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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.

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

1 Answer

$Ag(s)\leftrightharpoons Ag^+(C_1)+e^-$
$Ag^+(C_2)+e^-\leftrightharpoons Ag(s)$
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

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