Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Redox Reactions
0 votes

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

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

Can you answer this question?

1 Answer

0 votes
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 $rx_n,AgBr(s) \rightleftharpoons Ag^+(aq)+Br^-(aq)$
$\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

Related questions

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