Oxidation and Reduction (Redox Reactions)

What is Oxidation?

In chemistry, 'oxidation' and 'reduction' are two crucial concepts. They are opposites of each other, and almost every chemical reaction has them hidden somewhere.

Activity 1.11 — Oxidation of Copper

Take 1 g of copper powder in a china dish and heat it on a burner.

Observation: A black coating forms on the surface of the shiny brown copper powder.

What is this black coating?

$$2Cu(s) + O_2(g) \xrightarrow{\Delta} 2CuO(s)$$

This is copper oxide ($CuO$) — black in colour. Copper combined with oxygen.

Oxidation — Definition 1: When a substance gains oxygen, it is being oxidised.

Here, Cu gained O → Cu was oxidised.

Another Definition

Oxidation — Definition 2: When a substance loses hydrogen, it is being oxidised.

We will see an example of this shortly.

Other Examples of Oxidation

(i) Burning of magnesium — oxidation of Mg: $$2Mg + O_2 \rightarrow 2MgO$$ (O is added to Mg)

(ii) Burning of coal — oxidation of C: $$C + O_2 \rightarrow CO_2$$ (O is added to C)

(iii) Formation of water from hydrogen — oxidation of H₂: $$2H_2 + O_2 \rightarrow 2H_2O$$ (O is added to H₂)

[Board Tip] Remember — in oxidation, O is added or H is removed.

What is Reduction?

Reduction is exactly the opposite of oxidation.

A Clear Example — Copper Oxide + Hydrogen

If hydrogen gas is passed over hot copper oxide, the black coating turns brown — copper is recovered.

$$CuO(s) + H_2(g) \xrightarrow{\Delta} Cu(s) + H_2O(g)$$

Analysis:

• $CuO$ lost its O → $CuO$ was reduced.
• $H_2$ gained O → $H_2$ was oxidised.
• (Both happen at the same time!)

Reduction — Definition 1: When a substance loses oxygen, it is being reduced.

Reduction — Definition 2: When a substance gains hydrogen, it is being reduced.

Other Examples of Reduction

(i) Reduction of chlorine — by hydrogen: $$Cl_2 + H_2 \rightarrow 2HCl$$ (H is added to Cl₂ → Cl₂ is reduced)

(ii) Reduction of nitrogen — formation of ammonia: $$N_2 + 3H_2 \rightarrow 2NH_3$$ (H is added to N₂ → N₂ is reduced)

A Quick-Reference Table

When…	What is happening?
O is added	Oxidation
O is removed	Reduction
H is added	Reduction
H is removed	Oxidation

Memory trick: "OilRig" — Oxidation Is Loss (of electrons), Reduction Is Gain — but at Class 10 level, this becomes: oxidation = gain of O / loss of H; reduction = loss of O / gain of H.

What is a Redox Reaction?

A crucial point — oxidation and reduction always occur together. If one substance is gaining oxygen, another must be losing it.

Redox Reaction: A chemical reaction in which the oxidation of one reactant and the reduction of another happen simultaneously is called a redox reaction.

The word 'redox' is short for Reduction + Oxidation.

Pictorial Form

$$\text{Oxidation of A} \quad \xrightleftharpoons[\text{simultaneous}]{} \quad \text{Reduction of B}$$

Worked Example — 1

$$CuO(s) + H_2(g) \rightarrow Cu(s) + H_2O(l)$$

• CuO is reduced (O removed → Cu formed)
• H₂ is oxidised (O added → H₂O formed)
• This is a redox reaction.

Worked Example — 2

$$ZnO + C \rightarrow Zn + CO$$

• ZnO is reduced (O removed → Zn formed)
• C is oxidised (O added → CO formed)
• This is a redox reaction.

Worked Example — 3 (popular in board exams)

$$MnO_2 + 4HCl \rightarrow MnCl_2 + 2H_2O + Cl_2$$

• HCl is oxidised (H removed → Cl₂ formed)
• MnO₂ is reduced (O removed → MnCl₂ formed)
• This is a redox reaction.

