Standard electrode potential data are used for understanding the stability of an oxidant in a redox titration . Some half reactions and their standard potentials are given below :
`MnO_4^(-) (aq) +8H^(+)(aq)+5e^(-)rarrMn^(2+)(aq)+4H_2O(l)`
`" "E^@=1.51V`
`Cr_2O_7^(2-) (aq)+14H^(+)(aq)+6e^(-) rarr 2Cr^(3+)(aq)+7H_2O(l)`
`" "E^@=1.38V`
`Fe^(3+)(aq)+e^(-) rarr Fe^(2+)(aq) " "E^@=0.77V`
`Cl_2(g)+2e^(-) rarr2Cl^(-) (aq)" "E^@=1.40V`
Identify the only incorrect statement regarding the quantitative estirnation of aqueous `Fe(NO_3)_2`.

Options

A. `MnO_4^(-)` can be used in aqueous HCl
B. `Cr_2O_7^(2-)` can be used in aqueous HCl
C. `MnO_4^(-)` can be used in aqueous `H_2SO_4`
D. `Cr_2O_7^(2-)` can be used in aqueous `H_2SO_4`

Answer

A

Explanation

`MnO_4^(-)` will positive , the reaction is spontaneous .
`2KMnO_4+16H^(+)+16Cl^(-)rarr 2Mn^(2+)+8H_2O+5Cl_2`
The corresponding cell is
`Pt|Cl_2(g)|Cl^(-)(aq)||MnO_(4)^(-)(aq), Mn^(2+),H^(+)|Pt`
`E_("cell")^@=E^@` (cathode) `-E^@` (anode)
= 1.51 - 1.40 = 0.11 V
Since , `E_("cell")^@` is +ve , the above reaction is feasible . Therefore , `MnO_4^(-)` will oxidise both `Fe^(2+) and Cl^(-)` ions in aqueous medium . Thus, quantitative estimation of `Fe(NO_3)_2` is not suitable.