Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

For the following galvanic cell the emf will be at $25^{\large\circ}C$.$Ag/AgCl(s),KCl(0.2M\parallel KBr(0.001M).AgBr(s)/Ag$.The solubility product of $AgCl$ & $AgBr$ are $2.8\times 10^{-10}$ and $3.3\times 10^{-13}$ respectively.

$\begin{array}{1 1}(a)\;-0.03V&(b)\;-0.1V\\(c)\;-0.049V&(d)\;-0.2V\end{array}$

Can you answer this question?

1 Answer

0 votes
$Ag(s)\leftrightharpoons Ag^+(C_1)+e^-$
$Ag^+(C_2)+e^-\leftrightharpoons Ag(s)$
For concentration cell
$Ag/Ag(C_1)(s),KCl(0.2M)\parallel KBr(0.001M),AgBr(s)/Ag$
Net reaction is
$Ag^+(C_1)\rightarrow Ag^+(C_1)$
For this reaction in above concentration cell
$E^0_{cell}=\large\frac{0.0591}{n}$$\log 10 \large\frac{C_2}{C_1}$-----(1)
$AgCl\leftrightharpoons Ag^++Cl^-$
$k_{sp}$ of $agCl=[[Ag^+][Cl^-]$
$K_{sp}$ of AgCl=$2.8\times 10^{-10}$
$2.8\times 10^{-10}=[Ag^+]\times 0.2$
$[Ag^+]$ of AgCl=$C_1$
$C_1=14\times 10^{-10}$M
Now $AgBr\leftrightharpoons Ag^++Br^-$
$K_{sp}$ of $AgBr=[Ag^+][Br^-]$
$K_{sp}$ of $AgBr=3.3\times 10^{-13}$
$3.3\times 10^{-13}=[Ag^+][0.001]$
$[Ag^+]=\large\frac{3.3\times 10^{-13}}{0.001}$$=3.3\times 10^{-10}M$
This $[Ag^+]$ of AgBr=$C_2$
$\therefore C_2=3.3\times 10^{-10}M$
Substituting values of $C_1,C_2$ and $n(n-1)$ in equ(1)
$E^0_{cell}=\large\frac{0.0591}{1}$$\log \large\frac{3.3\times 10^{-10}}{14\times 10^{-16}}$
$\Rightarrow -0.03V$
Hence (a) is the correct answer.
answered Feb 18, 2014 by sreemathi.v

Related questions

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