[Board Important] In every redox reaction, clearly state which substance is oxidised and which is reduced — this is the most essential skill.

Oxidising and Reducing Agents

In redox reactions, two special roles appear —

Oxidising Agent (Oxidant)

The substance that causes the oxidation of another — but is itself reduced.

Behaviour: An oxidising agent gives oxygen (or takes hydrogen).

Reducing Agent (Reductant)

The substance that causes the reduction of another — but is itself oxidised.

Behaviour: A reducing agent takes oxygen (or gives hydrogen).

Clarifying with an Example

$$CuO + H_2 \rightarrow Cu + H_2O$$

• CuO gave O to H₂ → CuO oxidised H₂ → CuO is the oxidising agent
• CuO itself lost O → CuO was reduced
• H₂ took O from CuO → H₂ reduced CuO → H₂ is the reducing agent
• H₂ itself gained O → H₂ was oxidised

A Summary Table — Everything at a Glance

Role	What it does	What happens to it
Oxidising agent	Gives O (or takes H)	Reduced (itself decreases)
Reducing agent	Takes O (or gives H)	Oxidised (itself increases)

Some Famous Oxidising and Reducing Agents

Oxidising agents: $O_2, KMnO_4, K_2Cr_2O_7, HNO_3, H_2SO_4$ (concentrated), $Cl_2$

Reducing agents: $H_2, C, CO, Na, Mg, Al, KI, H_2S$

A Curious Question — Can One Substance Play Both Roles?

$H_2O_2$ (hydrogen peroxide) can act as both — sometimes oxidising, sometimes reducing — depending on the situation. This is a deeper Class 11 topic.

How to Identify a Redox Reaction — A Step-by-Step Approach

When given a reaction and asked whether it is redox, follow these steps —

Step 1 — Count O and H atoms in each substance

Think separately for the LHS and RHS.

Step 2 — Spot which substance gains/loses O or H

• O added → oxidation
• O removed → reduction
• H added → reduction
• H removed → oxidation

Step 3 — If both happen, it's redox

If one substance is being oxidised AND another reduced — both at once — then it is a redox reaction.

A Full Example

Reaction: $\quad 2H_2 + O_2 \rightarrow 2H_2O$

Analysis:

• $H_2$ had no O before; $H_2O$ has O → H₂ is oxidised
• $O_2$ had no H before; $H_2O$ has H → O₂ is reduced (O accepted H)
• Both at once → this is a redox reaction

Oxidising agent: $O_2$ (oxidised H₂; itself reduced) Reducing agent: $H_2$ (reduced O₂; itself oxidised)

Some More Important Redox Reactions

(i) Metal extraction from coal:

$$ZnO + C \xrightarrow{\Delta} Zn + CO$$

Zn is reduced, C is oxidised.

(ii) Decomposition of mercury oxide — a curious case:

$$2HgO \xrightarrow{\Delta} 2Hg + O_2$$

HgO is reduced (O released), O₂ — here oxygen alone forms.

(iii) Silver chloride in sunlight:

$$2AgCl \xrightarrow{\text{light}} 2Ag + Cl_2$$

AgCl is reduced (Ag formed — charge decreased); Cl is oxidised (Cl₂ formed).

[Board Important] This is one of the most essential question types in this section — given a reaction, you must clearly state what is oxidised/reduced.

🧠 Memory Capsule

A one-glance recap to revisit just before the board exam — every key idea about oxidation, reduction, and redox reactions in one place.

1. Oxidation and Reduction — Definitions Together

Situation	What is happening?
O added	Oxidation
O removed	Reduction
H added	Reduction
H removed	Oxidation

2. Definition of Redox

Oxidation and reduction together → redox reaction.

