Chapter 19.2, Problem 22PP

Interpretation Introduction

Interpretation:

The given redox reaction is to be balanced using oxidation number method.

Concept introduction:

Oxidation number method to balance ionic equation in basic solution involve:-

• Assign the oxidation number to each element.
• Check which element is oxidized and which element is reduced.
• Check the change in oxidation number for the elements which are oxidized and reduced.
• Balance the change in oxidation number by adjusting the coefficients in the equation.
• Then add enough hydroxyl ions to balance the charge and water molecules to the equation to balance hydrogen atoms on both sides.

Answer to Problem 22PP

$2Mn{O}_4{}^{-}+6I^{-1}+4H_{2}O\to 2Mn{O}_2+3I_2+8O{H}^{-}$

Explanation of Solution

The given redox reaction is:

$Mn{O}_4{}^{-}+I^{-1}\to Mn{O}_2+I_2$

Assigning the oxidation number-

$\stackrel{+7}{Mn}{\stackrel{-2}{O_4}}^{-}+\stackrel{-1}{I^{-1}}\to \stackrel{+4}{Mn}\stackrel{-2}{O_2}+\stackrel{0}{I_2}$

The oxidation numbers of atoms shows that

$I$ is oxidized and $Mn$ is reduced.

Change in oxidation number of $I=+1$

Change in oxidation number of $Mn=-3$

Thus adjusting the coefficients:

$Mn{O}_4{}^{-}+3I^{-1}\to Mn{O}_2+I_2$

Balancing the atoms:

$2Mn{O}_4{}^{-}+6I^{-1}\to 2Mn{O}_2+3I_2$

Now adding hydroxyl ions on right side of equation to balance negative charge

$2Mn{O}_4{}^{-}+6I^{-1}\to 2Mn{O}_2+3I_2+8O{H}^{-}$

Adding water to left side:

$2Mn{O}_4{}^{-}+6I^{-1}+4H_{2}O\to 2Mn{O}_2+3I_2+8O{H}^{-}$

Conclusion

Thus the balanced redox equation is:

$2Mn{O}_4{}^{-}+6I^{-1}+4H_{2}O\to 2Mn{O}_2+3I_2+8O{H}^{-}$