InterviewSolution
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InterviewSolution
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Standard electrode potenital data are useful for understanding the suitablity of an oxidant in a redox titration. Some half cell reactions and their standard potential are given below: `MnO_(4(aq))^(-) + 8H_((aq.))^(+) + 5e rarr Mn_((aq.))^(2+) + 4H_(2)O_(l),E^(@) = 1.51 V` `Cr_(2)O_(7(aq.))^(2-) + 14H_((aq.))^(+) + 6e rarr 2Cr_((aq.))^(3+) + 7H_(2)O_(l), E^(@) = 1.38 V` `Fe_(aq.)^(3+) 2e^(-) rarr Fe_((aq.))^(2), E^(@) = 0.77 V` `Cl_(2(g)) + 2e^(-) rarr 2Cl_((aq.))^(-), E^(@) = 1.40 V` Identify the only incorrect statement regarding the quantitative estimation of aqueous `Fe(NO_(3))_(2)`:A. `MnO_(4)^(-)` can be used in aqueous `HCl`B. `Cr_(2)O_(7)^(2-)` can be used in aqueous `HCl`C. `MnO_(4)^(-)` can be used in aqueous `H_(2)SO_(4)`D. `Cr_(2)O_(7)^(2-)` can be used in aqueous `H_(2)SO_(4)` |
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Answer» Correct Answer - A `MnO_(4)^(-)` will oxidise `Cl^(-)` ion according to equation. `Mn^(7+) + 5e rarr Mn^(2+)` `2Cl^(-) rarr Cl_(2) + 2e` Thus `E_(cell)^(@) = E_(OP_(Cl^(-)//Cl_(2)))^(@) + E_(RP_Mn^(7+)//Mn^(2))^(@)` `= -1.40 + 1.51 = 0.11 V` or reaction is feasible. `MnO_(4)^(-)` will oxidise `Fe^(2+)` to `Fe^(3+)` `Mn^(7+) + 5e rarr Mn^(2+)` `Fe^(2+) rarr Fe^(3+) + e` `E_(cell)^(@) = E_(OPFe^(2+)//Fe^(3+))^(@) + E_(RPMn^(7+)//Mn^(2+))^(@)` `= -0.77 + 1.51 = 0.74 V` or reaction is feasible. Thus `MnO_(4)^(-)` will not oxidise only `Fe^(2+)` to `Fe^(3+)` in aqueous `HCl` but it will also oxidise `Cl^(-)` to `Cl_(2)`. Suitable oxidant shoould not oxidise `Cl^(-)` to `Cl_(2)` and should oxidise only `Fe^(2+)` to `Fe^(3+)` in redox titration. |
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