Question

Permanganate ion reacts with bromide ion in the basic medium to give manganese dioxide and bromate ion. Write the balanced ionic equation for the reaction.

Answer 1

Hint: Write the chemical formulas of the compounds given, then, write and balance the half-cell equations of both the compounds. While balancing, balance the number of atoms of elements and charge on the elements. Then, obtain the net balanced equation. Finally, write the reaction in basic medium by adding $\left[ \text{O}{{\text{H}}^{-}} \right]$ ions in it.

Complete step by step answer:
The chemical formulas of permanganate ion is $\text{MnO}_{4}^{-}$, bromide ion is $\text{B}{{\text{r}}^{-}}$, manganese dioxide is $\text{Mn}{{\text{O}}_{2}}$ and bromate ion is $\text{BrO}_{3}^{-}$.
Let us first write the balanced individual equation of conversion between permanganate ion and bromide ion.
The reaction of permanganate ion to manganese dioxide. $\text{MnO}_{4}^{-}\to \text{Mn}{{\text{O}}_{2}}$. Now balance it.
Step (1): Left hand side has four oxygen atoms and the right hand side has two oxygen atoms.
To balance the number of oxygen atoms, add two water molecules at the right hand side. $\text{MnO}_{4}^{-}\to \text{Mn}{{\text{O}}_{2}}+2{{\text{H}}_{2}}\text{O}$.
Step (2): To balance hydrogen atoms, we need to add hydrogen ions.
Add four hydrogen ions to the left hand side. $\text{MnO}_{4}^{-}+4{{\text{H}}^{+}}\to \text{Mn}{{\text{O}}_{2}}+2{{\text{H}}_{2}}\text{O}$.
Step (3): The elements and their atoms are balanced, but the charge is still unbalanced.
To balance charge, add three electrons at the left hand side. The right hand side has no charge but the left hand has +3 charge.
$\text{MnO}_{4}^{-}+4{{\text{H}}^{+}}+3{{\text{e}}^{-}}\to \text{Mn}{{\text{O}}_{2}}+2{{\text{H}}_{2}}\text{O}$. This is the first equation.
Similarly, balance the reaction of bromide ion to bromate ion. $\text{B}{{\text{r}}^{-}}\to \text{BrO}_{3}^{-}$. Now balance it.
Step (1): Right hand side has three oxygen atoms and the left hand side has no oxygen atoms.
To balance the number of oxygen atoms, add three water molecules at the left hand side. $\text{B}{{\text{r}}^{-}}+\text{3}{{\text{H}}_{2}}\text{O}\to \text{BrO}_{3}^{-}$.
Step (2): To balance hydrogen atoms, we need to add hydrogen ions.
Add six hydrogen ions to the right hand side. $\text{B}{{\text{r}}^{-}}+\text{3}{{\text{H}}_{2}}\text{O}\to \text{BrO}_{3}^{-}+6{{\text{H}}^{+}}$.
Step (3): The elements and their atoms are balanced, but the charge is still unbalanced.
To balance charge, add six electrons at the right hand side. The left hand side has -1 charge but right hand has +5 charge.
$\text{B}{{\text{r}}^{-}}+\text{3}{{\text{H}}_{2}}\text{O}\to \text{BrO}_{3}^{-}+6{{\text{H}}^{+}}+6{{\text{e}}^{-}}$. This is the second equation.
To get the final and net equation, multiply the first equation by 2 and add the second equation to it. The reactions will be
$\text{2MnO}_{4}^{-}+8{{\text{H}}^{+}}+6{{\text{e}}^{-}}\to 2\text{Mn}{{\text{O}}_{2}}+4{{\text{H}}_{2}}\text{O}$ and $\text{B}{{\text{r}}^{-}}+\text{3}{{\text{H}}_{2}}\text{O}\to \text{BrO}_{3}^{-}+6{{\text{H}}^{+}}+6{{\text{e}}^{-}}$.
Add them, $2\text{MnO}_{4}^{-}+\text{B}{{\text{r}}^{-}}+2{{\text{H}}^{+}}\to 2\text{Mn}{{\text{O}}_{2}}+\text{BrO}_{3}^{-}+{{\text{H}}_{2}}\text{O}$.
Add 2 $\text{O}{{\text{H}}^{-}}$ ions both sides, the reaction will be $2\text{MnO}_{4}^{-}+\text{B}{{\text{r}}^{-}}+2{{\text{H}}^{+}}+2\text{O}{{\text{H}}^{-}}\to 2\text{Mn}{{\text{O}}_{2}}+\text{BrO}_{3}^{-}+{{\text{H}}_{2}}\text{O}+\text{2O}{{\text{H}}^{-}}$.
${{\text{H}}^{+}}+\text{O}{{\text{H}}^{-}}\to {{\text{H}}_{2}}\text{O}$, the net reaction is $2\text{MnO}_{4}^{-}+\text{B}{{\text{r}}^{-}}+2{{\text{H}}_{2}}\text{O}\to 2\text{Mn}{{\text{O}}_{2}}+\text{BrO}_{3}^{-}+{{\text{H}}_{2}}\text{O}+\text{2O}{{\text{H}}^{-}}$ or $2\text{MnO}_{4}^{-}+\text{B}{{\text{r}}^{-}}+{{\text{H}}_{2}}\text{O}\to 2\text{Mn}{{\text{O}}_{2}}+\text{BrO}_{3}^{-}+\text{2O}{{\text{H}}^{-}}$.
Permanganate ion reacts with bromide ion in the basic medium to give manganese dioxide and bromate ion. The balanced reaction is $2\text{MnO}_{4}^{-}+\text{B}{{\text{r}}^{-}}+{{\text{H}}_{2}}\text{O}\to 2\text{Mn}{{\text{O}}_{2}}+\text{BrO}_{3}^{-}+\text{2O}{{\text{H}}^{-}}$.

Note: This is the ion electron method to get the final reaction. There is another method also to balance reactions, that is oxidation number method. In that we assign oxidation numbers to elements which show variable oxidation states. Then, calculate the increment and decrement in oxidation number and make the increase equal to the decrease.