Chemical Reaction and Equations

Given: Magnesium ribbon burns in air (oxygen) to form magnesium oxide.

Reaction:
$$2\mathrm{Mg}(\mathrm{s}) + \mathrm{O}_2(\mathrm{g}) \rightarrow 2\mathrm{MgO}(\mathrm{s})$$

Concept: A substance is said to be oxidised if it gains oxygen during a reaction.

Conclusion: Magnesium gains oxygen to form magnesium oxide. Therefore, magnesium is being oxidised in this reaction.

Given: An iron nail is dipped in copper sulphate (CuSO₄) solution.

Concept: This is a displacement reaction. A more reactive metal displaces a less reactive metal from its salt solution. Iron is more reactive than copper.

Reaction:
$$\mathrm{Fe}(\mathrm{s}) + \mathrm{CuSO}_4(\mathrm{aq}) \rightarrow \mathrm{FeSO}_4(\mathrm{aq}) + \mathrm{Cu}(\mathrm{s})$$

Explanation: Iron displaces copper from copper sulphate solution. The blue colour of copper sulphate solution fades (or turns light green) because blue CuSO₄ is consumed and green FeSO₄ is formed. A reddish-brown deposit of copper metal appears on the iron nail.

Answer: The colour of copper sulphate solution changes because iron displaces copper from the solution, forming iron sulphate (FeSO₄) which is light green, and copper metal is deposited on the nail.

Concept: In a double displacement reaction, two compounds react by exchanging their ions (or groups of atoms) to form two new compounds.

Example: Reaction between sodium sulphate and barium chloride:
$$\mathrm{Na}_2\mathrm{SO}_4(\mathrm{aq}) + \mathrm{BaCl}_2(\mathrm{aq}) \rightarrow \mathrm{BaSO}_4(\mathrm{s})\downarrow + 2\mathrm{NaCl}(\mathrm{aq})$$

Explanation: Here, $\mathrm{Na}^+$ and $\mathrm{Ba}^{2+}$ ions exchange their respective anions. Barium sulphate ($\mathrm{BaSO}_4$) is formed as a white insoluble precipitate. This is also a precipitation reaction.

Concept:
- Oxidation = gain of oxygen OR loss of hydrogen.
- Reduction = loss of oxygen OR gain of hydrogen.

(i) $4\mathrm{Na}(\mathrm{s}) + \mathrm{O}_2(\mathrm{g}) \rightarrow 2\mathrm{Na}_2\mathrm{O}(\mathrm{s})$

- Sodium (Na) gains oxygen to form sodium oxide (Na₂O). Therefore, sodium (Na) is oxidised.
- Oxygen ($\mathrm{O}_2$) itself is the oxidising agent; it is being used up to oxidise Na. (In this reaction, only oxidation occurs — there is no separate substance being reduced in the classical sense, but $\mathrm{O}_2$ acts as the oxidising agent.)

Answer: Sodium (Na) is oxidised (gains oxygen).

(ii) $\mathrm{CuO(s)} + \mathrm{H}_2(\mathrm{g}) \rightarrow \mathrm{Cu(s)} + \mathrm{H}_2\mathrm{O(l)}$

- CuO loses oxygen to form Cu. Therefore, CuO (copper oxide) is reduced.
- H₂ gains oxygen to form H₂O. Therefore, hydrogen (H₂) is oxidised.

Answer:
- Substance oxidised → $\mathrm{H}_2$ (hydrogen gas gains oxygen).
- Substance reduced → $\mathrm{CuO}$ (copper oxide loses oxygen).

Correct Option: (i) (a) and (b)

Analysis of the reaction:
$$2\mathrm{PbO}(\mathrm{s}) + \mathrm{C}(\mathrm{s}) \rightarrow 2\mathrm{Pb}(\mathrm{s}) + \mathrm{CO}_2(\mathrm{g})$$

- PbO loses oxygen → PbO is reduced (statement (d) is correct).
- Pb is the product of reduction of PbO, so lead itself is not getting reduced — statement (a) is incorrect.
- Carbon (C) gains oxygen to form CO₂ → Carbon is oxidised (statement (c) is correct).
- CO₂ is the product of oxidation; it is not getting oxidised further — statement (b) is incorrect.

