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
B. `Cr_(2)O_(7)^(2-)` can be used in aqueous
C. `MNO_(4)^(-)` can be used in aqueous `H_(2)SO_(4)`
D. `Cr_(2)O_(7)^(2-)` can be used in aqueous `H_2SO_(4)`

Explanation

`M nO_(4)^(-)` will oxidize `Cl^(-)`ion according to equation
`M n^(7+)+5e to Mn^(2+)`
`2Cl^(-)to Cl_2+2e`
Thus, `E_(cell)^(@)=E_(Mn^(7+)//m_(m)^(2+))^(@)+E_(Cl_2//Cl^(-))^(@)`
or reaction is feasible. `M no_4^(-)` will oxidise
`Fe^(2+)` to `Fe^(3+)`
`M n^(7+)+5e to M n^(2+)`
`Fe^(2+)to Fe^(3+)+e therefore E_(cell)^(@)=E_(M n^(7+)//M n^(2+))^(@)-E_(Fe^(3+)//Fe^(2+))^(@)=1.51-0.77=0.74will oxidise`
Thus, `M no_4^(-)` will oxidise `Fe^(2+)` to `Fe^(3+)` aqueous HCl medium as well as it will also oxidise `Cl^(-)` to `Cl_2`. Suitable oxidant should not oxidise `Cl^(-)` to `Cl_2` and should only oxidise `Fe^(2+)` to `Fe^(3+)` in redox titrations.