3. Oxidising vs. Reducing Agent

Substance	Role	Itself
Gives O	Oxidising agent	Reduced
Takes O	Reducing agent	Oxidised

4. Must-Memorise Redox Reactions

Reaction	Oxidised	Reduced
$CuO + H_2 \rightarrow Cu + H_2O$	H₂	CuO
$ZnO + C \rightarrow Zn + CO$	C	ZnO
$MnO_2 + 4HCl \rightarrow MnCl_2 + 2H_2O + Cl_2$	HCl	MnO₂
$2H_2 + O_2 \rightarrow 2H_2O$	H₂	O₂
$C + O_2 \rightarrow CO_2$	C	O₂

5. Famous Oxidising and Reducing Agents

• Oxidising: $O_2, KMnO_4, K_2Cr_2O_7, HNO_3, H_2SO_4$ (conc.)
• Reducing: $H_2, C, CO, Na, Mg, Al, KI$

6. Memorable Experiment from Activity 1.11

• $2Cu + O_2 \xrightarrow{\Delta} 2CuO$ (black coating)
• $CuO + H_2 \xrightarrow{\Delta} Cu + H_2O$ (black to brown)
• Both are redox reactions.

7. Repeatedly Asked Board Questions

1. Define oxidation and reduction.
2. What is a redox reaction? Give an example.
3. Identify the oxidising and reducing agent in this reaction.
4. Why is a black coating formed in Activity 1.11?
5. Explain oxidation and reduction in terms of gain/loss of oxygen.

8. A Trick

• To identify oxidation/reduction in a reaction, just count O and H atoms in each substance and see what increased/decreased.

The Bottom Line: O up = oxidation, O down = reduction. Both at once = redox.

Solved Examples

Example 1: NCERT — Identification of Copper Oxidation

Identify the substances oxidised and reduced in the following reaction — $\quad 4Na(s) + O_2(g) \rightarrow 2Na_2O(s)$

Solution:

1. Analysis:

• Na had no O initially; now in $Na_2O$ it does → Na gained O → Na is oxidised
• $O_2$ gave its O to Na → went into $Na_2O$ — therefore $O_2$ underwent reduction (in the sense of gaining 'partners' of metal).

1. Answer:

• Substance oxidised: Na (sodium)
• Substance reduced: $O_2$ (oxygen)

1. Why is this redox? Both — oxidation and reduction — happen simultaneously.

[Board Important — NCERT textbook question]

Example 2: NCERT — Second Example

Identify the substances oxidised and reduced in — $\quad CuO(s) + H_2(g) \rightarrow Cu(s) + H_2O(l)$

Solution:

Analysis:

Substance	Initial	Final	Change	Conclusion
Cu	in $CuO$, with O	$Cu$ — O removed	O lost	Reduction
H₂	$H_2$ alone	in $H_2O$, with O	O gained	Oxidation

Answer:

• Oxidised substance: $H_2$ (hydrogen)
• Reduced substance: $CuO$ (copper oxide)

Oxidising and reducing agents:

• $CuO$ — oxidising agent (oxidised $H_2$)
• $H_2$ — reducing agent (reduced $CuO$)

What kind of reaction? Redox reaction.

Visible observation: The black $CuO$ coating turns brown ($Cu$).

[Repeatedly asked in board]

Example 3: Definition of a Redox Reaction

Define a redox reaction. Give two examples.

Solution:

Definition: A chemical reaction in which the oxidation of one reactant and the reduction of another happen simultaneously is called a redox reaction.

The word 'redox' is short for Reduction + Oxidation.

Key features

1. One substance gains O (or loses H) — oxidation.
2. The other substance loses O (or gains H) — reduction.
3. These two changes happen together.

