Question

Using the following Latimer diagram for bromine, pH = 0;
\[BrO_{4}^{-}\xrightarrow{1.82V}BrO_{3}^{-}\xrightarrow{1.50V}HBrO\xrightarrow{1.595V}B{{r}_{2}}\xrightarrow{1.06552}B{{r}^{-}}\]
The species undergoing disproportionation is:
a.) \[BrO_{4}^{-}\]
b.) \[BrO_{3}^{-}\]
c.) \[HBrO\]
d.) \[B{{r}_{2}}\]

Answer 1

Hint: Disproportionation reaction, sometimes called as dismutation. It is one type of redox reaction in which one compound formed in the intermediate oxidation state changes to two compounds, one of higher oxidation state and another of lower oxidation state.

Complete step by step answer:
The given equation is
\[BrO_{4}^{-}\xrightarrow{1.82V}BrO_{3}^{-}\xrightarrow{1.50V}HBrO\xrightarrow{1.595V}B{{r}_{2}}\xrightarrow{1.06552}B{{r}^{-}}\]

In the first step
\[\begin{align}
 & BrO_{4}^{-}\xrightarrow{1.82V}BrO_{3}^{-} \\
 & (+7)\text{ (+5)} \\
\end{align}\] here \[E_{1}^{o}\]= 1.82 V
In the above reaction the oxidation state of bromine is changing from +7 to +5.

In the second step
\[\begin{align}
 & BrO_{3}^{-}\xrightarrow{1.50V}HBrO \\
 & (+5)\text{ (+1)} \\
\end{align}\]here \[E_{2}^{o}\]= 1.50 V
In the above reaction the oxidation state of bromine is changing from +5 to +1.

In the third step
\[\begin{align}
 & HBrO\xrightarrow{1.595V}B{{r}_{2}} \\
 & (+1)\text{ (0)} \\
\end{align}\] here \[E_{3}^{o}\]= 1.595 V
In the above reaction the oxidation state of bromine is changing from +1 to 0.

In the fourth step
\[\begin{align}
 & B{{r}_{2}}\xrightarrow{1.06552}B{{r}^{-}} \\
 & (0)\text{ (-1)} \\
\end{align}\]here \[E_{4}^{o}\]= 1.0652 V
In the above reaction the oxidation state of bromine is changing from 0 to -1.

Now we have to find the disproportionation step among the above steps.
Now, we have to calculate the cell potential by taking two steps as one cell.
Step-1 and step-2 one cell, step-2 and step-3, step-3 and step-4. Totally there is a chance of formation of three cells.

\[\begin{align}
 & E_{cell}^{1}=E_{2}^{o}-E_{1}^{o} \\
 & \text{ }=\text{ }1.5-1.82 \\
 & \text{ }=\text{ }-0.32 \\
\end{align}\] (cell potential is negative means no reaction)
\[\begin{align}
 & E_{cell}^{2}=E_{3}^{o}-E_{2}^{o} \\
 & \text{ }=\text{ }1.595-1.5 \\
 & \text{ }=\text{ }0.095 \\
\end{align}\]
\[\begin{align}
 & E_{cell}^{3}=E_{4}^{o}-E_{3}^{o} \\
 & \text{ }=\text{ }1.0652-1.595 \\
 & \text{ }=-0.5298 \\
\end{align}\](cell potential is negative means no reaction)

The reaction is happening between step-2 and step-3 because of positive cell potential.
In step-2 and step-3 the common species is \[HBrO\].
Therefore \[HBrO\] is a disproportionation species.
So, the correct answer is “Option C”.

Note: Don’t be confused with the different oxidation states of the bromine in the given reaction. The various oxidation states of the bromine atom is going to depend on the neighboring atoms which are present with the bromine atom in the molecule.