Incorrect statements: (a) and (b) → Option (i)

Correct Option: (d) displacement reaction.

Justification: Aluminium (Al) is more reactive than iron (Fe). It displaces iron from iron(III) oxide (Fe₂O₃), taking its place. A single more reactive element displaces a less reactive element from its compound — this is the definition of a displacement reaction. This reaction is also known as the thermite reaction.

Correct Answer: (a) Hydrogen gas and iron chloride are produced.

Reaction:
$$\mathrm{Fe}(\mathrm{s}) + 2\mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{FeCl}_2(\mathrm{aq}) + \mathrm{H}_2(\mathrm{g})\uparrow$$

Justification: Iron is more reactive than hydrogen. It displaces hydrogen from dilute hydrochloric acid, forming iron(II) chloride (FeCl₂) and hydrogen gas. This is a displacement reaction.

Balanced Chemical Equation:

A balanced chemical equation is one in which the number of atoms of each element is equal on both the reactant side and the product side of the equation.

Example:
$$2\mathrm{H}_2 + \mathrm{O}_2 \rightarrow 2\mathrm{H}_2\mathrm{O}$$
Here, H: 4 on each side; O: 2 on each side — the equation is balanced.

Why should chemical equations be balanced?

Chemical equations must be balanced to satisfy the Law of Conservation of Mass, which states that matter can neither be created nor destroyed in a chemical reaction. This means the total mass of reactants must equal the total mass of products. Since atoms are neither created nor destroyed during a reaction, the number of atoms of each element must be the same on both sides of the equation. An unbalanced equation violates this fundamental law.

(a) Hydrogen gas combines with nitrogen to form ammonia.

Unbalanced: $\mathrm{H}_2 + \mathrm{N}_2 \rightarrow \mathrm{NH}_3$

Balancing:
- N: 2 on left, 1 on right → put 2 before NH₃: $\mathrm{H}_2 + \mathrm{N}_2 \rightarrow 2\mathrm{NH}_3$
- H: 2 on left, 6 on right → put 3 before H₂: $3\mathrm{H}_2 + \mathrm{N}_2 \rightarrow 2\mathrm{NH}_3$

Balanced equation:
$$3\mathrm{H}_2(\mathrm{g}) + \mathrm{N}_2(\mathrm{g}) \rightarrow 2\mathrm{NH}_3(\mathrm{g})$$

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(b) Hydrogen sulphide gas burns in air to give water and sulphur dioxide.

Unbalanced: $\mathrm{H}_2\mathrm{S} + \mathrm{O}_2 \rightarrow \mathrm{H}_2\mathrm{O} + \mathrm{SO}_2$

Balancing:
- S: 1 on each side ✓
- H: 2 on each side ✓
- O: 2 on left, 3 on right → Balance O by adjusting coefficients.
- Try: $2\mathrm{H}_2\mathrm{S} + 3\mathrm{O}_2 \rightarrow 2\mathrm{H}_2\mathrm{O} + 2\mathrm{SO}_2$
- S: 2 = 2 ✓; H: 4 = 4 ✓; O: 6 = 6 ✓

Balanced equation:
$$2\mathrm{H}_2\mathrm{S}(\mathrm{g}) + 3\mathrm{O}_2(\mathrm{g}) \rightarrow 2\mathrm{H}_2\mathrm{O}(\mathrm{l}) + 2\mathrm{SO}_2(\mathrm{g})$$

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(c) Barium chloride reacts with aluminium sulphate to give aluminium chloride and a precipitate of barium sulphate.

Unbalanced: $\mathrm{BaCl}_2 + \mathrm{Al}_2(\mathrm{SO}_4)_3 \rightarrow \mathrm{AlCl}_3 + \mathrm{BaSO}_4$

Balancing:
- Al: 2 on left → put 2 before AlCl₃
- SO₄: 3 on left → put 3 before BaSO₄
- Ba: 3 on right → put 3 before BaCl₂
- Cl: 6 on left (3×2) → 6 on right (2×3) ✓

Balanced equation:
$$3\mathrm{BaCl}_2(\mathrm{aq}) + \mathrm{Al}_2(\mathrm{SO}_4)_3(\mathrm{aq}) \rightarrow 2\mathrm{AlCl}_3(\mathrm{aq}) + 3\mathrm{BaSO}_4(\mathrm{s})\downarrow$$

Check: Ba=3, Cl=6, Al=2, S=3, O=12 — equal on both sides ✓

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(d) Potassium metal reacts with water to give potassium hydroxide and hydrogen gas.

