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  4. Copper reduces NO 3- into NO and NO 2 depending upon the concentration of HNO 3 in solution. (Assuming fixed [ Cu 2+] and . P NO = P NO 2), the HNO 3 concentration at which the thermodynamic tendency for reduction of NO 3- into NO and NO 2 by copper is same is 10 x M. The value of 2 x is (Rounded-off to the nearest integer) Given , Eo Cu 2+ / Cu =0.34 V , E NO 3- / NO o =0.96 V E NO 3- / NO 2 o =0.79 V and at 298 K , ( RT / F )(2.303)=0.059

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Q. Copper reduces $NO _{3}^{-}$ into $NO$ and $NO _{2}$ depending upon the concentration of $HNO _{3}$ in solution. (Assuming fixed $\left[ Cu ^{2+}\right]$ and $\left. P _{ NO }= P _{ NO _{2}}\right),$ the $HNO _{3}$ concentration at which the thermodynamic tendency for reduction of $NO _{3}^{-}$ into $NO$ and $NO _{2}$ by copper is same is $10^{ x } M$. The value of $2 x$ is_______ (Rounded-off to the nearest integer)
Given , $E^{o} _{ Cu ^{2+} / Cu }=0.34 V , E _{ NO _{3}^{-} / NO }^{ o }=0.96 V $
$E _{ NO _{3}^{-} / NO _{2}}^{ o }=0.79 V$ and at $ 298 K ,$
$\frac{ RT }{ F }(2.303)=0.059$

JEE MainJEE Main 2021Electrochemistry

Solution:

If the partial pressure of $NO$ and $NO_2$ gas is taken as 1 bar, then Answer is 4, else the question is bonus.

$NO_3^- + 4H^+ + 3e^- \to NO + 2H_2O$

$E^{o}_{NO^-_3 / NO} = 0.96 V$

$NO_3^- + 2H^+ + e^- \to NO_2 + H_2O$

$E^{o}_{NO_3^-/NO_2} = 0.79$

Let $\left[ HNO _{3}\right]= y \Rightarrow \left[ H ^{+}\right]= y$ and $\left[ NO _{3}\right]= y$

for same thermodynamic tendency

$E _{ NO^- _{3} / NO }= E _{ NO _{3}^{-} / NO _{2}}$

or, $E _{ NO _{3}^{-} / NO }^{\circ}-\frac{0.059}{3} \log \frac{ P _{ NO }}{ y \times y ^{4}}$

$= E _{ NO _{3}^{-} / NO _{2}}^{ o }-\frac{0.059}{1} \log \frac{ P _{ NO _{2}}}{ y \times y ^{2}}$

or, $0.96-\frac{0.059}{3} \log \frac{ P _{ NO }}{ y ^{5}}=0.79-\frac{0.059}{1} \log \frac{ P _{ NO _{2}}}{ y ^{3}}$

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

$0.17=-\frac{0.0591}{1} \log \frac{ P _{ NO _{2}}}{ y ^{3}}+\frac{0.0591}{3} \log \frac{ P _{ NO }}{ y ^{5}}$

$0.17=-\frac{0.0591}{3} \log \frac{ P _{ NO _{2}}^{3}}{ y ^{9}}+\frac{0.0591}{3} \log \frac{ P _{ NO }}{ y ^{5}}$

$0.17=\frac{0.0591}{3}\left[\log \frac{ P _{ NO }}{ y ^{5}}-\log \frac{ P _{ NO _{2}}^{3}}{ y ^{9}}\right]$

$0.17=\frac{0.0591}{3}\left[\log \frac{ P _{ NO }}{ y ^{5}} \times \frac{ y ^{9}}{ P _{ NO _{2}}^{3}}\right]$

Assume $ P_{ NO } \simeq P _{ NO _{2}}=1 bar$

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

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

$\log y=2.161$

$y=10^{2.1 6}$

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