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Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Redox Reactions
0 votes

$E_{AgI/Ag/I^-}=$?,$E^o_{Ag^+/Ag}=0.8V,K_{spAg I}=10^{-18}$

$\begin{array}{1 1}(a)\;0.8V&(b)\;-0.28V\\(c)\;0.4V&(d)\;-0.56\end{array}$

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1 Answer

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By Nernst equation for metal-metal ppt electrode
$E_{Ag^+/Ag}=E^o_{ag^+/Ag}+\large\frac{0.06}{\eta}$$\log K_{sp}$
$\qquad\;\;\;=0.8+\large\frac{0.06}{1}$$\log 10^{-18}$
$\qquad\;\;\;=0.8-1.08$
$\qquad\;\;\;=-0.28$
Hence (b) is the correct answer.
answered Dec 11, 2013 by sreemathi.v
 

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