Unbalanced: $\mathrm{K} + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{KOH} + \mathrm{H}_2$

Balancing:
- K: 1 on each side ✓
- O: 1 on each side ✓
- H: 2 on left, 3 on right → multiply through:
- $2\mathrm{K} + 2\mathrm{H}_2\mathrm{O} \rightarrow 2\mathrm{KOH} + \mathrm{H}_2$
- K: 2=2 ✓; O: 2=2 ✓; H: 4 left = 2+2=4 right ✓

Balanced equation:
$$2\mathrm{K}(\mathrm{s}) + 2\mathrm{H}_2\mathrm{O}(\mathrm{l}) \rightarrow 2\mathrm{KOH}(\mathrm{aq}) + \mathrm{H}_2(\mathrm{g})\uparrow$$

(a) $\mathrm{HNO}_3 + \mathrm{Ca(OH)}_2 \rightarrow \mathrm{Ca(NO}_3)_2 + \mathrm{H}_2\mathrm{O}$

- $\mathrm{Ca(NO}_3)_2$ has 2 NO₃ groups → need 2 HNO₃ on left.
- H on left: 2(from HNO₃) + 2(from Ca(OH)₂) = 4 → need 2 H₂O on right.

Balanced:
$$2\mathrm{HNO}_3(\mathrm{aq}) + \mathrm{Ca(OH)}_2(\mathrm{aq}) \rightarrow \mathrm{Ca(NO}_3)_2(\mathrm{aq}) + 2\mathrm{H}_2\mathrm{O}(\mathrm{l})$$
Check: H=4, N=2, O=8, Ca=1 — equal on both sides ✓

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(b) $\mathrm{NaOH} + \mathrm{H}_2\mathrm{SO}_4 \rightarrow \mathrm{Na}_2\mathrm{SO}_4 + \mathrm{H}_2\mathrm{O}$

- Na₂SO₄ needs 2 Na → 2 NaOH on left.
- H on left: 2(NaOH) + 2(H₂SO₄) = 4 → 2 H₂O on right.

Balanced:
$$2\mathrm{NaOH}(\mathrm{aq}) + \mathrm{H}_2\mathrm{SO}_4(\mathrm{aq}) \rightarrow \mathrm{Na}_2\mathrm{SO}_4(\mathrm{aq}) + 2\mathrm{H}_2\mathrm{O}(\mathrm{l})$$
Check: Na=2, O=6, H=4, S=1 — equal on both sides ✓

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(c) $\mathrm{NaCl} + \mathrm{AgNO}_3 \rightarrow \mathrm{AgCl} + \mathrm{NaNO}_3$

- All atoms are already balanced: Na=1, Cl=1, Ag=1, N=1, O=3 on each side.

Balanced:
$$\mathrm{NaCl}(\mathrm{aq}) + \mathrm{AgNO}_3(\mathrm{aq}) \rightarrow \mathrm{AgCl}(\mathrm{s})\downarrow + \mathrm{NaNO}_3(\mathrm{aq})$$
(Already balanced as written.)

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(d) $\mathrm{BaCl}_2 + \mathrm{H}_2\mathrm{SO}_4 \rightarrow \mathrm{BaSO}_4 + \mathrm{HCl}$

- Cl: 2 on left → 2 HCl on right.
- H: 2 on left (H₂SO₄) → 2 HCl on right ✓

Balanced:
$$\mathrm{BaCl}_2(\mathrm{aq}) + \mathrm{H}_2\mathrm{SO}_4(\mathrm{aq}) \rightarrow \mathrm{BaSO}_4(\mathrm{s})\downarrow + 2\mathrm{HCl}(\mathrm{aq})$$
Check: Ba=1, Cl=2, H=2, S=1, O=4 — equal on both sides ✓

(a) Calcium hydroxide + Carbon dioxide → Calcium carbonate + Water

Unbalanced: $\mathrm{Ca(OH)}_2 + \mathrm{CO}_2 \rightarrow \mathrm{CaCO}_3 + \mathrm{H}_2\mathrm{O}$

