic tendency

${\mathrm{E}}_{{\mathrm{NO}}_3^{-}/\mathrm{NO}}={\mathrm{E}}_{{\mathrm{NO}}_3^{-}/{\mathrm{NO}}_2}$

or, ${\mathrm{E}}_{{\mathrm{NO}}_3^{-}/\mathrm{NO}}^{\mathrm{o}}-\frac{0.059}{3}\log \frac{{\mathrm{P}}_{\mathrm{NO}}}{\mathrm{y}\times {\mathrm{y}}^4}={\mathrm{E}}_{{\mathrm{NO}}_3^{-}/{\mathrm{NO}}_2}^{\mathrm{o}}-\frac{0.059}{1}\log \frac{{\mathrm{P}}_{{\mathrm{NO}}_2}}{\mathrm{y}\times {\mathrm{y}}^2}$

or, $0.96-\frac{0.059}{3}\log \frac{{\mathrm{P}}_{\mathrm{N}0}}{{\mathrm{y}}^5}=0.79-\frac{0.059}{1}\log \frac{{\mathrm{P}}_{{\mathrm{NO}}_2}}{{\mathrm{y}}^3}$

or, $0.17=-\frac{0.059}{1}\log \frac{{\mathrm{P}}_{{\mathrm{NO}}_2}}{{\mathrm{y}}^3}+\frac{0.059}{3}\log \frac{{\mathrm{P}}_{\mathrm{NO}}}{{\mathrm{y}}^5}$
$0.17=-\frac{0.0591}{1}\log \frac{{\mathrm{P}}_{{\mathrm{NO}}_2}}{{\mathrm{y}}^3}+\frac{0.0591}{3}\log \frac{{\mathrm{P}}_{\mathrm{NO}}}{{\mathrm{y}}^5}$

$0.17=-\frac{0.0591}{3}\log \frac{{\mathrm{P}}_{{\mathrm{NO}}_2}^3}{{\mathrm{y}}^9}+\frac{0.0591}{3}\log \frac{{\mathrm{P}}_{\mathrm{NO}}}{{\mathrm{y}}^5}$

$0.17=\frac{0.0591}{3}\left[\log \frac{{\mathrm{P}}_{\mathrm{NO}}}{{\mathrm{y}}^5}-\log \frac{{\mathrm{P}}_{{\mathrm{NO}}_2}^3}{{\mathrm{y}}^9}\right]$

$0.17=\frac{0.0591}{3}\left[\log \frac{{\mathrm{P}}_{\mathrm{NO}}}{{\mathrm{y}}^5}\times \frac{{\mathrm{y}}^9}{{\mathrm{P}}_{{\mathrm{NO}}_2}^3}\right]$

Assume ${\mathrm{P}}_{\mathrm{NO}}\simeq {\mathrm{P}}_{{\mathrm{NO}}_2}=1 \mathrm{bar}$

$\frac{0.17\times 3}{0.059}={\mathrm{logy}}^4=8.644$

$\log y=\frac{8.644}{4}$

$\log y=2.161$

$y={10}^{2.1}$

$\therefore 2\mathrm{x}=2\times 2.161=4.322$

Answer (4)