InterviewSolution
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InterviewSolution
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The electrochemical cell shown below is a concentration cell. `M|M^(2+)(` saturated solution of sparingly soluble salt, `MX_(2))||M^(2+)(0.001 mol dm^(-3))|M` The `emf` of the cell depends on the difference in the concentration of `M^(2+)` ions at the two electrodes. The `emf` of the cell at `298` is `0.059V`. The solubility product `(K_(sp),mol^(3) dm^(-9))` of `MX_(2)` at 298 based on the information available the given concentration cell is `(` Take `2.303xxRxx298//F=0.059V)`A. `1xx10^(-15)`B. `4xx10^(-15)`C. `1xx10^(-15)`D. `4xx10^(12)` |
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Answer» Correct Answer - B For concentration cell, `E_(cell)=(0.0591)/(n)log.(C_(2(RHS)))/(C_(1(LHS)))` `E_(cell)=0.059V,C_(2(RHS))=0.001` `0.059=(0.0591)/(2)log.(0.001)/(C_(1))` `or (2xx0.059)/(0.0591)=log.(0.001)/(C_(1))" or anti log2"=(0.001)/(C_(1))` `thereforeC_(1)=(0.001)/(100)=10^(-5)` `C_(1)"=concentration or solubility of M"^(2+)=10^(-5)` `MX_(2)hArrM_(s)^(2+)+2underset(2S)X^(-)` `K_(sp)=S(2S)^(2)=4S^(3)` `K_(sp)=4xx(10^(-5))^(3)=4xx10^(-15)mol^(3)dm^(-9)` |
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