Two examples

(i) Copper oxide + hydrogen:

$$CuO + H_2 \rightarrow Cu + H_2O$$

• $H_2$ is oxidised (gains O in $H_2O$)
• $CuO$ is reduced (loses O, $Cu$ formed)

(ii) Reduction of zinc oxide:

$$ZnO + C \rightarrow Zn + CO$$

• $C$ is oxidised (gains O to form $CO$)
• $ZnO$ is reduced (loses O, $Zn$ formed)

[Board Important — 3-mark question]

Example 4: Decomposition of Mercury Oxide

Analyse the reaction — $\quad 2HgO(s) \xrightarrow{\Delta} 2Hg(l) + O_2(g)$

Solution:

1. Analysis:

• In $HgO$, Hg was bonded to oxygen; the final product $Hg$ is pure metal (no O) → HgO is reduced
• The oxygen in $HgO$ leaves to form $O_2$ — it goes from a 'bonded' state to free $O_2$ → in simple terms, this is not just an oxidation of O₂ (this is a special case of decomposition where oxygen is liberated).

1. A more detailed view:

• This is a decomposition reaction (one → two products).
• Hg²⁺ is reduced — it becomes Hg⁰.
• O²⁻ (in oxide) is oxidised — it becomes O⁰ (in $O_2$).
• (Here we have implicitly used the language of oxidation states — fully clarified in Class 11.)

1. Class 10 level answer:

• This is a decomposition reaction.
• Mercury ($Hg$) was reduced (from $HgO$ to $Hg$).
• Technically a redox decomposition.

1. Special note: This reaction is historically tied to the discovery of oxygen by Priestley.

Lesson: Not every decomposition is redox (e.g., $CaCO_3 \rightarrow CaO + CO_2$ is not redox). But the decomposition of $HgO$ is a redox decomposition.

Example 5: NCERT Class 10 — A Short Question

Identify the substances oxidised and reduced in the following reactions: (i) $4Na(s) + O_2(g) \rightarrow 2Na_2O(s)$ (ii) $CuO(s) + H_2(g) \rightarrow Cu(s) + H_2O(l)$

Solution:

For (i):

• Na gained O (Na → Na₂O) → Na is oxidised
• O₂ gave O to Na, itself reduced → O₂ is reduced

For (ii):

• CuO lost O (CuO → Cu) → CuO is reduced
• H₂ gained O (H₂ → H₂O) → H₂ is oxidised

Table:

Reaction	Oxidised	Reduced
$4Na + O_2 \rightarrow 2Na_2O$	Na	$O_2$
$CuO + H_2 \rightarrow Cu + H_2O$	$H_2$	CuO

Both are redox reactions.

[NCERT textbook question — verbatim answer]

Example 6: Identifying the Oxidising and Reducing Agent

Identify the oxidising and reducing agent in — $\quad MnO_2 + 4HCl \rightarrow MnCl_2 + 2H_2O + Cl_2$

Solution:

1. Analysis of substances:

• $MnO_2$ had Mn bonded to oxygen; in $MnCl_2$, Mn now has chloride; O has been removed → MnO₂ is reduced
• $HCl$ had H; the final $Cl_2$ is pure chlorine, with H now in $H_2O$; HCl lost H → HCl is oxidised

1. Roles:

• $MnO_2$ oxidised $HCl$ → $MnO_2$ is the oxidising agent
• $HCl$ reduced $MnO_2$ → $HCl$ is the reducing agent

1. It is a redox reaction, because oxidation and reduction happen simultaneously.
2. Daily-life use: This reaction is an old laboratory method for producing $Cl_2$ gas.

[Board Important]

Example 7: Extraction of Zinc Using Coal

When zinc oxide is heated with carbon, zinc metal is obtained. Write the balanced reaction and identify the oxidising/reducing agent.

