Redox Reactions And Electrochemistry Question 461
Question: Following cell has EMF 0.7995V. $ Pt|H_2(1,atm)|HNO_3(1M)||AgNO_3(1M)|Ag $ If we add enough $ KCl $ to the Ag cell so that the final $ C{l^{-}} $ is 1M. Now the measured emf of the cell is 0.222V. The $ K_{sp} $ of $ AgCl $ would be -
Options:
A) $ 1\times {10^{-9.8}} $
B) $ 1\times {10^{-19.6}} $
C) $ 2\times {10^{-10}} $
D) $ 2.64\times {10^{-14}} $
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Answer:
Correct Answer: A
Solution:
[a] $ 2A{g^{+}}+H_2\xrightarrow{{}}2{H^{+}}+2Ag $
$ E=E{}^\circ -\frac{0.0591}{2}\log \frac{{{[{H^{+}}]}^{2}}}{{P_{H_2}}\times {{[A{g^{+}}]}^{2}}} $ $ 0.222=0.7995-\frac{0.0591}{2}log\frac{1}{{{[A{g^{+}}]}^{2}}} $
$ [A{g^{+}}]={10^{-9.8}} $ $ K_{sp}=[A{g^{+}}][C{l^{-}}]=({10^{-9.8}})\times (1)={10^{-9.8}} $
BETA
