InterviewSolution
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InterviewSolution
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Standard electrode potential data are useful for understanding the suitability of an oxidant in a redox titration. Some half cell reaction and their standard potentials are given below: `MnO_(4)^(-)(aq) +8H^(+)(aq) +5e^(-) rarr Mn^(2+)(aq) +4H_(2)O(l) E^(@) = 1.51V` `Cr_(2)O_(7)^(2-)(aq) +14H^(+) (aq) +6e^(-) rarr 2Cr^(3+)(aq) +7H_(2)O(l), E^(@) = 1.38V` `Fe^(3+) (aq) +e^(-) rarr Fe^(2+) (aq), E^(@) = 0.77V` `CI_(2)(g) +2e^(-) rarr 2CI^(-)(aq), E^(@) = 1.40V` Identify the only incorrect statement regarding quantitative estimation of aqueous `Fe(NO_(3))_(2)`A. `MnO_(4)^(-)` can be used in aqueous `HCI`B. `Cr_(2)O_(7)^(2-)` can be used in aqueous `HCI`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 `2MnO_(4)^(Theta) +16H^(o+) +10CI^(Theta) rarr 5CI_(2)(g) +2Mn^(2+) 1+8H_(2)O` Cell constituted is `Pt|CI_(2)|CI^(Theta) ||MnO_(4)^(-) |Mn^(2+), H^(+)|Pt` `E_(cell) = E_(MnO_(4)^(Theta)//Mn^(2+))^(Theta) -E_(CI_(2)//CI^(Theta))` `= 1.51-1.40 = 0.11V` Since the cell voltage is +ve, hence reaction will take place. `MnO_(4)^(Theta)` will oxidise both `Fe^(+2)` and `HCI`. |
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