Copper reduces NO _ 3 ^ - into NO and NO _ 2 depending upon the...

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

${NO}_{3}^{-} + 4{H}^{+} + 3{e}^{-} ⟶ NO + 2{H}_{2}O {E}_{{NO}_{3}^{-} / NO}^{o} = 0 . 96V$

${NO}_{3}^{-} + 2{H}^{+} + {e}^{-} ⟶ {NO}_{2} + {H}_{2}O {E}_{{NO}_{3}^{-} / {NO}_{2}}^{o} = 0 . 79$

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

for same thermodynamic tendency

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

or, ${E}_{{NO}_{3}^{-} / NO}^{o} - \dfrac{0 . 059}{3}log\dfrac{{P}_{NO}}{y \times {y}^{4}} = {E}_{{NO}_{3}^{-} / {NO}_{2}}^{o} - \dfrac{0 . 059}{1}log\dfrac{{P}_{{NO}_{2}}}{y \times {y}^{2}}$

or, $0 . 96 - \dfrac{0 . 059}{3}log\dfrac{{P}_{N0}}{{y}^{5}} = 0 . 79 - \dfrac{0 . 059}{1}log\dfrac{{P}_{{NO}_{2}}}{{y}^{3}}$

or, $0 . 17 = - \dfrac{0 . 059}{1}log\dfrac{{P}_{{NO}_{2}}}{{y}^{3}} + \dfrac{0 . 059}{3}log\dfrac{{P}_{NO}}{{y}^{5}}$
$0 . 17 = - \dfrac{0 . 0591}{1}log\dfrac{{P}_{{NO}_{2}}}{{y}^{3}} + \dfrac{0 . 0591}{3}log\dfrac{{P}_{NO}}{{y}^{5}}$

$0 . 17 = - \dfrac{0 . 0591}{3}log{P}_{{NO}_{2}}^{3}{y}^{9} + \dfrac{0 . 0591}{3}log\dfrac{{P}_{NO}}{{y}^{5}}$

$0 . 17 = \dfrac{0 . 0591}{3}\left[log\dfrac{{P}_{NO}}{{y}^{5}} - log{P}_{{NO}_{2}}^{3}{y}^{9}\right]</p><p>$0 . 17 = $\dfrac{0 . 0591}{3}\left[log\dfrac{{P}_{NO}}{{y}^{5}} \times {y}^{9}{P}_{{NO}_{2}}^{3}\right]</p><p>Assume${P}_{NO} ≃ {P}_{{NO}_{2}} = 1bar$</p><p>\dfrac{0 . 17 \times 3}{0 . 059} ={\log y}^{4} = 8 . 644$

$\log y = \dfrac{8 . 644}{4}$

$\log y = 2 . 161$

$y = {10}^{2 . 1}$

$∴ 2x = 2 \times 2 . 161 = 4 . 322$

Answer (4)