Chapter 19, Problem 87A

Interpretation Introduction

Interpretation:

The given half reaction is to be balanced

Concept introduction:

Half cell reaction represents either oxidation or reduction.

Oxidation half reaction represents loss of electrons

Reduction half reaction represents gain of electrons

Answer to Problem 87A

Half reaction:

$N{O}_2{}^{-}\to N{H}_4{}^{+}$

On left side electrons should be added.

$N{O}_2{}^{-}+6e^{-}\to N{H}_4{}^{+}$

In acidic medium:

$N{O}_2{}^{-}+6e^{-}+8H^{+}\to N{H}_4{}^{+}+2H_{2}O$

In basic medium:

$N{O}_2{}^{-}+6e^{-}+6H_{2}O\to N{H}_4{}^{+}+8O{H}^{-}$

Explanation of Solution

Half reaction:

$N{O}_2{}^{-}\to N{H}_4{}^{+}$

Assigning oxidation states:

$\stackrel{+3}{N}{O}_2{}^{-}\to \stackrel{-3}{N}{H}_4{}^{+}$

Therefore, on left side electrons should be added.

$N{O}_2{}^{-}+6e^{-}\to N{H}_4{}^{+}$

In acidic medium:

$N{O}_2{}^{-}+6e^{-}+8H^{+}\to N{H}_4{}^{+}+2H_{2}O$

In basic medium:

$N{O}_2{}^{-}+6e^{-}+6H_{2}O\to N{H}_4{}^{+}+8O{H}^{-}$

Conclusion

In acidic medium:

$N{O}_2{}^{-}+6e^{-}+8H^{+}\to N{H}_4{}^{+}+2H_{2}O$

In basic medium:

$N{O}_2{}^{-}+6e^{-}+6H_{2}O\to N{H}_4{}^{+}+8O{H}^{-}$