Chapter 19.2, Problem 24PP

Interpretation Introduction

Interpretation:

The given redox reaction is to be balanced using half reaction method.

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 24PP

$3M{n}^{+2}+5Bi{O}_3{}^{-}+6H^{+}\to 3Mn{O}_4{}^{-}+3H_{2}O+5B{i}^{+2}$

Explanation of Solution

The given redox reaction is:

$M{n}^{+2}+Bi{O}_3{}^{-1}\to Mn{O}_4{}^{-}+B{i}^{+2}$

Assigning the oxidation number-

$\stackrel{+2}{M{n}^{+2}}+\stackrel{+5}{Bi}{\stackrel{-2}{O}}_3{}^{-1}\to \stackrel{+7}{Mn}{\stackrel{-2}{O}}_4{}^{-}+\stackrel{+2}{B{i}^{+2}}$

The oxidation numbers of atoms shows that

$Mn$ is oxidized and $Bi$ is reduced.

Change in oxidation number of $Mn=+5$

Change in oxidation number of $Bi=-3$

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

$Bi{O}_3{}^{-}+3e^{-}\to B{i}^{+2}(\mathrm{Re}duction)$

Balancing the atoms and charges:

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

$Bi{O}_3{}^{-}+3e^{-}+6H^{+}\to B{i}^{+2}+3H_{2}O(\mathrm{Re}duction)$

Now adding both reactions after multiplying oxidation half cell reaction with 3 and reduction half reaction with 5

$3M{n}^{+2}+5Bi{O}_3{}^{-}+6H^{+}\to 3Mn{O}_4{}^{-}+3H_{2}O+5B{i}^{+2}$

Conclusion

Thus the balanced redox equation is:

$3M{n}^{+2}+5Bi{O}_3{}^{-}+6H^{+}\to 3Mn{O}_4{}^{-}+3H_{2}O+5B{i}^{+2}$