Question

The balanced equation in the acidic medium is:
${{N}_{2}}{{O}_{4}}+BrO_{3}^{-}\xrightarrow[{{H}_{2}}O]{{{H}^{+}}}NO_{3}^{-}+B{{r}^{-}}$
A. $3{{N}_{2}}{{O}_{4}}+BrO_{3}^{-}+3{{H}_{2}}O\xrightarrow[{{H}_{2}}O]{{{H}^{+}}}6NO_{3}^{-}+B{{r}^{-}}+6{{H}^{+}}$
B. $3{{N}_{2}}{{O}_{4}}+5BrO_{3}^{-}+3{{H}_{2}}O\xrightarrow[{{H}_{2}}O]{{{H}^{+}}}6NO_{3}^{-}+4B{{r}^{-}}+6{{H}^{+}}$
C. $3{{N}_{2}}{{O}_{4}}+3BrO_{3}^{-}+3{{H}_{2}}O\xrightarrow[{{H}_{2}}O]{{{H}^{+}}}6NO_{3}^{-}+B{{r}^{-}}+6{{H}^{+}}$
D. None of these

Answer 1

In a chemical reaction the number of all the atoms on the right side is equal to the number of all the atoms on the left side then it is called a balanced reaction. If we are going to balance the redox reaction in an acidic medium we are supposed to balance by using hydrogens.

Complete Solution :
- In the question it is given that balances the given chemical reaction in an acidic medium.
- The given chemical reaction is as follows.
${{N}_{2}}{{O}_{4}}+BrO_{3}^{-}\xrightarrow[{{H}_{2}}O]{{{H}^{+}}}NO_{3}^{-}+B{{r}^{-}}$

- First we should assign the oxidation numbers to all the atoms individually in the given chemical reaction and it is as follows.
${{\overset{+4}{\mathop{N}}\,}_{2}}{{O}_{4}}+\overset{+5}{\mathop{Br}}\,O_{3}^{-}\xrightarrow[{{H}_{2}}O]{{{H}^{+}}}\overset{+5}{\mathop{N}}\,O_{3}^{-}+\overset{-1}{\mathop{B}}\,{{r}^{-}}$
- In the above chemical reaction the oxidation state of nitrogen changes from +4 to +5. The change in the oxidation state of nitrogen is +1.
- The oxidation state of bromine changes from +5 to -1. The change in the oxidation state of bromine is 6.
- The increase in the oxidation state is balanced with the decrease in the oxidation state by multiplying the ${{N}^{4+}}$ and ${{N}^{5+}}$ with 3 because there are two nitrogen atoms in ${{N}_{2}}{{O}_{4}}$ .
- The chemical reaction by multiplying with 3 is as follows.
$3{{N}_{2}}{{O}_{4}}+BrO_{3}^{-}\xrightarrow[{{H}_{2}}O]{{{H}^{+}}}6NO_{3}^{-}+B{{r}^{-}}$

- In the above chemical reaction oxygen atoms are balanced by adding three water molecules on the left side of the equation.
- The chemical reaction after adding three water molecules is as follows.
$3{{N}_{2}}{{O}_{4}}+BrO_{3}^{-}+3{{H}_{2}}O\xrightarrow[{{H}_{2}}O]{{{H}^{+}}}6NO_{3}^{-}+B{{r}^{-}}$

- The hydrogen atoms are balanced by adding $6{{H}^{+}}$ atoms on the right hand side of the equation and it is as follows.
$3{{N}_{2}}{{O}_{4}}+BrO_{3}^{-}+3{{H}_{2}}O\xrightarrow[{{H}_{2}}O]{{{H}^{+}}}6NO_{3}^{-}+B{{r}^{-}}+6{{H}^{+}}$

- Therefore the balanced chemical reaction of the given reaction in acidic medium is as follows.
$3{{N}_{2}}{{O}_{4}}+BrO_{3}^{-}+3{{H}_{2}}O\xrightarrow[{{H}_{2}}O]{{{H}^{+}}}6NO_{3}^{-}+B{{r}^{-}}+6{{H}^{+}}$
So, the correct answer is “Option A”.

Note: If we have to balance the chemical reaction in the basic medium we are supposed to add $O{{H}^{-}}$ to balance both hydrogen and oxygen atoms in the chemical reaction. In an acidic medium we have to add oxygen atoms and ${{H}^{+}}$ atoms to balance the chemical reaction.