If the partial pressure of NO and $\mathrm{NO}_2$ gas is taker as 1 bar, then Answer is 4, else the question is bonus. $$ \begin{aligned} & \mathrm{NO}_3^{-}+4 \mathrm{H}^{+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{NO}+2 \mathrm{H}_2 \mathrm{O} \\ & \mathrm{E}_{\mathrm{NO}_3 / \mathrm{NO}}^{\circ}=0.96 \mathrm{~V} \\ & \mathrm{NO}_3^{-}+2 \mathrm{H}^{+}+\mathrm{e}^{-} \longrightarrow \longrightarrow \mathrm{NO}_2+\mathrm{H}_2 \mathrm{O} \\ & \mathrm{E}_{\mathrm{NO}_3 / \mathrm{NO}_2=0.79}^{\circ} \end{aligned} $$ 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 $$ \begin{aligned} & \mathrm{E}_{\mathrm{NO}_3 / \mathrm{NO}}=\mathrm{E}_{\mathrm{NO}_3 / \mathrm{NO}_2} \\ & \text { or, } \mathrm{E}_{\mathrm{NO}_3 / \mathrm{NO}}^{\prime}-\frac{0.059}{3} \log \frac{\mathrm{P}_{\mathrm{NO}^4}}{\mathrm{y} \times \mathrm{y}^4} \\ & =\mathrm{E}_{\mathrm{NO}_3 / \mathrm{NO}_3}^{\circ}-\frac{0.059}{1} \log \frac{\mathrm{P}_{\mathrm{NO}_2}}{\mathrm{y} \times \mathrm{y}^2} \\ & \text { or, } 0.96-\frac{0.059}{3} \log \frac{\mathrm{P}_{\mathrm{NO}^2}}{\mathrm{y}^5}=0.79-\frac{0.059}{1} \log \frac{\mathrm{P}_{\mathrm{NO}_3}}{\mathrm{y}^3} \\ & \text { or, } 0.17=\frac{0.059}{1} \log \frac{\mathrm{P}_{\mathrm{NO}_3}}{\mathrm{y}^3}+\frac{0.059}{3} \log \frac{\mathrm{P}_{\mathrm{NO}^5}}{\mathrm{y}^5} \end{aligned} $$ $$ \begin{aligned} & 0.17=\frac{0.0591}{1} \log \frac{P_{\mathrm{NO}_2}}{y^3}+\frac{0.0591}{3} \log \frac{P_{\mathrm{NO}}}{y^5} \\ & 0.17=\frac{0.0591}{3} \log \frac{P_{\mathrm{NO}_2}^3}{y^9}+\frac{0.0591}{3} \log \frac{P_{\mathrm{NO}}}{y^5} \\ & 0.17=\frac{0.0591}{3}\left[\log \frac{P_{\mathrm{NO}}}{y^5}-\log \frac{P_{\mathrm{NO}_2}^3}{y^9}\right] \end{aligned} $$ $$ 0.17=\frac{0.0591}{3}\left[\log \frac{\mathrm{P}_{\mathrm{NO}}}{\mathrm{y}^5} \times \frac{\mathrm{y}^9}{\mathrm{P}_{\mathrm{NO}_2}^3}\right] $$ Assume $\mathrm{P}_{\mathrm{NO}} \square \mathrm{P}_{\mathrm{NO} 2}=1$ bar $$ \begin{aligned} & \frac{0.17 \times 3}{0.059}=\log y^4=8.644 \\ & \log y=\frac{8.644}{4} \\ & \log y=2.161 \\ & \quad y=10^{2.16} \\ & \therefore \quad 2 x=2 \times 2.161=4.322 \end{aligned} $$