Chapter 19, Problem 86A

Interpretation Introduction

Interpretation:

The reaction of potassium permanganate with chloride ions in acid solution needs to be balanced.

Concept introduction:

Half reaction method to balance ionic equation in acidic solution involve:-

• 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 86A

Balanced equation in ionic form

$2Mn{O}_4{}^{-}+10C{l}^{-}+16H^{+}\to 5C{l}_2+2M{n}^{+2}+8H_{2}O$

Explanation of Solution

Reaction in ionic form:

$Mn{O}_4{}^{-}+C{l}^{-}\to C{l}_2+M{n}^{+2}$

Assigning the oxidation number-

$\stackrel{+7}{Mn}{O}_4{}^{-}+\stackrel{-1}{C{l}^{-}}\to \stackrel{0}{C{l}_2}+\stackrel{+2}{M{n}^{+2}}$

The oxidation numbers of atoms shows that

$Cl$ is oxidized and $Mn$ is reduced.

Change in oxidation number of $Cl=+1$

Change in oxidation number of $Mn=-5$

$C{l}^{-}\to C{l}_2+e^{-}$

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

Balancing the atoms and charges:

$2C{l}^{-}\to C{l}_2+2e^{-}$

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

Now adding both reactions after multiplying reduction half cell reaction with 2 and oxidation half cell reaction with 5

$2Mn{O}_4{}^{-}+10C{l}^{-}+16H^{+}\to 5C{l}_2+2M{n}^{+2}+8H_{2}O$

Conclusion

Balanced equation in ionic form

$2Mn{O}_4{}^{-}+10C{l}^{-}+16H^{+}\to 5C{l}_2+2M{n}^{+2}+8H_{2}O$