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 a sparingly soluble salt, `MX_(2))||M^(2+)(0.001 mol dm^(-3))|M` The emf of the cell depends on the difference in concentrations of `Mn^(2+)` ions at the two electrodes. The emf of the cell at `298 K` is `0.059 V`. The solublity product `(K_(SP) , "mol"^(3)"dm"^(-9))` of `MX_(2)` at `298 K` based on the information available for the given concentration cell is : (take `2.303 xx R xx 298//F = 0.059V`)A. `1 xx 10^(-15)`B. `4 xx 10^(-15)`C. `1 xx 10^(-12)`D. `4 xx 10^(-12)` |
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Answer» Correct Answer - B For concentration cell, `E_(cell) = (0.059)/(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 `(2 xx 0.59)/(0.0591) = "log"(0.001)/(C_(1))` or antilog `2 = (0.001)/(C_(1))` `:. C_(1) = (0.001)/(100) = 10^(-5)` `C_(1) =` concentration or solubility of `M^(2+) = 10^(-5)` `MX_(2) hArr underset(S)(M^(2+))+underset(2S)(2X^(-))` `K_(SP) = S(2S)^(2) = 4S^(3)` `K_(SP) = 4 xx (10^(-5))^(3) = 4 xx 10^(-15) "mol"^(3) "dm"^(-9)` |
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