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InterviewSolution
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Calculate the concentration of silver ions in the cell constructed by using `0.1` M concentration of `Cu^(2+) and Ag^(+)` ions. Cu and Ag metals are used as electrodes. The cell potential is `0.422` V. `[E _(Ag^(2+)//Ag)=0.80V,E_(Cu^(2+)//Cu)=+0.34V]` |
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Answer» Given `E^(0)of Ag^(+)//Ag =0.80V` `E^(0) of Cu^(+2)//Cu=0.34V` `E^(0)` of cell `E_(RHS) -E_(LHS)=0.80-0.34=0.46V` Cell is `Cu|Cu_(""(0.1M))^(+2)||Ag^(+)|Ag` ? `E_(cell ) =E_(cell)^(0) -(0.059)/(2)log ""(0.1)/([Ag^(+)]^(2))=-0.038=-0.295log ""(0.1)/([Ag ^(+)]^(2))` `log ""(0.1)/([Ag ^(+)]^(2))=(0.038)/(0.0295)=1.288` `[Ag+]^(2)=(0.1)/(19.32)=5. 176 xx10^(-3)` `[Ag^(+)]=sqrt(51. 76 xx10^(-4))=7.16xx10^(-2)M` |
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