InterviewSolution
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InterviewSolution
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The standard reduction potential of the `Ag^(o+)|Ag` electrode at `298K` is `0.799V`. Given that for `AgI,K_(sp)=8.7xx10^(-17)`, evaluate the potential of the `Ag^(o+)|Ag` electrode in a saturated solution of `AgI`. Also calculate the standard reduction potential of the `I^(c-)`|Agl|Ag` electrode. |
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Answer» We know, `becauseE_(Ag^(+)//Ag)=E_(Ag^(+)//Ag)+(0.059)/(1)log_(10)[Ag^(+)]`...(1) Also `K_(SP_(Agl))=[Ag^(+)][I^(-)]` `because[Ag^(+)]=[I^(-)]` (for a saturated solution) `therefore[Ag^(+)]=sqrt(K_(SP_(Agl)))sqrt(8.7xx10^(-17))=9.32xx10^(-9)`...(2) `therefore` By Eq. (1) `E_(Ag^(+)//Ag)=0.799+(0.059)/(1)log_(10)(9.32xx10^(-9))` `=0.799-0.474=0.32V` Also `AgrarrAg^(+)+e,E_(OP)^(0)=-0.799V` `(AgI_((S))+erarrAg+I^(-))/(AgIrarrAg^(+)+I^(-))` `thereforeE_(cell)=E_(OP_(Ag//Ag^(+)))^(0)-(0.059)/(1)log[Ag^(+)]+E_(RP_(I^(-)//AgI//Ag))^(0)+(0.059)/(1)log.(1)/([I^(-)])`...(3) `becauseE_(cell)=0` at equilibrium, thus, from Eq. (3) `E_(O_(Ag//Ag^(+)))^(0)+E_(RP_(I^(-)//AgI//Ag))^(0)=(0.059)/(1)log[Ag^(+)][I^(-)]` `=(0.059)/(1)logK_(SP_(AgI))` `-0.799+E_(RP_(I^(-)//AgI//Ag))^(0)=(0.059)/(1)log8.7xx10^(-17)` `orE_(RP_(I^(-)//AgI//Ag))^(0)=-0.948+0.799=-0.149V` |
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