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 correct statement regarding quantitative estimation of aqueous `Fe(NO_(3))_(2)`

Options

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)`

Answer

A

Explanation

The oxidation of `Cl^(-)` ion by `MnO_(4)^(-)` can be represented as `2MnO_(4)^(-) + 16 H+ +10Cl^(-) to 2Mn^(2+) + 8H_(2)O + 5Cl_(2(g))`
The cell corresponding to the above equation is
`Pt, Cl_(2) ` (1 atm)/`Cl^(-)|| MnO_(4)^(-) , Mn^(2+) , H^(+)//Pt`
`E_("cell")^(@)` is positive so `DeltaG^(@)(DeltaG^(@)=-nFE^(@))` must be negative and so this cell is feasible
`MnO_(4)^(-) ` will oxidise both `Fe^(2+)` ions and `Cl^(-)` ions simultaneously and so it cannot be used for the quantitative estimation of aq. `Fe(NO_(3))_(2)`