Solution:

1. Reaction:

$$ZnO(s) + C(s) \xrightarrow{\Delta} Zn(s) + CO(g)$$

2. Analysis:

• ZnO lost O → ZnO is reduced → ZnO is the oxidising agent (gave O to C)
• C gained O (to form CO) → C is oxidised → C is the reducing agent (reduced ZnO)

2. Type of reaction:

• Redox reaction
• Also a kind of displacement (C displaced Zn)

2. Industrial significance:

• This is the famous method for extracting zinc.
• Using coal as a 'reducing agent' is an old and inexpensive technique.
• The same principle is used to extract iron in a blast furnace: $Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2$

[Board + Preview of Class 10 Chapter 3]

Example 8: Numerical — Quantities in a Redox Reaction

8 g of $CuO$ is reduced by $H_2$ gas. Find the masses of Cu and $H_2O$ formed. (Cu=63.5, O=16, H=1)

Solution:

1. Reaction:

$$CuO(s) + H_2(g) \rightarrow Cu(s) + H_2O(l)$$

2. Molecular masses:

• $CuO = 63.5 + 16 = 79.5$ g
• $Cu = 63.5$ g
• $H_2O = 18$ g

2. Mole ratio: 1 mol $CuO$ → 1 mol Cu and 1 mol H₂O
3. Moles of $CuO$:

$$n_{CuO} = \frac{8}{79.5} \approx 0.1006 \text{ mol}$$

5. Products:

• Cu = $0.1006 \times 63.5 \approx 6.39$ g
• $H_2O$ = $0.1006 \times 18 \approx 1.81$ g

5. Verify mass conservation: $CuO + H_2 = 8 + 0.2 \approx 8.2$ g; $Cu + H_2O = 6.39 + 1.81 = 8.2$ g ✓

Answer:

• Mass of Cu ≈ 6.39 g
• Mass of $H_2O$ ≈ 1.81 g

Lesson: Conservation of mass applies to redox reactions as well.

Example 9: A Famous Board Question

A shiny, brown-coloured element 'X' becomes black when heated in air. Name 'X' and the black compound.

Solution:

This refers to the famous Activity 1.11.

1. Identification of 'X': A shiny brown element that turns black when heated in air — this is copper (Cu).
2. Identification of the black compound: Copper (II) oxide, i.e., $CuO$ — black in colour.
3. Reaction:

$$2Cu(s) + O_2(g) \xrightarrow{\Delta} 2CuO(s)$$

4. Type of reaction:

• Combination reaction (two → one)
• Oxidation / redox (Cu gained O)

4. Visible observation:

• Initially — shiny brown copper powder.
• Finally — a black coating ($CuO$) on the surface.

4. Bonus: This also explains the principle behind the corrosion of copper — although in atmospheric conditions a green coating of $CuCO_3 \cdot Cu(OH)_2$ tends to form instead.

[NCERT textbook question Q.17 — frequently asked]

Example 10: The 'Dual' Nature of Redox

Why must one substance always be oxidised and another reduced in a redox reaction?

Solution:

The Principle:

1. Oxygen can only move from one place to another — it cannot disappear.
2. When one substance loses oxygen (reduced), that oxygen must go to some other substance (oxidised).
3. This is a direct consequence of conservation of mass.

Illustrating with an example:

$$CuO + H_2 \rightarrow Cu + H_2O$$

• If $CuO$ is being reduced, $CuO$ is giving its O to something else.
• That 'something else' is $H_2$ — it took the O and became $H_2O$.
• Hence the oxidation of $H_2$ is necessary.

An analogy from real life: Like a game where if one team passes the ball, another team must catch it. Similarly — if one substance loses oxygen, another must take it.

Conclusion: This is why oxidation and reduction always happen together — and that combined reaction is what we call redox.

[Board conceptual question]

Example 11: Oxidation/Reduction by Loss/Gain of H

Which substance is oxidised/reduced in — $\quad H_2 + Cl_2 \rightarrow 2HCl$?

Solution:

Notice — there is no oxygen here, so we must reason in terms of H.

