Copper reduces 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 , $P_{\mathrm{NO}}=P_{N{O}_2}),$the ${\mathrm{HNO}}_3$ concentration at which the thermodynamic tendency for reduction of into $\mathrm{NO}$ and ${\mathrm{NO}}_2$ by copper is same is${10}^{\mathrm{x}}\mathrm{M}$

The value of $2\mathrm{x}$ is _____. (Rounded-off to the nearest integer $\mathrm{Cu}\left(\mathrm{s}\right)\to {\mathrm{Cu}}^{2+}+2{\mathrm{e}}^{-}$)

$$\begin{gathered} [\mathrm{Given}:{\mathrm{E}}_{{\mathrm{Cu}}^{2+}/\mathrm{Cu}}^0=0.34\mathrm{V},{\mathrm{E}}_{{{\mathrm{NO}}_3}^{-}/\mathrm{NO}}^0=0.96\mathrm{V},{\mathrm{E}}_{{{\mathrm{NO}}_3}^{-}/\mathrm{N}}^0\mathrm{and}\mathrm{at}298\mathrm{K} \\ \frac{\mathrm{RT}}{\mathrm{F}}(2.303)=0.059] \end{gathered}$$

Answer: The value of $2x$$=$$4$.

From the equations:

Step 1:

Anode:

$\mathrm{Cu}\left(\mathrm{s}\right)\to {\mathrm{Cu}}^{+2}+2{\mathrm{e}}^{-}$

Cathode:

$3{\mathrm{e}}^{-}+4{\mathrm{H}}^{+}+{\mathrm{NO}}_3^{-}\to \mathrm{NO}+2{\mathrm{H}}_2\mathrm{O}$

Full cell reaction

$8{\mathrm{H}}^{-}+2{\mathrm{NO}}_3^{-}+3{\mathrm{Cu}}^{+2}+2\mathrm{NO}+4{\mathrm{H}}_2\mathrm{O}$

Step 2:

Applying Nerst equation for the reaction quotient :

$$\begin{gathered} \mathrm{Q}=\frac{{\left[{\mathrm{Cu}}^{+2}\right]}^3\times {\left({\mathrm{p}}_{\mathrm{NO}}\right)}^2}{{\left[{\mathrm{NO}}_3^{-}\right]}^2{\left[{\mathrm{H}}^{+}\right]}^8} \\ {\in }_{\mathrm{cell}}^0=1.3 \\ {\in }_{\mathrm{cell}}=1.3-\frac{0.059}{6}\log \frac{{\left[{\mathrm{Cu}}^{+2}\right]}^3\times {\left({\mathrm{p}}_{\mathrm{NO}}\right)}^2}{{\left[{\mathrm{NO}}_3^{-}\right]}^2{\left[{\mathrm{H}}^{+}\right]}^8} \end{gathered}$$

Step 3:

Similarly,

Anode:

$\mathrm{Cu}\left(\mathrm{s}\right)\to {\mathrm{Cu}}^{+2}+2{\mathrm{e}}^{-}$

Cathode:

${\mathrm{e}}^{-}+2{\mathrm{H}}^{+}+{{\mathrm{NO}}_3}^{-}\to {\mathrm{NO}}_2+{\mathrm{H}}_2\mathrm{O}$

Step 4:

For full cell reaction:

$$\begin{gathered} \mathrm{Cu}\left(\mathrm{s}\right)+4{\mathrm{H}}^{+}+2{\mathrm{NO}}_3^{-}\to 2{\mathrm{NO}}_2+2{\mathrm{H}}_2\mathrm{O}+{\mathrm{Cu}}^{+2} \\ {\in }_{\mathrm{cell}}^0=1.3 \\ \mathrm{Q}=\frac{\left[{\mathrm{Cu}}^{+2}\right]\times {\left({\mathrm{p}}_{{\mathrm{NO}}_2}\right)}^2}{{\left[{\mathrm{NO}}_3^{-}\right]}^2{\left[{\mathrm{H}}^{+}\right]}^4} \\ {\in }_{\mathrm{cell}}=1.3-\frac{0.059}{2}\log \frac{\left[{\mathrm{Cu}}^{+2}\right]\times {\left({\mathrm{p}}_{{\mathrm{NO}}_2}\right)}^2}{{\left[{\mathrm{NO}}_3^{-}\right]}^2{\left[{\mathrm{H}}^{+}\right]}^4} \end{gathered}$$

Step 5:

According to the question requirement:

${\in }_{{\mathrm{cell}}_{\mathrm{T}}}={\in }_{{\mathrm{cell}}_2}$

$$\begin{gathered} 1.3-\frac{0.059}{6}\log \left({\mathrm{Q}}_1\right)=1.13-\frac{0.059}{2}\log \left({\mathrm{Q}}_2\right) \\ 0.17=\frac{0.059}{6}\left\{\log \left({\mathrm{Q}}_1\right)-3\log \left({\mathrm{Q}}_2\right)\right\} \\ =\frac{0.059}{6}\left\{\log \frac{{\left({\mathrm{Cu}}^{+2}\right)}^3\times {\left({\mathrm{p}}_{\mathrm{NO}}\right)}^2\times {\left({{\mathrm{NO}}_3}^{-}\right)}^6{\left({\mathrm{H}}^{+}\right)}^{12}}{\left({\mathrm{NO}}_3^{-}\right){\left({\mathrm{H}}^{+}\right)}^8\times {\left({\mathrm{Cu}}^{+2}\right)}^3\times {\left({\mathrm{P}}_{{\mathrm{NO}}_2}\right)}^6}\right\} \\ =\frac{0.059}{6}\left\{\log \frac{\left[{\mathrm{NO}}_3^{-}\right]{\left[{\mathrm{H}}^{+}\right]}^4}{{\left({\mathrm{p}}_{{\mathrm{NO}}_2}\right)}^4}\right\} \\ 0.17=\frac{0.059}{6}\times 8\log \left({\mathrm{HNO}}_3\right) \\ \log \left({\mathrm{HNO}}_3\right)=2.16 \\ \left[{\mathrm{HNO}}_3\right]={10}^{2.16}={10}^{\mathrm{x}} \\ \mathrm{x}=2.16\Rightarrow 2\mathrm{x}=4.32\approx 4 \end{gathered}$$

Therefore, the thermodynamic tendency for reduction of into $\mathrm{NO}$ and ${\mathrm{NO}}_2$ by copper is same is${10}^{\mathrm{x}}\mathrm{M}$ where $2\mathrm{x}$ is $4$.