Question

Copper reduces $N{O}_3^{-}$ 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 $10^{x}M$. The value of $2x$ is_______ (Rounded-off to the nearest integer)
Given , $E_{C{u}^{2+}/Cu}^o=0.34V,E_{N{O}_3^{-}/NO}^o=0.96V$
$E_{N{O}_3^{-}/N{O}_2}^o=0.79V$ and at $298K,$
$\frac{RT}{F}(2.303)=0.059$

Answer 1

Answer: 1

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

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

$E_{N{O}_3^{-}/NO}^o=0.96V$

$N{O}_3^{-}+2H^{+}+e^{-}\to N{O}_2+H_{2}O$

$E_{N{O}_3^{-}/N{O}_2}^o=0.79$

Let $\left[HN{O}_3\right]=y\Rightarrow \left[H^{+}\right]=y$ and $\left[N{O}_3\right]=y$

for same thermodynamic tendency

$E_{N{O}_3^{-}/NO}=E_{N{O}_3^{-}/N{O}_2}$

or, $E_{N{O}_3^{-}/NO}^{\circ }-\frac{0.059}{3}\log \frac{P_{NO}}{y\times y^4}$

$=E_{N{O}_3^{-}/N{O}_2}^o-\frac{0.059}{1}\log \frac{P_{N{O}_2}}{y\times y^2}$

or, $0.96-\frac{0.059}{3}\log \frac{P_{NO}}{y^5}=0.79-\frac{0.059}{1}\log \frac{P_{N{O}_2}}{y^3}$

or, $0.17=-\frac{0.059}{1}\log \frac{P_{N{O}_2}}{y^3}+\frac{0.059}{3}\log \frac{P_{NO}}{y^5}$

$0.17=-\frac{0.0591}{1}\log \frac{P_{N{O}_2}}{y^3}+\frac{0.0591}{3}\log \frac{P_{NO}}{y^5}$

$0.17=-\frac{0.0591}{3}\log \frac{P_{N{O}_2}^3}{y^9}+\frac{0.0591}{3}\log \frac{P_{NO}}{y^5}$

$0.17=\frac{0.0591}{3}\left[\log \frac{P_{NO}}{y^5}-\log \frac{P_{N{O}_2}^3}{y^9}\right]$

$0.17=\frac{0.0591}{3}\left[\log \frac{P_{NO}}{y^5}\times \frac{y^9}{P_{N{O}_2}^3}\right]$

Assume $P_{NO}\simeq P_{N{O}_2}=1bar$

$\frac{0.17\times 3}{0.059}=\log y^4=8.644$

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

$\log y=2.161$

$y=10^{2.16}$

$\therefore 2x=2\times 2.161=4.322$