1. Analysis of $H_2$:

• $H_2$ was alone; now in $HCl$, H is paired with Cl.
• H itself didn't change quantity; but Cl gained an H bond.
• Switch viewpoint: $H_2$ → $HCl$ — H₂ now has a 'partner' (Cl).
• Alternative view: $H_2$ → $HCl$ — H is oxidised (will be clearer in Class 11 — based on charge).

1. Analysis of $Cl_2$:

• $Cl_2$ had no H; in $HCl$, H is now bonded.
• $Cl_2$ gained H → Cl₂ is reduced

1. Answer (Class 10 level):

• $Cl_2$ is reduced (H added)
• $H_2$ is oxidised (technically — based on charge)
• This is a redox reaction.

1. Special note: Redox reactions can occur even without oxygen — just gain/loss of H is enough.

Lesson: Oxidation/reduction is not limited to oxygen — also look at gain/loss of hydrogen.

Example 12: Identifying Five Redox Reactions

In the following reactions, state which substances are oxidised and reduced: (i) $2H_2 + O_2 \rightarrow 2H_2O$ (ii) $C + O_2 \rightarrow CO_2$ (iii) $Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2$ (iv) $2KI + Cl_2 \rightarrow 2KCl + I_2$ (v) $4Al + 3O_2 \rightarrow 2Al_2O_3$

Solution:

#	Reaction	Oxidised	Reduced
(i)	$2H_2 + O_2 \rightarrow 2H_2O$	$H_2$ (gains O)	$O_2$
(ii)	$C + O_2 \rightarrow CO_2$	C (gains O)	$O_2$
(iii)	$Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2$	$CO$ (gains O → CO₂)	$Fe_2O_3$ (loses O → Fe)
(iv)	$2KI + Cl_2 \rightarrow 2KCl + I_2$	$I^-$ (in KI) → $I_2$ — (Class 11 — change in charge)	$Cl_2$ → $Cl^-$
(v)	$4Al + 3O_2 \rightarrow 2Al_2O_3$	Al (gains O)	$O_2$

Special significance of reaction (iii):

• This is the reaction in the Blast Furnace (iron extraction).
• $Fe_2O_3$ ore + $CO$ as reducing agent → pure Fe.
• This is the foundation of industrial steel production.

[Board + JEE/NEET level]

Example 13: Oxidising and Reducing Agents — A Concise List

Give two examples each of oxidising and reducing agents, and state one use of each.

Solution:

Oxidising agents (two examples):

(i) Oxygen ($O_2$):

• The most common oxidising agent.
• The basis of all combustion.
• Use: Fuel combustion (LPG, petrol), respiration (in our body), rocket engines.

(ii) Potassium permanganate ($KMnO_4$):

• Deep purple, a strong oxidising agent.
• Use: Standard solution in titrations, water treatment, antiseptic for wounds.

Reducing agents (two examples):

(i) Hydrogen ($H_2$):

• The most common reducing agent.
• Use: $CuO$ → Cu (lab demonstration), metal extraction, ammonia synthesis ($N_2 + H_2 \rightarrow NH_3$).

(ii) Carbon (coal, C):

• Cheap, very powerful.
• Use: Extraction of iron/zinc/copper ($Fe_2O_3 + C \rightarrow Fe + CO_2$), thermite-like reactions.

[Board Important — practical context]

Example 14: A Challenging Question — Respiration and Combustion

Respiration and combustion are both redox reactions. Explain.

Solution:

(1) Combustion (of methane):

$$CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O + \text{Energy}$$

Analysis:

• C in $CH_4$ gained O (to form CO₂) → C is oxidised
• H in $CH_4$ gained O (to form H₂O) → H is oxidised
• $O_2$ gave O → O₂ is reduced
• It is a redox + exothermic reaction.

(2) Respiration (in the body):

$$C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O + \text{Energy}$$

Analysis:

• C in glucose gained O → C is oxidised
• $O_2$ gave O → O₂ is reduced
• It is also a redox + exothermic reaction.

