A) \[1.39\,V\]
B) \[0.829\,V\]
C) \[-\,0.7705\,V\]
D) \[0.21\,V\]
Correct Answer: D
Solution :
\[{{K}_{sp}}\]for \[AgCl={{10}^{-9}}\times (0.1)={{10}^{-10}}\] \[\therefore \,E_{C{{l}^{-}}/AgCl/Ag}^{0}=+0.80+\frac{0.059}{1}{{\log }^{{{10}^{-10}}}}\] \[=0.21\,V\]You need to login to perform this action.
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