At 298 K, the standard reduction potentials are 1.51 V for `MnO_(4)^(-) | Mn^(2+), 1.36 V` for `Cl^(2) | Cl^(-)`, 1.07 V for `Br_(2)|Br^(-),` and 0.54 V for `I_(2)|I^(-)`. At pH=3, permanganate is expected to oxidize `((RT)/(F) = 0.059V)`:

Options

A. `CI^(-) Br^(-) and T^(-)`
B. `Br^(-)` and `I^(-)`
C. `CI^(-)` and `Br^(-)`
D. only `I^(-)`

Answer

B

Explanation

At `pH=3, MnO_(2)^(-)` will oxidise only those species which have lower reduction potential than reduction potential of `MnO_(2)^(-)`
From given equation `MnO_(4)^(-)|Mn^(2+)=1.51 V`
`CI_(2)|CI^(-)=1.36 BV , Br_(2)|Br^(-)=0.54 v`
So in acidic medium
from nernst equation of elctrode potentential
`E=1.51 -(0.0591)/(5) log [(Mn^(2+))]/[(MnO_(4))][(H^(+))]^(6)`
Taking `Mn^(2+)` and `MnO_(4)^(-)` in standard state i.e 1 M
`E=1.51 -(0.0591)/(5)xx8log (1)/(H^(+))`
`E= 1.51 -(0.059)/(5)xx8xx3=1.2268` V
Hence at this pH `MnO_(4)^(-)` will oxidise only `Br^(-)` and `l^(-)` as standard reduction potential of `CI_(2)//CI^(-)` is 1.36 V which is greater than that for `MnO_(4)^(-)//Mn^(2+)`