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. `C1^(-),Br^(-)and I^(-)`
B. `C1^(-) and Br^(-)`
C. `Br^(-)and I^(-)`
D. `I(-) only`

Explanation

`M no_4^(-)+8H^(+)+5e^(-)to M n^(2+)+4H_2o`
`E=E^(0)-(0,059)/nlog.([M n^(2)])/([M no_4^(-)][H^(+)]^(3))`
Taking `[M n^(2+)]` and` [M no_4^(-)]` in standard state i.e , 1M.
`E=E^(0)-(0.059)/(n)log. 1/([H^(+)]^(3))=1.51-(0.059)/nxx8xxpH=1.51-(0.059)/5xx8xx3=1.227 V`
So, `MnO_4^(-)` eill oxidise only `Br^(-)` and `I^(-)` as standard reducing potential of `Cl_2//Cl^(-)` (1.26V) is greater than that of `M nO_4^(-)//M n^(2+)`Since,the value of `E^(@)`is -ve ,therefore the reaction is non sponteneous,
Alternate method:
`M n^(2)+2e^(-)to M n ,E^(@)=-1.18`....(i)
`DeltaG ^(@)-nFE^(@)` (here, n= number of `e^(-)` involved in the reaction )`DeltaG_1^(@)=-2F(-1.18)=2.36F`
`2M n^(3+)+2e^(-) to 2M n^(2+),E^(@)=+1.51V`....(ii)
`DeltaG_2^(@)=-2F[=1.51]=-3.03 F`
Substracting Eq. (ii) from Eq. (i), we get
`M n^(2+) to M n+ @M n^(3+), [n=2]`
`DeltaG_3^(@)=DeltaG_1^(@)-DeltaG_2^(@)=-5.38 F implies -2FE^(@)=5.38FimpliesE^(@)=-2.69V`