Q.In the reactions given below, identify the species undergoing oxidation and reduction:

Concept: Oxidation–Reduction (Redox)

Oxidation is the loss of electrons (increase in oxidation number), reduction is the gain of electrons (decrease in oxidation number).

Step 1 — Assign oxidation numbers to each element in the reactants and products.

(i) $\text{H}_2\text{S} + \text{Cl}_2 \rightarrow 2\text{HCl} + \text{S}$

• In $\text{H}_2\text{S}$: H = +1, S = –2. In $\text{Cl}_2$: Cl = 0.
• In $\text{HCl}$: H = +1, Cl = –1. In $\text{S}$: S = 0.

Step 2 — Identify changes.

• S: from –2 (in $\text{H}_2\text{S}$) to 0 (in S) → oxidation (loss of electrons).
• Cl: from 0 (in $\text{Cl}_2$) to –1 (in $\text{HCl}$) → reduction (gain of electrons).

(ii) $3\text{Fe}_3\text{O}_4 + 8\text{Al} \rightarrow 9\text{Fe} + 4\text{Al}_2\text{O}_3$

• In $\text{Fe}_3\text{O}_4$: O = –2, so Fe has mixed oxidation states (+2 and +3); average = +8/3. Al = 0.
• In products: Fe = 0, Al in $\text{Al}_2\text{O}_3$ = +3.

Step 2 — Identify changes.

• Fe: from +8/3 (average) to 0 → reduction. …

The Core Idea: Tracking Electrons Through Oxidation Numbers

Before we touch a single equation, let's get the principle straight. Oxidation and reduction always happen together — that's why we call them redox reactions. The trick is to follow the electrons. But electrons are invisible, so we use oxidation numbers as a bookkeeping tool.

The species that causes oxidation by accepting electrons is the oxidising agent (it gets reduced itself). The species that causes reduction by donating electrons is the reducing agent (it gets oxidised itself). For each reaction, we'll assign oxidation numbers to every atom, spot the changes, and name the winners and losers.

(i) H₂S (g) + Cl₂ (g) → 2 HCl (g) + S (s)

1. Assign oxidation numbers to every atom.

• In H₂S: Hydrogen is always +1 (except in metal hydrides, which this isn't). So H = +1. Since the molecule is neutral, S must be −2.
• In Cl₂: A pure element always has oxidation number 0. So Cl = 0.
• In HCl: H = +1, so Cl = −1 (to balance).
• In S (s): Pure element, so S = 0.

2. Identify the changes.

Sulphur goes from −2 to 0 — it loses two electrons. Chlorine goes from 0 to −1 — each Cl atom gains one electron.

3. Write the conclusion.

• Oxidised: H₂S (specifically the S in H₂S)
• Reduced: Cl₂ (specifically the Cl atoms in Cl₂)
• Oxidising agent: Cl₂ (it accepts electrons and gets reduced)
• Reducing agent: H₂S (it donates electrons and gets oxidised)

(ii) 3 Fe₃O₄ (s) + 8 Al (s) → 9 Fe (s) + 4 Al₂O₃ (s)

1. Assign oxidation numbers — this one is trickier.

• Al (s): Pure element → 0.
• Al in Al₂O₃: Oxygen is always −2 (except in peroxides). So 2(Al) + 3(−2) = 0 → Al = +3.
• Fe₃O₄: This is magnetite, a mixed oxide. Oxygen = −2, so total from O = 4 × (−2) = −8. For the three Fe atoms to balance: 3(Fe) + (−8) = 0 → Fe = +8/3.

• Fe (s): Pure element → 0.

2. Identify the changes.

Iron's oxidation number drops — it gains electrons. Aluminium's rises — it loses electrons.

3. Write the conclusion.

• Oxidised: Al (aluminium metal)
• Reduced: Fe₃O₄ (the iron in magnetite)
• Oxidising agent: Fe₃O₄ (it accepts electrons from Al)
• Reducing agent: Al (it donates electrons to Fe₃O₄) …