logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Redox Reactions
0 votes

$E_{cell}$ for reduction $NO_3^-\rightarrow NO$ by $Ag(s)$?

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

$[Ag^+]=0.2M$

$[HNO_3]=1.5M$

$\begin{array}{1 1}(a)\;0.24v\\(b)\;0.25v\\(c)\;0.3v\\(d)\;0.27v\end{array}$

Can you answer this question?
 
 

1 Answer

0 votes
$Ag^++e^-\rightarrow Ag$
$E^o=0.79$
$E_{Ag^+/Ag}=0.79-\large\frac{0.06}{1}$$\log\large\frac{1}{1.5}$
$E_{Ag^+/Ag}=0.80$
$Ag+NO_3^-\rightarrow Ag^++NO$
At cathode
$3e^-+NO_3^-+4H^+\rightarrow NO+2H_2O$
$(E_{red})_{cathode}=1-\large\frac{0.06}{3}$$\log\large\frac{P_{NO}}{[NO_3^-][H^+]^4}$
$(E_{red})_{cathode}=1-0.02\log\large\frac{106{-3}}{1.5^5}$
$\Rightarrow 1.07v$
$E_{cell}=1.07-0.80$
$\quad\;\;\;=0.27v$
Hence (d) is the correct answer.
answered Dec 12, 2013 by sreemathi.v
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...