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InterviewSolution
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A copper-silver cell is set up. The copper ion concentration in it is 0.10 M. The concentration of silver ions is not known.The cell potential measured is 0.422 V. Determine the concentration of silver ions in the cell. [Given `E_(Ag^(+)//Ag)^(@)=0.80,E_(Cu^(2+)//Cu)^(@)=+0.34 V]` |
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Answer» Half cell reactions : At cathod :`2Ag^(+)(aq)+2e^(-)to2Ag(s)` At anode : `Cu(s)toCu^(2+)(aq)+2e^(-)` The cell reaction is `Cu(s)+2Ag^(+)(aq)toCu^(2+)(aq)+2Ag(s)` The cell can represented as `Cu(s)|Cu^(2+)(aq)||Ag^(+)(aq)|Ag(s)` `:.E_("cell")^(@)=E_("cathode")^(@)-E_("anone")^(@)` `=0.80=0.34=0`.46V` `:.E_("cell")^(@)=E_("cell")^(@)-(0.059)/(n)"log " ([Cu^(2+)(aq)])/([Ag^(+)(aq)]^(2))` `0.422-0.46=(-0.059)/(2)"log" [-1-2[Ag^(+)]]` `-0.038=0.95+0.059" log" [Ag^(+)]` `-log[Ag^(+)]=(0.0675)/(-0.059)` `log[Ag^(+)]=1.14407` `:.[Ag^(+)]="Antilog" (1.144)` `[Ag^(+)]=13.93M` |
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