Following cell has EMF $0.7995 \mathrm{~V}$. $\mathrm{Pt}\left|\mathrm{H}_{2}(1 \mathrm{~atm})ight|…

Following cell has EMF $0.7995 \mathrm{~V}$.
$\mathrm{Pt}\left|\mathrm{H}_{2}(1 \mathrm{~atm})ight| \mathrm{HNO}_{3}(1 \mathrm{M}) \| \mathrm{AgNO}_{3}(1 \mathrm{M}) \mid \mathrm{Ag}$
If we add enough $\mathrm{KCl}$ to the $\mathrm{Ag}$ cell so that the final $\mathrm{Cl}^{-}$ is $1 \mathrm{M}$. Now the measured emf of the cell is $0.222 \mathrm{~V}$. The $\mathrm{K}_{s p}$ of $\mathrm{AgCl}$ would be-
  1. $1 \times 10^{-9.8}$
  2. $1 \times 10^{-19.6}$
  3. $2 \times 10^{-10}$
  4. $2.64 \times 10^{-14}$

Solution

$2 \mathrm{Ag}^{+}+\mathrm{H}_{2} \longrightarrow 2 \mathrm{H}^{+}+2 \mathrm{Ag}$
$E=E^{\circ}-\frac{0.0591}{2} \log \frac{\left[\mathrm{H}^{+}ight]^{2}}{\mathrm{P}_{\mathrm{H}_{2}} \times\left[\mathrm{Ag}^{+}ight]^{2}}$
$0.222=0.7995-\frac{0.0591}{2} \log \frac{1}{\left[\mathrm{Ag}^{+}ight]^{2}}$
$\left[\mathrm{Ag}^{+}ight]=10^{-9.8}$
$K_{s p}=\left[\mathrm{Ag}^{+}ight]\left[\mathrm{Cl}^{-}ight]=\left(10^{-9.8}ight) \times(1)=10^{-9.8}$ ,

Asked in: JEE-TOPICTESTS-CHEMISTRY

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