JEE Main 2021 — Chemistry Physical Chemistry

2021integerhard

Copper reduces ${\mathrm{NO}}_{3}^{-}$ into $\mathrm{NO}$ and ${\mathrm{NO}}_{2}$ depending upon the concentration of ${\mathrm{HNO}}_{3}$ in solution. (Assuming fixed $[{\mathrm{Cu}}^{2+}]$ and ${P}_{\mathrm{NO}}={P}_{{\mathrm{NO}}_{2}}$ ), 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 ${10}^{x}M$ . The value of $2x$ is ___. (Rounded-off to the nearest integer)

[Given, ${E}_{{\mathrm{Cu}}^{2+}/\mathrm{Cu}}^{o}=0.34V,{E}_{{\mathrm{NO}}_{3}^{-}/\mathrm{NO}}^{o}=0.96V,{E}_{{\mathrm{NO}}_{3}^{-}/{\mathrm{NO}}_{2}}^{o}=0.79V$ and at $298K$ , $\frac{\mathrm{RT}}{F}(2.303)=0.059$ ]

Official previous-year question

Held on 25 Feb 2021 · Verified 6 Jul 2026.

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Solution

If the partial pressure of $\mathrm{NO}$ and ${\mathrm{NO}}_{2}$ gas is taken as $1$ bar, then Answer is $4$ , else the question is bonus.

${\mathrm{NO}}_{3}^{-}+4{H}^{+}+3{e}^{-}\longrightarrow \mathrm{NO}+2{H}_{2}O$ ${E}_{{\mathrm{NO}}_{3}^{-}/\mathrm{NO}}^{o}=0.96V$

${\mathrm{NO}}_{3}^{-}+2{H}^{+}+{e}^{-}\longrightarrow {\mathrm{NO}}_{2}+{H}_{2}O$ ${E}_{{\mathrm{NO}}_{3}^{-}/{\mathrm{NO}}_{2}}^{o}=0.79$

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

for same thermodynamic tendency

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

or, ${E}_{{\mathrm{NO}}_{3}^{-}/\mathrm{NO}}^{o}-\frac{0.059}{3}\mathrm{log}\frac{{P}_{\mathrm{NO}}}{y\times {y}^{4}}={E}_{{\mathrm{NO}}_{3}^{-}/{\mathrm{NO}}_{2}}^{o}-\frac{0.059}{1}\mathrm{log}\frac{{P}_{{\mathrm{NO}}_{2}}}{y\times {y}^{2}}$

or, $0.96-\frac{0.059}{3}\mathrm{log}\frac{{P}_{N0}}{{y}^{5}}=0.79-\frac{0.059}{1}\mathrm{log}\frac{{P}_{{\mathrm{NO}}_{2}}}{{y}^{3}}$

or, $0.17=-\frac{0.059}{1}\mathrm{log}\frac{{P}_{{\mathrm{NO}}_{2}}}{{y}^{3}}+\frac{0.059}{3}\mathrm{log}\frac{{P}_{\mathrm{NO}}}{{y}^{5}}$

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

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

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

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

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

$\frac{0.17\times 3}{0.059}={\mathrm{logy}}^{4}=8.644$

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

$\mathrm{log}y=2.161$

$y={10}^{2.1}$

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

Answer (4)

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