Similarities:

1. Both are redox reactions.
2. Both release energy.
3. The final products in both are $CO_2$ and $H_2O$.

Differences:

1. Combustion — fast, at high temperature.
2. Respiration — slow, at body temperature, with the help of enzymes.

Lesson: Life is essentially a controlled combustion — a beautiful intersection of biology and chemistry.

[Board Important — science-philosophy question]

Example 15: Identify the Oxidising Agent — A Challenging Question

Identify the oxidising agent in: (i) $Pb_3O_4 + 8HCl \rightarrow 3PbCl_2 + Cl_2 + 4H_2O$ (ii) $Cl_2 + 2KBr \rightarrow 2KCl + Br_2$ (iii) $2Mg + O_2 \rightarrow 2MgO$ (iv) $Fe + CuSO_4 \rightarrow FeSO_4 + Cu$

Solution:

(i) $Pb_3O_4 + 8HCl$:

• O leaves $Pb_3O_4$ to combine with H, forming $H_2O$ → $Pb_3O_4$ reduced.
• $HCl$'s $Cl^-$ → $Cl_2$ (lost H) → HCl oxidised.
• Oxidising agent: $Pb_3O_4$ (oxidised HCl)

(ii) $Cl_2 + 2KBr$:

• $Br^-$ → $Br_2$ (lost electron) → Br oxidised.
• $Cl_2$ → $Cl^-$ (gained electron) → Cl reduced.
• Oxidising agent: $Cl_2$

(iii) $2Mg + O_2$:

• Mg gained O → Mg oxidised.
• O₂ took electrons from Mg → O₂ reduced.
• Oxidising agent: $O_2$

(iv) $Fe + CuSO_4$:

• Fe → Fe²⁺ (lost electron) → Fe oxidised.
• Cu²⁺ → Cu (gained electron) → Cu²⁺ reduced.
• Oxidising agent: $CuSO_4$ (Cu²⁺ took electrons from Fe)

[Board + entrance level]

Remark: The language of charge change becomes clearer in Class 11, but at this level you can resolve such questions just by tracking gain/loss of O or H.

Example 16: A Tabular Question — All Types of Reactions

Classify the following reactions by 'type of reaction' and 'is it redox?':

(a) $CaO + H_2O \rightarrow Ca(OH)_2 + \text{Heat}$ (b) $2KClO_3 \rightarrow 2KCl + 3O_2$ (c) $BaCl_2 + Na_2SO_4 \rightarrow BaSO_4\downarrow + 2NaCl$ (d) $Zn + CuSO_4 \rightarrow ZnSO_4 + Cu$ (e) $C + O_2 \rightarrow CO_2$ (f) $NaOH + HCl \rightarrow NaCl + H_2O$

Solution:

#	Reaction	Type	Redox?
(a)	$CaO + H_2O \rightarrow Ca(OH)_2$	Combination	No (O attaches but CaO didn't 'lose' O in oxidation sense)
(b)	$2KClO_3 \rightarrow 2KCl + 3O_2$	Decomposition	Yes ($O^{2-} \rightarrow O_2$)
(c)	$BaCl_2 + Na_2SO_4 \rightarrow BaSO_4 + 2NaCl$	Double displacement	No
(d)	$Zn + CuSO_4 \rightarrow ZnSO_4 + Cu$	Displacement	Yes (Zn oxidised, Cu reduced)
(e)	$C + O_2 \rightarrow CO_2$	Combination	Yes (C oxidised)
(f)	$NaOH + HCl \rightarrow NaCl + H_2O$	Double displacement (neutralisation)	No

Main conclusions:

1. Double displacement and neutralisation reactions are usually not redox (only ion exchange).
2. Displacement reactions are always redox.
3. Combustion and many combinations/decompositions are redox.
4. Some combinations (like $CaO + H_2O$) are not redox — because no element's oxidation state truly changed.

[A short version of a 5-mark board question]