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Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Redox Reactions
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Calculate the solubility product of $AgBr$ in water at $25^{\large\circ}$ from the cell: $Ag,Ag^+Br^-$(Saturated solution). The standard potentials are $E^{\large\circ}_{AgBr,Ag}=0.07V$, $E^{\large\circ}_{Ag^+,Ag}=0.80V$. The $K_{SP}$ Value is

$\begin{array}{1 1}(A)\;4.81\times 10^{-9}&(B)\;4.81\times 10^{-11}\\(C)\;3.42\times 10^{-11}&(D)\;4.1\times 10^{-10}\end{array}$

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1 Answer

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Here,
R.H.E:$AgBr(s)+e^-\rightarrow Ag(s)+Br^-(aq);E^{\large\circ}_R=0.07V$
L.H.E:$Ag(s)\rightarrow Ag^+(aq)+e^-;E^{\large\circ}+L=0.80V$
Hence,for the overall $rx_n,AgBr(s) \rightleftharpoons Ag^+(aq)+Br^-(aq)$
$E^{\large\circ}=E^{\large\circ}_R-E^{\large\circ}_L=-0.73V$
$\Rightarrow \large\frac{0.0591}{1}$$\log\big((Ag^+)(Br^-)\big)$
$\Rightarrow 0.0591\log K_{SP}$
$\log K_{SP}=\large\frac{E^{\large\circ}}{0.0591}$
$\Rightarrow \large\frac{-0.73}{0.0591}$
$K_{SP}=4.81\times 10^{-11}$
Hence (B) is the correct answer.
answered Dec 10, 2013 by sreemathi.v
 

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