Question

Match the half-reactions in Column I with the corresponding characteristic(s) in column II.

Answer 1

(A) $N_2{H}_5^{+}⟶N_2{H}^{+}$

Loss of hydrogen is taking place and therefore, it is an oxidation reaction.

$N_2{H}_5^{+}\to N_2{H}^{+}+4H^{+}+4e^{-}$

Reactant is giving $4H^{+}$ and therefore, equivalent weight $=\frac{molarmass}{4}$

There is exchange of $4$ electrons.

(B) $N_2{H}_4+2O{H}^{-}⟶N_2+2H_{2}O$

Loss of hydrogen is taking place and therefore, it is is an oxidation reaction for $N_2{H}_4$.

$N_2{H}_4\to 4H^{+}+4e^{-}$

Reactant is giving $4H^{+}$, therefore, equivalent weight $=\frac{molarmass}{4}$

There is an exchange of $4$ electrons.

(C) ${2N_3}^{-}⟶3N_2$

Loss of electron is an oxidation reaction.

$2N_3^{-}\to 3N_2+2e^{-}$

Equivalent weight $=\frac{molarmass}{2}$

Two electrons are exchanged in the reaction.

(D) $N{H}_{2}OH⟶N{H}_3$

Loss of oxygen is a reduction reaction.

Oxidation state of $N$ in $N{H}_{2}OH=-1$ which is changed to $-3$ in $N{H}_3$

Therefore there is an exchange of two electrons.

Equivalent weight $=\frac{molarmass}{2}$

(E) $N{H}_{2}OH⟶N_{2}O$

Loss of $H$ is an oxidation reaction.

Oxidation state of $N$ in $N{H}_{2}OH=-1$ changes to $+1$ in $N_{2}O$. There is an exchange of $2$ electrons.

Equivalent weight $=\frac{molarmass}{2}$