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. `Cl^(-) , Br^(-)` and `I^(-)`
B. `Br^(-)` and `I^(-)`
C. `Cl^(-)` and `Br^(-)`
D. `I^(-)` only

Answer

B

Explanation

`MnO_(4)^(-) + 8 H^(+) + 5e^(-) to Mn^(2+) + 4 H_(2)O`
` E = 1.51 - (0.059)/(5) "log" ([Mn^(2+)])/([MnO_(4)^(-)][H^(+)]^(8))`
Taking `Mn^(2+)` and `Mn O_(4)^(-)` in standard state i.e. 1M , `E = 1.51 - (0.059)/(5) xx 8 log (1)/([H^(+)])`
`=1.51 - (0.059)/(5) xx 8 xx 3 = 1.2268 V `
Hence at this pH , `MnO_(4)^(-)` will oxidise only `Br^(-)` and `I^(-)` as SRP of `Cl_(2)//Cl^(-)` is 1.36 V which is greater than that for `MnO_(4)^(-) // Mn^(2+)`.