logo

Ask Questions, Get Answers

X
 
Home  >>  JEEMAIN and NEET  >>  Chemistry  >>  Electrochemistry

The EMF of cell $Ag/AgCl(K_{sp})(saturated\; sol)\parallel Cl^-(C_1)/AgCl(K_{sp}/Ag)$ is given by

$\begin{array}{1 1}(a)\;E=-\large\frac{0.059}{1}\normalsize\log \large\frac{\sqrt{K_{sp}}}{C_1}\\(b)\;E=-\large\frac{0.059}{1}\normalsize\log \large\frac{C_1}{\sqrt{K_{sp}}}\\(c)\;E=-\large\frac{0.059}{1}\normalsize\log \large\frac{\sqrt{K_{sp}}}{C_1^2}\\(d)\;E=\large\frac{0.059}{1}\normalsize\log \large\frac{C_1}{\sqrt{K_{sp}}}\end{array}$

Download clay6 mobile app

1 Answer

The cell can be reduced to $Ag/AgCl$(saturated solution)$\parallel Ag^+(\large\frac{K_{sp}}{[Cl^-]})$$/Ag$
The reactions occuring at 2 electrodes are
At anode : $Ag\rightarrow Ag^+_A+e^-$
At cathode : $Ag^+_C+e^-\rightarrow Ag$
Net reaction : $Ag^+_C+e^-\rightarrow Ag^+_A$
$\therefore E_{cell}=E^0_{cell}-0.059\log \large\frac{[Ag_A^+]}{[Ag^+_C]}$
$\Rightarrow 0.059\log \large\frac{[Ag^+_C]}{[Ag_A^+]}$
Since it is a concentration cell for which $E^0_{cell}=0$
$\therefore E_{cell}=0.059\log \large\frac{K_{sp}}{[Cl^+]\sqrt{K_{sp}}}$
$\Rightarrow -\large\frac{0.059}{1}\normalsize\log \large\frac{C_1}{\sqrt{K_{sp}}}$
Hence (b) is the correct answer.
answered Feb 17, 2014 by sreemathi.v
 

Related questions

Ask Question
Tag:MathPhyChemBioOther
...
X