Find the solubility product (K_{sp}) of a sat | vedclass

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(D) The cell reaction is: $Ag_{(s)} + Ag^{+}(0.1 \ M) \rightarrow Ag^{+}(C_1) + Ag_{(s)}$.Using the Nernst equation: $E_{cell} = E^{\circ}_{cell} - \f

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$5 	imes 10^{-34} \ M^3$

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(D) The cell reaction is: $Ag_{(s)} + Ag^{+}(0.1 \ M) ightarrow Ag^{+}(C_1) + Ag_{(s)}$.Using the Nernst equation: $E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log \frac{[Ag^{+}]_{anode}}{[Ag^{+}]_{cathode}}$.Since $E^{\circ}_{cell} = 0$,we have $0.591 = 0 - 0.0591 \log \frac{C_1}{0.1}$.$-10 = \log \frac{C_1}{0.1} \implies \frac{C_1}{0.1} = 10^{-10} \implies C_1 = 10^{-11} \ M$.For $Ag_2CrO_4 ightleftharpoons 2Ag^{+} + CrO_4^{2-}$,$[Ag^{+}] = 10^{-11} \ M$.Solubility $S = \frac{[Ag^{+}]}{2} = 0.5 	imes 10^{-11} \ M$.$K_{sp} = [Ag^{+}]^2 [CrO_4^{2-}] = (10^{-11})^2 	imes (0.5 	imes 10^{-11}) = 0.5 	imes 10^{-33} = 5 	imes 10^{-34} \ M^3$.

Find the solubility product $(K_{sp})$ of a saturated solution of $Ag_2CrO_4$ in water at $298 \ K$,if the $emf$ of the cell $Ag | Ag^{+} (\text{satd. } Ag_2CrO_4 \text{ solution}) || Ag^{+} (0.1 \ M) | Ag$ is $0.591 \ V$ at $298 \ K$.

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