Chapter 19, Problem 70A

Interpretation Introduction

Interpretation: The given redox reaction is to be balanced using a half-reaction method.

Concept Introduction :

A half-reaction method to balance the ionic equation in acidic solution involves:-

• Assign the oxidation number to each element.
• Check which element is oxidized and which element is reduced. Write oxidation and reduction half-reactions.
• Check the change in oxidation number for the elements which are oxidized and reduced.
• Then add enough hydrogen ions and water molecules to the equation to balance hydrogen atoms on both sides.
• Adjust the coefficients so that number of electrons lost in oxidation is equal to electrons gained in reduction.
• Add the balanced half-reactions.

Answer to Problem 70A

A balanced equation in ionic form

  $\text{M}\text{n}{\text{O}}_4{}^{-}+5\text{F}{\text{e}}^{2+}+8{\text{H}}^{+}\to 5\text{F}{\text{e}}^{+3}+\text{M}{\text{n}}^{+2}+4{\text{H}}_2\text{O}$

A balanced equation in given form

  $2\text{K}\text{M}\text{n}{\text{O}}_4+10\text{F}\text{e}\text{S}{\text{O}}_4+8{\text{H}}_2\text{S}{\text{O}}_4\to 5\text{F}{\text{e}}_2{\left(\text{S}{\text{O}}_4\right)}_3+2\text{M}\text{n}\text{S}{\text{O}}_4+{\text{K}}_2\text{S}{\text{O}}_4+8{\text{H}}_2\text{O}$

Explanation of Solution

The given redox reaction is:

  $\text{K}\text{M}\text{n}{\text{O}}_4+\text{F}\text{e}\text{S}{\text{O}}_4+{\text{H}}_2\text{S}{\text{O}}_4\to \text{F}{\text{e}}_2{\left(\text{S}{\text{O}}_4\right)}_3+\text{M}\text{n}\text{S}{\text{O}}_4+{\text{K}}_2\text{S}{\text{O}}_4+{\text{H}}_2\text{O}$

A reaction in the ionic form:

  $\text{M}\text{n}{\text{O}}_4{}^{-}+\text{F}{\text{e}}^{2+}\to \text{F}{\text{e}}^{+3}+\text{M}{\text{n}}^{+2}$

Assigning the oxidation number-

  $\stackrel{+7}{\text{M}\text{n}}{\text{O}}_4{}^{-}+\stackrel{+2}{\text{F}{\text{e}}^{2+}}\to \stackrel{+3}{\text{F}{\text{e}}^{+3}}+\stackrel{+2}{\text{M}{\text{n}}^{+2}}$

The oxidation numbers of atoms show that

$\text{F}\text{e}$ is oxidized and $\text{M}\text{n}$ is reduced.

Change in oxidation number of $\text{F}\text{e}=+1$

Change in oxidation number of $\text{M}\text{n}=-5$

  $\text{F}{\text{e}}^{2+}\to \text{F}{\text{e}}^{+3}+{\text{e}}^{-}$

  $\text{M}\text{n}{\text{O}}_4{}^{-}+5{\text{e}}^{-}\to \text{M}{\text{n}}^{+2}$

Balancing the atoms and charges:

  $\text{F}{\text{e}}^{2+}\to \text{F}{\text{e}}^{+3}+{\text{e}}^{-}$

  $\text{M}\text{n}{\text{O}}_4{}^{-}+5{\text{e}}^{-}+8{\text{H}}^{+}\to \text{M}{\text{n}}^{+2}+4{\text{H}}_2\text{O}$

Now, adding both reactions after multiplying reduction half cell reaction with 2

  $\text{M}\text{n}{\text{O}}_4{}^{-}+5\text{F}{\text{e}}^{2+}+8{\text{H}}^{+}\to 5\text{F}{\text{e}}^{+3}+\text{M}{\text{n}}^{+2}+4{\text{H}}_2\text{O}$

Conclusion

A balanced equation in ionic form

  $\text{M}\text{n}{\text{O}}_4{}^{-}+5\text{F}{\text{e}}^{2+}+8{\text{H}}^{+}\to 5\text{F}{\text{e}}^{+3}+\text{M}{\text{n}}^{+2}+4{\text{H}}_2\text{O}$

A balanced equation in given form

  $2\text{K}\text{M}\text{n}{\text{O}}_4+10\text{F}\text{e}\text{S}{\text{O}}_4+8{\text{H}}_2\text{S}{\text{O}}_4\to 5\text{F}{\text{e}}_2{\left(\text{S}{\text{O}}_4\right)}_3+2\text{M}\text{n}\text{S}{\text{O}}_4+{\text{K}}_2\text{S}{\text{O}}_4+8{\text{H}}_2\text{O}$