
Equivalent weight of $ {\text{B}}{{\text{r}}_2} $ is 96 in the following disproportionation reaction:
(A) $ {\text{B}}{{\text{r}}_2} + {\text{O}}{{\text{H}}^ - } \to {\text{B}}{{\text{r}}^ - } + {{\text{H}}_2}{\text{O}} + ?\left( {{\text{Oxidized Product}}} \right) $
The oxidation state of Br in the oxidized product is $ \left( {{\text{Br}} = 80} \right) $ .
Answer
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Hint: We will use the relation between the molecular weight and equivalent weight. The n factor thus comes out will be the change in oxidation state taking place.
Formula Used: $ {\text{Equivalent weight}} = \dfrac{{{\text{Molecular weight}}}}{{{\text{n factor}}}} $
Complete step by step solution:
To calculate the oxidation state we first need to calculate the n factor. N factor gives us the value of the number of electrons that are exchanged. We will use the relation between the equivalent weight and the molecular weight. The formula between molecular weight and equivalent weight is:
$ {\text{Equivalent weight}} = \dfrac{{{\text{Molecular weight}}}}{{{\text{n factor}}}} $
We have been given the value of equivalent weight that is 96. The molecular weight of bromine is 80. There are 2 bromine atoms in 1 molecule of bromine gas. The n factor will come out to be:
$ {\text{n factor}} = \dfrac{{{\text{3}} \times {\text{2}} \times {\text{80}}}}{{{\text{96}}}} = 5 $
That means there is a change of plus 5 oxidation state of bromine in the oxidised product.
The balanced chemical reaction is as follow:
$ {\text{6B}}{{\text{r}}_2} + 12{\text{O}}{{\text{H}}^ - } \to 10{\text{B}}{{\text{r}}^ - } + 6{{\text{H}}_2}{\text{O}} + 2{\text{Br}}{{\text{O}}_3}^ - $ .
Additional information
Oxidation and reduction are complementary to each other. That means that it is not possible that only oxidation or only reduction is occurring. If one substance is getting oxidised then the other will definitely get reduced and the reverse is also true. The oxidation and reduction depends on their respective electrode potential. The one with the higher standard reduction potential will have more tendencies to get reduced and the one with the higher standard oxidation potential will have more tendencies to oxidise.
Note:
Disproportination reaction is that reaction in which single substances get oxidise and get reduced. In our case the bromine is getting oxidised as well as getting reduced.
Formula Used: $ {\text{Equivalent weight}} = \dfrac{{{\text{Molecular weight}}}}{{{\text{n factor}}}} $
Complete step by step solution:
To calculate the oxidation state we first need to calculate the n factor. N factor gives us the value of the number of electrons that are exchanged. We will use the relation between the equivalent weight and the molecular weight. The formula between molecular weight and equivalent weight is:
$ {\text{Equivalent weight}} = \dfrac{{{\text{Molecular weight}}}}{{{\text{n factor}}}} $
We have been given the value of equivalent weight that is 96. The molecular weight of bromine is 80. There are 2 bromine atoms in 1 molecule of bromine gas. The n factor will come out to be:
$ {\text{n factor}} = \dfrac{{{\text{3}} \times {\text{2}} \times {\text{80}}}}{{{\text{96}}}} = 5 $
That means there is a change of plus 5 oxidation state of bromine in the oxidised product.
The balanced chemical reaction is as follow:
$ {\text{6B}}{{\text{r}}_2} + 12{\text{O}}{{\text{H}}^ - } \to 10{\text{B}}{{\text{r}}^ - } + 6{{\text{H}}_2}{\text{O}} + 2{\text{Br}}{{\text{O}}_3}^ - $ .
Additional information
Oxidation and reduction are complementary to each other. That means that it is not possible that only oxidation or only reduction is occurring. If one substance is getting oxidised then the other will definitely get reduced and the reverse is also true. The oxidation and reduction depends on their respective electrode potential. The one with the higher standard reduction potential will have more tendencies to get reduced and the one with the higher standard oxidation potential will have more tendencies to oxidise.
Note:
Disproportination reaction is that reaction in which single substances get oxidise and get reduced. In our case the bromine is getting oxidised as well as getting reduced.
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