Question

Copper reduces nitrate into $NO$ and $N{O}_2$ depending upon the concentration of $HN{O}_3$ in solution (Assuming fixed $\left[C{u}^{2+}\right]$ and $P_{NO}=P_{N{O}_2}$). The $HN{O}_3$ concentration at which the thermodynamic tendency for reduction of $N{O}_3^{-}$ into $NO$ and $N{O}_2$ by copper is same is ${10}^x]\text{M}$. The value of $2x$ is
(Rounded-off to the nearest integer)
[Given : $E_{C{u}^{2+}/cu}^0=0.34V,E_{N{O}_3^{-}/NO}^0=0.96V,E_{N{O}_3^{-}/N{O}_2}^0=0.79V$ and at $\frac{RT}{F}\left(2.303\right)=0.059$]

Answer 1

Anode
$Cu\left(s\right)\to C{u}^{2+}+2e^{-}$
Cathode (1)
For reduction of nitric acid to $NO$
$\frac{3e^{-}+4H^{+}+N{O}_3^{-}\to NO+2H_{2}O}{8H^{-}+2N{O}_3^{-}+3Cu\left(s\right)\to 3C{u}^{+2}+2NO+4H_{2}O}$
$Q=\frac{[C{u}^{+2}{]}^3\times (P_{NO}{)}^2}{\left[N{O}_3^{-}{]}^2\right[H^{+}{]}^8}$
$E_{\text{cell}}^0=0.62$
$E_{\text{cell}}=0.62-\frac{0.059}{6}\log \frac{[C{u}^{+2}{]}^3\times (P_{NO}{)}^2}{\left[N{O}_3^{-}{]}^2\right[H^{+}{]}^8}.......\left(1\right)$
Anode $Cu\left(s\right)\to C{u}^{2+}+2e^{-}$
$\text{Cathode}\text{For the reduction of nitric acid to}N{O}_2\frac{e^{-}+2H^{+}+N{O}_3^{-}\to N{O}_2+H_{2}O}{Cu\left(s\right)+4H^{+}+2N{O}_3^{-}\to 2N{O}_2+2H_{2}O+C{u}^{+2}}$
$E_{\text{cell}}^0=0.45$
$Q=\frac{\left(C{u}^{2+}\right)(P_{N{O}_2}{)}^2}{\left(N{O}_3^{-}{)}^2\right(H^{+}{)}^4}$
$E_{\text{cell}}=0.45-\frac{0.059}{2}\text{log}\frac{\left(C{u}^{+2}\right)(P_{N{O}_2}{)}^2}{\left(N{O}_3^{-}{)}^2\right(H^{+}{)}^4}$
Given that the thermodynamic tendency to reduce nitric acid to $NO$ and $N{O}_2$ is same hence,
$E_{\text{cell (1)}}=E_{\text{cell (2)}}$
$0.62-\frac{0.059}{6}\text{log}\left(Q_1\right)=0.49-\frac{0.059}{2}\text{log}\left(Q_2\right)$
$\Rightarrow 0.17=\frac{0.059}{6}${$\text{log}\left(Q_1\right)-3\text{log}\left(Q_2\right)$}
$\Rightarrow \frac{0.059}{6}\left\{\text{log}\frac{(C{u}^{+2}{)}^3\times (P_{NO}{)}^2\times \left(N{O}_3^{-}{)}^6\right(H^{+}{)}^{12}}{\left(N{O}_3^{-}{)}^2\right(H^{+}{)}^8\times (C{u}^{+2}{)}^3\times (P_{N{O}_2}{)}^6}\right\}$
$\Rightarrow \frac{0.059}{6}\left\{\text{log}\frac{\left(N{O}_3^{-}{)}^4\right(H^{+}{)}^4}{(P_{N{O}_2}{)}^4}\right\}$
$\Rightarrow 0.17=\frac{0.059}{6}\times 8\text{log}\left(HN{O}_3\right)$
$\text{log}\left(HN{O}_3\right)=2.16$
$\left[HN{O}_3\right]={10}^{2.16}={10}^x$
$x=2.16\Rightarrow 2x=4.32\approx 4$