Derive expression to calculate emf (reduction) of the following half c – InterviewSolution

1.	Derive expression to calculate emf (reduction) of the following half cells at 25^@C Hg_2Cl_2 (s), Cl^(-)\|Hg
Answer» Solution :`1/2Hg_2 Cl_2(s)+E leftrightarrow HG(L)+CL^(-)` (reduction) `E_(Hg_2Cl_2Cl^-)=E_(Hg_2Cl_2)^@ -0.0591/1 log""([Hg][Cl^-])/([Hg_2Cl_2])` `=E_(Hg_2Cl_2)^@ -0.0591 log[Cl^-]`

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