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Following cell has EMF $0.7995\: V$ .

$Pt \left |\: H_{2}(1\:atm) \right\:\: |\left | \:HNO_{3}(1M)\: \right\: |\left | AgNO_{3}(1M) \right |Ag$

we add enough $KCl$ to the Ag cell so that the final $Cl^{-}$ is $1M$ . Now the measured emf of the cell is $0.222 \:V$ . The Ksp of $AgCl$ would be :
Option: 1
$1\times 10^{-9.8}$

Option: 2
$1\times 10^{-19.6}$

Option: 3
$2\times 10^{-10}$

Option: 4
$2.64\times 10^{-14}$

Answers (1)

best_answer

$2Ag^{+}+H_{2}\rightarrow \:\:2H^{+}+2Ag$

$E=E^{o}-\frac{0.0591}{2}\:log\frac{(H^{+})^{2}}{P_{H_{2}}\times (Ag^{+})^{2}}$

$0.222=0.7995-\frac{0.0591}{2}\:log\frac{1}{(Ag^{+})^{2}}$

$(Ag^{+})=10^{-9.8}$

$Ksp=(Ag^{+})(Cl^{-})=(10^{-9.8})\times (1)=10^{-9.8}$

Posted by

sudhir kumar

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