Chapter 19.2, Problem 25PP

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 basic 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 hydroxl 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 25PP

$N_{2}O+2Cl{O}^{-1}+2O{H}^{-}\to 2N{O}_2{}^{-}+2C{l}^{-}+H_{2}O$

Explanation of Solution

The given redox reaction is:

$N_{2}O+Cl{O}^{-}\to N{O}_2{}^{-}+C{l}^{-}$

Assigning the oxidation number-

${\stackrel{+1}{N}}_2\stackrel{-2}{O}+\stackrel{+1}{Cl}{\stackrel{-2}{O}}^{-}\to \stackrel{+3}{N}{\stackrel{-2}{O}}_2{}^{-}+\stackrel{-1}{C{l}^{-}}$

The oxidation numbers of atoms shows that

$N$ is oxidized and $Cl$ is reduced.

Change in oxidation number of $Nperatom=+4$

Change in oxidation number of $Cl=-2$

$N_{2}O\to N{O}_2{}^{-}+4e^{-}(Oxidation)$

$Cl{O}^{-}+2e^{-}\to C{l}^{-}(\mathrm{Re}duction)$

Balancing the atoms and charges:

$N_{2}O+6O{H}^{-}\to 2N{O}_2{}^{-}+4e^{-}+3H_{2}O(Oxidation)$

$Cl{O}^{-}+H_{2}O+2e^{-}\to C{l}^{-}+2O{H}^{-}(\mathrm{Re}duction)$

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

$N_{2}O+2Cl{O}^{-1}+2O{H}^{-}\to 2N{O}_2{}^{-}+2C{l}^{-}+H_{2}O$

Conclusion

Thus the balanced redox equation is:

$N_{2}O+2Cl{O}^{-1}+2O{H}^{-}\to 2N{O}_2{}^{-}+2C{l}^{-}+H_{2}O$