1.

Derive expression to calculate emf (reduction) of the following half cells at 25^@C AgCl(s),Cl^(-)|Ag

Answer»

SOLUTION :`AgCl(s)+e^(-) leftrightarrow Ag(s)+CL^(-) ` (REDUCTION)
`E_(AgCl,Cl^-)=E_(AgClCl^-)^@- 0.0591/1 LOG"" ([Ag][Cl^-])/([AgCl])`
`=E_(AgCl,Cl^-)^@ -0.0591 log[Cl^-]`
For Ag and AgCl, both being solids `[Ag]=[AgCl]=1`


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