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1. |
Following cell has EMF 0.7995V. \[Pt|{{H}_{2}}(1\,atm)|HN{{O}_{3}}(1M)||AgN{{O}_{3}}(1M)|Ag\] 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 - |
A. | \[1\times {{10}^{-9.8}}\] |
B. | \[1\times {{10}^{-19.6}}\] |
C. | \[2\times {{10}^{-10}}\] |
D. | \[2.64\times {{10}^{-14}}\] |
Answer» B. \[1\times {{10}^{-19.6}}\] | |