Check: Ca=1, C=1, O=4 left (2+2), O=4 right (3+1), H=2 on each side ✓

Balanced:
$$\mathrm{Ca(OH)}_2(\mathrm{aq}) + \mathrm{CO}_2(\mathrm{g}) \rightarrow \mathrm{CaCO}_3(\mathrm{s})\downarrow + \mathrm{H}_2\mathrm{O}(\mathrm{l})$$

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(b) Zinc + Silver nitrate → Zinc nitrate + Silver

Unbalanced: $\mathrm{Zn} + \mathrm{AgNO}_3 \rightarrow \mathrm{Zn(NO}_3)_2 + \mathrm{Ag}$

- Zn(NO₃)₂ needs 2 NO₃ → 2 AgNO₃ on left → 2 Ag on right.

Balanced:
$$\mathrm{Zn}(\mathrm{s}) + 2\mathrm{AgNO}_3(\mathrm{aq}) \rightarrow \mathrm{Zn(NO}_3)_2(\mathrm{aq}) + 2\mathrm{Ag}(\mathrm{s})$$
Check: Zn=1, Ag=2, N=2, O=6 — equal on both sides ✓

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(c) Aluminium + Copper chloride → Aluminium chloride + Copper

Unbalanced: $\mathrm{Al} + \mathrm{CuCl}_2 \rightarrow \mathrm{AlCl}_3 + \mathrm{Cu}$

- AlCl₃ needs 3 Cl; CuCl₂ provides 2 Cl each → LCM of 3 and 2 = 6
- 2 AlCl₃ needs 6 Cl → 3 CuCl₂ on left → 3 Cu on right → 2 Al on left.

Balanced:
$$2\mathrm{Al}(\mathrm{s}) + 3\mathrm{CuCl}_2(\mathrm{aq}) \rightarrow 2\mathrm{AlCl}_3(\mathrm{aq}) + 3\mathrm{Cu}(\mathrm{s})$$
Check: Al=2, Cu=3, Cl=6 — equal on both sides ✓

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(d) Barium chloride + Potassium sulphate → Barium sulphate + Potassium chloride

Unbalanced: $\mathrm{BaCl}_2 + \mathrm{K}_2\mathrm{SO}_4 \rightarrow \mathrm{BaSO}_4 + \mathrm{KCl}$

- K: 2 on left → 2 KCl on right → Cl: 2 on right, 2 on left ✓

Balanced:
$$\mathrm{BaCl}_2(\mathrm{aq}) + \mathrm{K}_2\mathrm{SO}_4(\mathrm{aq}) \rightarrow \mathrm{BaSO}_4(\mathrm{s})\downarrow + 2\mathrm{KCl}(\mathrm{aq})$$
Check: Ba=1, Cl=2, K=2, S=1, O=4 — equal on both sides ✓

(a) Potassium bromide + Barium iodide → Potassium iodide + Barium bromide

Unbalanced: $\mathrm{KBr} + \mathrm{BaI}_2 \rightarrow \mathrm{KI} + \mathrm{BaBr}_2$

- Ba: 1 on each side ✓; BaBr₂ needs 2 Br → 2 KBr on left; 2 KI on right.

Balanced:
$$2\mathrm{KBr}(\mathrm{aq}) + \mathrm{BaI}_2(\mathrm{aq}) \rightarrow 2\mathrm{KI}(\mathrm{aq}) + \mathrm{BaBr}_2(\mathrm{s})$$
Check: K=2, Br=2, Ba=1, I=2 — equal on both sides ✓

Type of Reaction: Double Displacement Reaction (two compounds exchange their ions; BaBr₂ is precipitated).

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(b) Zinc carbonate → Zinc oxide + Carbon dioxide

$$\mathrm{ZnCO}_3(\mathrm{s}) \xrightarrow{\Delta} \mathrm{ZnO}(\mathrm{s}) + \mathrm{CO}_2(\mathrm{g})$$

Check: Zn=1, C=1, O=3 on each side ✓ (Already balanced)

Type of Reaction: Decomposition Reaction (a single compound breaks down into two simpler substances on heating).

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(c) Hydrogen + Chlorine → Hydrogen chloride

$$\mathrm{H}_2(\mathrm{g}) + \mathrm{Cl}_2(\mathrm{g}) \rightarrow 2\mathrm{HCl}(\mathrm{g})$$

Check: H=2, Cl=2 on each side ✓

Type of Reaction: Combination Reaction (two elements combine to form a single product).

