If an element can exist in several oxidation states, it is convenient to display the reduction potentials corresponding to the various half reactions in diagrammatic from, known as Latimer diagram. The Latimer diagram for chlorine in acid solution is
`Cl_(4)^(-)overset(+1.20V)toClO_(3)^(-)overset(+1.60V)toClO_(2)^(-)overset(+1.60V)to`
`ClO^(-)overset(+1.67V)toCl_(2)overset(+1.36V)toCl^(-)`
In basic solution is :
`ClO_(4)^(-)overset(+0.37V)toClO_(3)^(-)overset(+0.30V)toClO_(2)^(-)overset(+0.68)to`
`ClO^(-)overset(+0.42V)toCl_(2)overset(+1.36)toCl^(-)`. The standard potentials for two nonadjacent species can also be calculated by using the concept that `DeltaG^(@)` is an additive property but using potential is not an assitive property and `DeltaG^(@)=-nFx^(0),` If a given oxidation state is a the next higher oxidation state disproportionation can occur. The reverse of relative stabilities of the oxidation state can also be understood by drawing a graph of `DeltaG^(@)//F` against oxidation state, known as Frost diagram, Choosing the stability of zero oxidation state arbitrery as zero. The most stable oxidation state of a apecies lies lowest in the digram Disproportionation is spontaneous if the species lies above a straight line joining its two product species.
Which of the following statement s correct ?

Options

A. `Cl_(2)` undergoes disproportionation into `Cl^(-0` and `ClO^(-)` at pH = 0 and pH = 14
B. `Cl_(2)` undergoes disproportionation into `Cl^(-)` and `ClO^(-)` at pH = 14 and pH = 0
C. `Cl_(2)` undergoes disproportionation into `Cl^(-)` and `ClO^(-)` at pH = 0 not at pH = 14
D. None of the above

Answer

B