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Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Redox Reactions
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$Cd(Hg).Cd5O_4.\large\frac{8}{3}$$H_2O(s)\parallel Cd5O_4(sat)Hg_25O_4(s)Hg$ and cell reaction is $Cd(Hg)+Hg_25O_4(s)+\large\frac{8}{3}$$H_2O(l)\rightarrow Cd5O_4.\large\frac{8}{3}$$H_2O(s)+2Hg(l)$.EMF of cell is 1.1V at $25^oC$.Then $\Delta H^o$ will be

$\begin{array}{1 1}(a)\;-210.86KJ\\(b)\;-208.85KJ\\(c)\;-212.89KJ\\(d)\;-211.89KJ\end{array}$

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$\Delta G^o=-\eta F E^o$
$\qquad=-2\times 96500\times 1.1$
$\Delta S^o=-\eta F \big(\large\frac{\partial E^o}{\partial T}\big)_P$
$\Delta G^o=\Delta H^o-T\Delta S^o$
$-212300+298\times 11.58=\Delta H^o$
$\Delta H^o=-208849.16J=-208.85KJ$
Hence (b) is the correct answer.
answered Dec 11, 2013 by sreemathi.v

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