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Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Redox Reactions
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$Cu/CuSO_4\parallel AgNO_3/Ag$, Find $K_{eq}$ for the reaction. Given : $E^o_{Cu^{+2}/Cu}=0.34, E^o_{Ag^+/Ag}=0.8, T=298K$

$\begin{array}{1 1}(a)\;10^{42.71}&(b)\;10^{39.62}\\(c)\;20^{-42.71}&(d)\;20^{-39.62}\end{array}$

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1 Answer

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$\Delta G^o=-\eta F \large\epsilon^o$
$\Delta^o=-RTl_nKeq$
$-\eta F\large \epsilon^o$$=-RTl_nKeq$
$E^o_{cell}==\large\frac{RT}{\eta F}$$\times 2.3031\log Keq$
At 298K
$E^o_{cell}=\large\frac{0.059}{2}$$\log Keq\qquad\eta=2$
$E^o=2E^o_{Ag^+/Ag}-e^o_{Cu^{+2}/Cu}$
$\;\;\;\;\;=2\times 0.8-0.34$
$\;\;\;\;\;=1.6-0.34=1.26V$
So $1.26=\large\frac{0.059}{2}$$\log Keq$
$Keq=10^{42.71}$
Hence (a) is the correct answer.
answered Dec 11, 2013 by sreemathi.v
 

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