Copper reduces NO_{3}^{-} into NO and NO_{2} | vedclass

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(B) For the reduction of $NO_{3}^{-}$ to $NO$:$NO_{3}^{-} + 4H^{+} + 3e^{-} \longrightarrow NO + 2H_{2}O$$E_{1} = E_{NO_{3}^{-} / NO}^{\circ} - \frac{

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$4$

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(B) For the reduction of $NO_{3}^{-}$ to $NO$:$NO_{3}^{-} + 4H^{+} + 3e^{-} \longrightarrow NO + 2H_{2}O$$E_{1} = E_{NO_{3}^{-} / NO}^{\circ} - \frac{0.059}{3} \log \frac{P_{NO}}{[NO_{3}^{-}][H^{+}]^{4}}$For the reduction of $NO_{3}^{-}$ to $NO_{2}$:$NO_{3}^{-} + 2H^{+} + e^{-} \longrightarrow NO_{2} + H_{2}O$$E_{2} = E_{NO_{3}^{-} / NO_{2}}^{\circ} - \frac{0.059}{1} \log \frac{P_{NO_{2}}}{[NO_{3}^{-}][H^{+}]^{2}}$Given $[HNO_{3}] = y$,so $[H^{+}] = y$ and $[NO_{3}^{-}] = y$. Also $P_{NO} = P_{NO_{2}} = 1 \ bar$.Equating $E_{1} = E_{2}$:$0.96 - \frac{0.059}{3} \log \frac{1}{y \cdot y^{4}} = 0.79 - 0.059 \log \frac{1}{y \cdot y^{2}}$$0.17 = \frac{0.059}{3} \log (y^{-5}) - 0.059 \log (y^{-3})$$0.17 = -\frac{0.059}{3} \cdot 5 \log y + 0.059 \cdot 3 \log y$$0.17 = 0.059 \log y (3 - \frac{5}{3}) = 0.059 \log y (\frac{4}{3})$$\log y = \frac{0.17 	imes 3}{0.059 	imes 4} = \frac{0.51}{0.236} \approx 2.16$$y = 10^{2.16}$,so $x = 2.16$.$2x = 4.32$.

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