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At what $[HNO_3]$ thermodynamic tendency or reduction of $NO_3^-$ into NO and $NO_2$ by $Ag$ is same?

$E^o_{Ag^+/Ag}=0.79$volt

$E^o_{NO_3^-/NO}=1$volt

$E^o_{NO_3^-/NO_2}=0.82$volt

$298\large\frac{RT}{F}$$=0.06$volt

$P_{NO_2}=P_{NO}=10^{-3}$

T=298K

$[Ag^+]=0.2M$

$[HNO_3]=1.5M$

$\begin{array}{1 1}(a)\;10^{10}\\(b)\;10^{\Large\frac{1}{4}}\\(c)\;10^{\Large\frac{3}{4}}\\(d)\;10^{\Large\frac{7}{4}}\end{array}$

1 Answer

$e^-+NO_3^-+2H^+\rightarrow NO_2+H_2O$
$E_1=0.82-\large\frac{0.06}{1}$$\log\large\frac{P_{NO_2}}{[NO_3^-][H^+]^2}$
Let $[HNO_3]$ be $x$
$E_1=0.82-0.06\log\large\frac{10^{-3}}{x^3}$
$3^-+NO_3^-+4H^+\rightarrow NO+2H_2O$
$E_2=1-\large\frac{0.06}{3}$$\log\large\frac{P_{NO}}{[NO_3^-][H^+]^4}$
$E_2=1-0.02\log\large\frac{10^{-3}}{x^5}$
For same tendency
$E_1=E_2$
$0.82-0.06\log\large\frac{10^{-3}}{x^2}$$=1-0.02\log\large\frac{10^{-3}}{x^5}$
$x=10^{\large\frac{3}{4}}$
Hence (c) is the correct answer.
answered Dec 12, 2013 by sreemathi.v
 

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