# Solve the following

Question:

Copper reduces $\mathrm{NO}_{3}^{\text {-into }} \mathrm{NO}$ 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 $2 \mathrm{x}$ is_______________ . (Rounded-off to the nearest integer)

[Given: $E_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{0}=0.34 \mathrm{~V}, E_{\mathrm{NO}_{3} / \mathrm{NO}}^{0}=0.96 \mathrm{~V}, E_{\mathrm{NO}_{3} / \mathrm{NO}_{2}}^{0}=0.79 \mathrm{~V}$ and at $298 \mathrm{~K}, \frac{R T}{F}(2.303)=0.0597$

Solution:

(1)

Anode

$\mathrm{Cu}(\mathrm{s}) \rightarrow \mathrm{Cu}^{+2}+2 \mathrm{e}^{-}$

Cathode (1) $\frac{3 e^{-}+4 H^{+}+N O_{3}^{-} \rightarrow N O+2 H_{2} O}{8 H^{-}+2 N 0_{3}^{-}+3 C u(s) \rightarrow 3 C u^{+2}+2 N O+4 H_{2} O}$

$\mathbf{Q}=\frac{\left[\mathrm{Cu}^{+2}\right]^{3} \times\left(p_{\mathrm{NO}}\right)^{2}}{\left[\mathrm{NO}_{3}^{-}\right]^{2}\left[\mathrm{H}^{+}\right]^{8}}$

$\epsilon_{\text {cell }}^{0}=1.3$

$\epsilon_{\text {cell }}=1.3-\frac{0.059}{6} \log \frac{\left({ }^{\mathrm{Cu}^{+2}}\right)^{3}\left(p_{\mathrm{NO}}\right)^{2}}{\left(\mathrm{NO}_{3}^{-}\right)^{2} \times\left(\mathrm{H}^{+}\right)^{8}} \ldots(1)$

Anode $\mathrm{Cu}(\mathrm{s}) \rightarrow \mathrm{Cu}^{+2}+2 \mathrm{e}$

Cathode $\frac{e^{-}+2 n^{+}+N O_{3}^{-} \rightarrow N O_{2}+H_{2} O}{\mathrm{Cu}(s)+4 \mathrm{H}^{+}+2 \mathrm{NO}_{3}^{-} \rightarrow 2 \mathrm{NO}_{2}+2 \mathrm{H}_{2} \mathrm{O}+\mathrm{Cu}^{+2}}$

$\epsilon_{\text {cell }}^{0}=1.13$

$Q=\frac{\left(\mathrm{Cu}^{+2}\right)\left(p_{\mathrm{NO}_{2}}\right)^{2}}{\left(\mathrm{NO}_{3}^{-}\right)^{2}\left(\mathrm{H}^{+}\right)^{4}}$

$\epsilon_{\text {cell }}=1.13-\frac{0.059}{2} \log \frac{\left(\mathrm{Cu}^{+2}\right)\left(p_{\mathrm{NO}_{2}}\right)^{2}}{\left(N O_{3}^{-}\right)^{2}\left(H^{+}\right)^{4}}$

$\epsilon_{\text {cell }_{-2}}=\epsilon_{\text {cell }_{2}}$

$1.3-\frac{0.059}{6} \log \left(Q_{1}\right)=1.13-\frac{0.059}{2} \log \left(Q_{2}\right)$

$0.17=\frac{0.059}{6}\left\{\log \left(Q_{1}\right)-3 \log \left(Q_{2}\right)\right\}$

$=\frac{0.059}{6}\left\{\log \frac{\left(\mathrm{Cu}^{+2}\right)^{3} \times\left(p_{N O}\right)^{2} \times\left(N O_{3}^{-}\right)^{6}\left(H^{+}\right)^{12}}{\left(N O_{3}^{-}\right)^{2}\left(H^{+}\right)^{8} \times\left(C u^{+2}\right)^{3} \times\left(p_{N O_{2}}\right)^{6}}\right\}$

$=\frac{0.059}{6}\left\{\log \frac{\left[\mathrm{NO}_{3}^{-}\right]^{4}\left[\mathrm{H}^{+}\right]^{4}}{\left(P_{N O}\right)^{4}}\right\}$

$0.17=\frac{0.059}{6} \times 8 \log \left(\mathrm{HNO}_{3}\right)$

$\log \left(\mathrm{HNO}_{3}\right)=2.16$

$\left[\mathrm{HNO}_{3}\right]=10^{2.16}=10^{\mathrm{x}}$

$x=2.16 \Rightarrow 2 x=4.32 \approx 4$