1.

For the measurement of the solubility product of AgCl the following cell is constructed : Ag|AgCl||KCl (0.1 M)||AgNO_(3) (0.1 M)|Ag The emf of the cell is 0.45 volt. In the cell, KCl is dissociated to the extent of 83% and AgNO_(3) is dissociated to the extent of 86%. Calculate the solubility product of AgCl at 298 K.

Answer»


SOLUTION :`E_(CELL)=E_(cell)^(@)-0.0591 "log "([Ag^(+)]_("ANODE"))/([Ag^(+)]_("CATHODE"))`


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