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The standard reduction potentials of `Cu^(2+)//Cu` and `Ag^(+)//Ag` electrodes are 0.337 volt and 0.799 volt respectively. Construct a galvanic cell using these electrols so that its standard e.m.f. is positive. For what concentration of `Ag^(+)`, will the e.m.f. of the cell at `25^(@)C` be zero if the concentration of `Cu^(2+)` is 0.01 M.

Answer» `E_(cell)^(@)=E_(cathode)^(@)-E_(anode)^(@)`
=0.799-0.337=0.462 V
Applying Nernst equation ,
`E_(cell)=E_(cell)^(@)-(0.0591)/(n)"log"([Anode])/([Cathode])`
Given `" " E_(cell)^(@)=0.462 V , E_(cell)=zero , n=2, [Cu^(2+)]=0.01 M`
`0=0.462-(0.0591)/(2)"log"([Cu^(2+)])/([Ag^(+)]^(2))=0.462-0.02955" log"([Cu^(2+)])/([Ag^(+)]^(2)`
`"log"([Cu^(2+)])/([Ag^(+)]^(2))=(0.462)/(0.02955)=15.6345`
`([Cu^(2+)])/([Ag^(+)]^(2)) ="Antilog" 15.6345=4.3102xx10^(15)`
`[Ag^(+)]^(2)=(0.01)/(4.3102xx10^(15))=0.2320xx10^(-17)`
`[Ag^(+)]=(0.2320xx10^(-17))^(1//2)=(2.320xx10^(-18))^(1//2)=1.523xx10^(-9) M`.


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