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(d) Magnesium + Hydrochloric acid → Magnesium chloride + Hydrogen

Unbalanced: $\mathrm{Mg} + \mathrm{HCl} \rightarrow \mathrm{MgCl}_2 + \mathrm{H}_2$

- MgCl₂ needs 2 Cl → 2 HCl on left; H: 2 on left, 2 on right ✓

Balanced:
$$\mathrm{Mg}(\mathrm{s}) + 2\mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{MgCl}_2(\mathrm{aq}) + \mathrm{H}_2(\mathrm{g})\uparrow$$
Check: Mg=1, H=2, Cl=2 — equal on both sides ✓

Type of Reaction: Displacement Reaction (magnesium, being more reactive, displaces hydrogen from hydrochloric acid).

Exothermic Reactions:

Reactions in which energy (heat) is released along with the formation of products are called exothermic reactions.

Example 1: Burning of natural gas (methane):
$$\mathrm{CH}_4(\mathrm{g}) + 2\mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{CO}_2(\mathrm{g}) + 2\mathrm{H}_2\mathrm{O}(\mathrm{g}) + \text{Heat}$$

Example 2: Respiration:
$$\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6(\mathrm{aq}) + 6\mathrm{O}_2(\mathrm{aq}) \rightarrow 6\mathrm{CO}_2(\mathrm{aq}) + 6\mathrm{H}_2\mathrm{O}(\mathrm{l}) + \text{Energy}$$

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Endothermic Reactions:

Reactions in which energy (heat) is absorbed from the surroundings are called endothermic reactions.

Example 1: Decomposition of calcium carbonate (limestone):
$$\mathrm{CaCO}_3(\mathrm{s}) \xrightarrow{\text{Heat}} \mathrm{CaO}(\mathrm{s}) + \mathrm{CO}_2(\mathrm{g})$$

Example 2: Photosynthesis:
$$6\mathrm{CO}_2(\mathrm{g}) + 6\mathrm{H}_2\mathrm{O}(\mathrm{l}) \xrightarrow{\text{Sunlight}} \mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6(\mathrm{aq}) + 6\mathrm{O}_2(\mathrm{g})$$

Given: The process of respiration in living organisms.

Explanation:

During respiration, glucose (food) that we eat is broken down in the presence of oxygen inside our body cells. This process releases energy in the form of heat, which is used by the body for various life processes.

Reaction:
$$\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6(\mathrm{aq}) + 6\mathrm{O}_2(\mathrm{aq}) \rightarrow 6\mathrm{CO}_2(\mathrm{aq}) + 6\mathrm{H}_2\mathrm{O}(\mathrm{l}) + \text{Energy}$$

Since energy (heat) is released during this reaction, respiration is considered an exothermic reaction.

The energy released during respiration is used by our body to maintain body temperature and to carry out other activities.

Combination Reaction:

In a combination reaction, two or more substances combine to form a single new substance.

$$\mathrm{A} + \mathrm{B} \rightarrow \mathrm{AB}$$

Example:
$$2\mathrm{Mg}(\mathrm{s}) + \mathrm{O}_2(\mathrm{g}) \rightarrow 2\mathrm{MgO}(\mathrm{s})$$

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Decomposition Reaction:

In a decomposition reaction, a single substance breaks down into two or more simpler substances.

$$\mathrm{AB} \rightarrow \mathrm{A} + \mathrm{B}$$

Example:
$$2\mathrm{MgO}(\mathrm{s}) \xrightarrow{\text{Heat}} 2\mathrm{Mg}(\mathrm{s}) + \mathrm{O}_2(\mathrm{g})$$

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Why are they called opposites?

In a combination reaction, reactants combine to give one product (many → one). In a decomposition reaction, one reactant breaks down to give many products (one → many). The direction of the process is exactly reversed. Hence, decomposition reactions are called the opposite of combination reactions.

(i) Decomposition by Heat (Thermolysis):

$$2\mathrm{Pb(NO}_3)_2(\mathrm{s}) \xrightarrow{\text{Heat}} 2\mathrm{PbO}(\mathrm{s}) + 4\mathrm{NO}_2(\mathrm{g}) + \mathrm{O}_2(\mathrm{g})$$

Lead nitrate decomposes on heating to give lead oxide, nitrogen dioxide (brown fumes), and oxygen.

