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$\mathrm{Cd}_{(s)}+\mathrm{Hg}_{2} \mathrm{SO}_{4(s)}+\frac{9}{5} \mathrm{H}_{2} \mathrm{O}_{(l)} \rightleftharpoons \mathrm{CdSO}_{4} \cdot \frac{9}{5} \mathrm{H}_{2} \mathrm{O}_{(s)}+2 \mathrm{Hg}_{(l)}$
The value of $\mathrm{E}_{\text {cell }}^{0}$ is $4.315\, \mathrm{~V}$ at $25^{\circ} \mathrm{C}$. If $\Delta \mathrm{H}^{\circ}=-825.2\, \mathrm{~kJ} \,\mathrm{~mol}^{-1}$, the standard entropy change $\Delta \mathrm{S}^{\circ}$ in $\mathrm{J} \,\mathrm{K}^{-1}$ is ........ . (Nearest integer) [Given : Faraday constant $=96487\, \mathrm{C}\, \mathrm{mol}^{-1}$ ]
$N _{2} O _{4}( g ) \rightleftharpoons 2 NO _{2}( g ) ; \Delta H ^{0}=+58 kJ$
For each of the following cases $(a, b),$ the direction in which the equilibrium shifts is:
$(a)$ Temperature is decreased
$(b)$ Pressure is increased by adding $N _{2}$ at constant $T$