If an element can exist in several oxidation states, it is convernient to display the reduction potentials correspondingg to the various half reactions in diagrammatic form, know as latimer diagram the latimer diagram for chlorine in acid solutio is
`CIO_(4)^(-)overset(+1.20V)toCiO_(3)^(-)overset(+1.18V)toHClO_(2)overset(+1.60V)toHClOoverset(1.67V)toCl_(2)overset(1.36V)toCl^(-)` in basic solution.
`ClO_(4)^(-)overset(0.37V)toClO_(3)^(-)overset(0.30V)toClO_(2)^(-)overset(0.68V)toClO^(-)overset(0.42V)toCl_(2)overset(1.36V)toCl^(-)` The standard potentials for two nonadjacent species can also be calculateed by using the concept that `triangleG^(@)` as an additive property but potential is not an additive property and `triangleG^(@)=-nFx^(0)`. if a given oxidation stateis a stronger oxidising agent than the next higher oxidation state, disproportionation can occur. The reverse of disproportionation is called comproportionation. The relative stabilities of the oxidation state can also be understood by drawing a graph of `triangleG^(@)//F` against oxidation state, known as frost diagram, choosing the stability of zero oxidation state arbitrarily as zero. The most stable oxidation state of a species lies lowest in the diagram, disproportionation is spontaneous if the species lies above a straight line joining its two product species.
Q. What is the potential couple `(ClO^(-))/(Cl^(-))` at `pH=14`?

Options

A. `1.78V`
B. `-0.94V`
C. `0.89V`
D. `-0.89V`

Explanation

To find the potential couple \((ClO^(-))/(Cl^(-))\) at \(pH = 14\), we can follow these steps:

### Step 1: Identify the relevant half-reactions
From the Latimer diagram provided, we can identify the half-reactions involving \(ClO^-\) and \(Cl^-\). The relevant half-reactions are:
1. \(ClO^-(aq) + 2e^- \rightarrow Cl^-(aq)\)

### Step 2: Determine the standard reduction potentials
From the Latimer diagram for basic solution, we have the following standard reduction potentials:
- \(ClO^-(aq) \rightarrow ClO_2^-(aq)\) with \(E^0 = 0.68 V\)
- \(ClO_2^-(aq) \rightarrow ClO_3^-(aq)\) with \(E^0 = 0.30 V\)
- \(ClO_3^-(aq) \rightarrow ClO_4^-(aq)\) with \(E^0 = 0.37 V\)
- \(ClO^-(aq) \rightarrow Cl^-(aq)\) with \(E^0 = 0.42 V\)
- \(Cl^-(aq) \rightarrow Cl_2(g)\) with \(E^0 = 1.36 V\)

### Step 3: Calculate the potential for the couple \((ClO^-(aq))/(Cl^-(aq))\)
Using the formula for calculating the potential of non-adjacent species:
\[ E^0 = \frac{E^0_1 + E^0_2}{2} \]
where \(E^0_1\) is the potential for \(ClO^-\) and \(E^0_2\) is the potential for \(Cl^-\).

From the Latimer diagram:
- \(E^0(ClO^-) = 0.42 V\)
- \(E^0(Cl^-) = 1.36 V\)

Now, substitute these values into the formula:
\[ E^0 = \frac{0.42 + 1.36}{2} = \frac{1.78}{2} = 0.89 V \]

### Step 4: Conclusion
The potential couple \((ClO^-(aq))/(Cl^-(aq))\) at \(pH = 14\) is \(0.89 V\).

### Final Answer:
\[ \text{The potential couple } (ClO^-(aq))/(Cl^-(aq)) \text{ at } pH = 14 \text{ is } 0.89 V. \]

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