Q.Suggest a scheme of classification of the following redox reactions

Concept: Oxidation–Reduction (Redox) Reactions

The key idea is to classify each reaction based on how oxidation numbers change. The main types are: combination, decomposition, displacement, and disproportionation.

Step 1 – Reaction (a):

$N_2 + O_2 \rightarrow 2NO$

Oxidation numbers: $N$ goes from $0$ to $+2$, $O$ goes from $0$ to $-2$. Two elements combine and both change oxidation number.

Type: Combination redox reaction.

Step 2 – Reaction (b):

$2Pb(NO_3)_2 \rightarrow 2PbO + 4NO_2 + O_2$

$N$ in $NO_3^-$ is $+5$, in $NO_2$ it is $+4$ (reduction); $O$ in $NO_3^-$ is $-2$, in $O_2$ it is $0$ (oxidation). A single compound breaks into three products with redox changes.

Type: Decomposition redox reaction.

Step 3 – Reaction (c):

$NaH + H_2O \rightarrow NaOH + H_2$

$H$ in $NaH$ is $-1$, in $H_2O$ it is $+1$; both become $0$ in $H_2$. One species ($H^-$) is oxidised, another ($H^+$) is reduced — hydrogen is both oxidised and reduced. …

The first step in any redox classification is to spot which elements change oxidation state. Without that, you’re guessing. So let’s assign oxidation numbers to every atom in each reaction, then see the pattern of change.

1. Reaction (a): $N_2(g) + O_2(g) \rightarrow 2NO(g)$

• In $N_2$, each N has oxidation number $0$. In $O_2$, each O has $0$.
• In $NO$, O is $-2$ (usual), so N must be $+2$.

N goes from $0$ to $+2$ — it is oxidised (loses electrons).

O goes from $0$ to $-2$ — it is reduced (gains electrons).

Two elements combine into a single product. This is a combination reaction (also called synthesis). Both oxidation and reduction happen in the same step as the elements join.

Classification: Combination redox.

2. Reaction (b): $2Pb(NO_3)_2(s) \rightarrow 2PbO(s) + 4NO_2(g) + O_2(g)$

Assign oxidation numbers carefully:

• In $Pb(NO_3)_2$: Pb is $+2$ (common for lead). Each $NO_3^-$ has N at $+5$ (since O is $-2$, total $-6$, so N must be $+5$ to give $-1$ charge on the ion).
• In $PbO$: Pb stays $+2$, O is $-2$ — no change for Pb.
• In $NO_2$: O is $-2$ each (total $-4$), so N is $+4$.
• In $O_2$: O is $0$.

So N changes from $+5$ (in nitrate) to $+4$ (in $NO_2$) — that’s reduction (gain of electrons).

O changes from $-2$ (in nitrate) to $0$ (in $O_2$) — that’s oxidation (loss of electrons).

One compound breaks down into three simpler substances, with simultaneous oxidation and reduction of different atoms within it. This is a decomposition reaction.

Classification: Decomposition redox.

3. Reaction (c): $NaH(s) + H_2O(l) \rightarrow NaOH(aq) + H_2(g)$

Oxidation numbers:

• In $NaH$: Na is $+1$, so H must be $-1$ (hydride ion).
• In $H_2O$: H is $+1$, O is $-2$.
• In $NaOH$: Na $+1$, O $-2$, H $+1$.
• In $H_2$: H is $0$.

Which atoms change?

The H in $NaH$ goes from $-1$ to $0$ (in $H_2$) — oxidation.

The H in $H_2O$ goes from $+1$ to $0$ (in $H_2$) — reduction. …