Copper reduces $N O_{3}^{-}$ into NO and NO₂ depending upon the concentration of HNO₃ in solution. (Assuming fixed $[C u^{2 +}] a n d P_{N O} = P_{N O_{2}}$ ), the HNO₃ concentration at which the thermodynamic tendency for reduction of $N O_{3}^{-} i n t o$ NO and NO₂by copper is same is 10^x M. The value of 2x is…….. (rounded-off to the nearest integer).

$[G i v e n, E_{C u^{2 +} / C u}^{0} = 0.34 V, E_{N O_{3}^{-} / N O}^{0} = 0.96 V, E_{N O_{3}^{-} / N O_{2}}^{0} = 0.79 V a n d a t 298 K, \frac{R T}{F} (2.303) = 0.059]$

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Payal Gupta | Contributor-Level 10

6 months ago

$N O_{3}^{-} + 4 H^{+} + 3 e^{-} \to N O + 2 H_{2} O; E_{N O_{3}^{-}}^{0} = 0.96 V$

$L e t [H N O_{3}] = y M$

For same thermodynamic tendency of reduction

$E_{N O_{3}^{-} / N O} = E_{N O_{3}^{-} / N O_{2}}$

$0.96 - \frac{0.059}{3} l o g \frac{p_{N O}}{y^{5}} = 0.79 - 0.059 l o g \frac{p_{N O_{2}}}{y^{3}}$

Assume $(p_{N O} = p_{N O_{2}} = 1 b a r)$

$0.17 = - \frac{0.0591}{1} l o g \frac{p_{N O_{2}}}{y^{3}} + \frac{0.059}{3} l o g \frac{p_{N O}}{y^{5}}$

$= \frac{0.059}{3} [l o g \frac{p_{N O}}{y^{5}} - l o g \frac{{p^{3}}_{N O_{2}}}{y^{9}}] = \frac{0.059}{3} * l o g [\frac{p_{N O}}{y^{5}} * \frac{y^{9}}{{p^{3}}_{N O_{2}}}] = \frac{0.059}{3} l o g y^{4}$

$l o g y^{4} = \frac{0.17 * 3}{0.059} \Rightarrow y = 1 0^{2.16}$

Comparing x = 2.16 and 2x = 4.32

Nearest integer is 4.

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