# 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.

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

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