Question

Find the solubility product of a saturated solution of $A{g}_{2}Cr{O}_4$ in water at $298K$, if the emf of the cell $Ag|A{g}^{+}$ (Saturated $A{g}_{2}Cr{O}_4$ solution.) $||A{g}^{+}(0.1M)|$ $Ag$ is $0.164V$ at $298K$.

Answer 1

$E=0-\frac{0.0592}{1}\log \frac{[A{g}^{+}{]}_{\text{anode}}}{[A{g}^{+}{]}_{\text{cathode}}}$
$\Rightarrow 0.164=-0.0592\log \frac{[A{g}^{+}]{+}_{\text{anode}}}{0.10}$
$\Rightarrow [A{g}^{+}{]}_{\text{anode}}=1.7\times 10^{-4}M$
In saturated $A{g}_{2}Cr{O}_4$ solution present in anode chamber :
$A{g}_{2}Cr{O}_4(s)\rightleftharpoons 2A{g}^{+}+Cr{O}_4^{2-}$
$1.7\times 10^{-4}M\frac{1.7}{2}\times 10^{-4}M$
$K_{sp}=[A{g}^{+}{]}^2[Cr{O}_4^{2-}]$
$=(1.7\times 10^{-4}{)}^2(\frac{1.7}{2}\times 10^{-4})$
$=2.45\times 10^{-12}$