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(ii) Decomposition by Light (Photolysis):

$$2\mathrm{AgCl}(\mathrm{s}) \xrightarrow{\text{Sunlight}} 2\mathrm{Ag}(\mathrm{s}) + \mathrm{Cl}_2(\mathrm{g})$$

Silver chloride decomposes in the presence of sunlight to give silver metal and chlorine gas. (This reaction is used in black-and-white photography.)

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(iii) Decomposition by Electricity (Electrolysis):

$$2\mathrm{H}_2\mathrm{O}(\mathrm{l}) \xrightarrow{\text{Electric current}} 2\mathrm{H}_2(\mathrm{g}) + \mathrm{O}_2(\mathrm{g})$$

Water decomposes into hydrogen and oxygen gases when electric current is passed through it.

Displacement Reaction:

A reaction in which a more reactive element displaces a less reactive element from its compound (salt solution, acid, etc.).

$$\mathrm{A} + \mathrm{BC} \rightarrow \mathrm{AC} + \mathrm{B}$$

Example:
$$\mathrm{Fe}(\mathrm{s}) + \mathrm{CuSO}_4(\mathrm{aq}) \rightarrow \mathrm{FeSO}_4(\mathrm{aq}) + \mathrm{Cu}(\mathrm{s})$$

Iron displaces copper from copper sulphate solution.

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Double Displacement Reaction:

A reaction in which two compounds react by exchanging their ions (or groups of atoms) to form two new compounds. Usually one of the products is a precipitate, gas, or water.

$$\mathrm{AB} + \mathrm{CD} \rightarrow \mathrm{AD} + \mathrm{CB}$$

Example:
$$\mathrm{Na}_2\mathrm{SO}_4(\mathrm{aq}) + \mathrm{BaCl}_2(\mathrm{aq}) \rightarrow \mathrm{BaSO}_4(\mathrm{s})\downarrow + 2\mathrm{NaCl}(\mathrm{aq})$$

Both compounds exchange their ions; barium sulphate precipitates out.

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Key Difference:

| Feature | Displacement | Double Displacement |
|---|---|---|
| Number of reactants | One element + one compound | Two compounds |
| Exchange | One element replaces another | Two groups of ions are exchanged |
| Example | Fe + CuSO₄ | Na₂SO₄ + BaCl₂ |

Given: Copper metal displaces silver from silver nitrate solution.

Concept: Copper is more reactive than silver, so it displaces silver from silver nitrate solution. This is a displacement reaction.

Reaction:
$$\mathrm{Cu}(\mathrm{s}) + 2\mathrm{AgNO}_3(\mathrm{aq}) \rightarrow \mathrm{Cu(NO}_3)_2(\mathrm{aq}) + 2\mathrm{Ag}(\mathrm{s})$$

Explanation: Copper displaces silver from silver nitrate solution. Copper nitrate $[\mathrm{Cu(NO}_3)_2]$ is formed in solution and silver metal is deposited.

Precipitation Reaction:

A reaction in which an insoluble solid (precipitate) is formed when two solutions are mixed is called a precipitation reaction. The insoluble product that settles down is called a precipitate.

Example 1:
$$\mathrm{Na}_2\mathrm{SO}_4(\mathrm{aq}) + \mathrm{BaCl}_2(\mathrm{aq}) \rightarrow \mathrm{BaSO}_4(\mathrm{s})\downarrow + 2\mathrm{NaCl}(\mathrm{aq})$$

When sodium sulphate solution is mixed with barium chloride solution, a white insoluble precipitate of barium sulphate ($\mathrm{BaSO}_4$) is formed.

Example 2:
$$\mathrm{NaCl}(\mathrm{aq}) + \mathrm{AgNO}_3(\mathrm{aq}) \rightarrow \mathrm{AgCl}(\mathrm{s})\downarrow + \mathrm{NaNO}_3(\mathrm{aq})$$

When sodium chloride solution is mixed with silver nitrate solution, a white insoluble precipitate of silver chloride ($\mathrm{AgCl}$) is formed.

Precipitation reactions are a type of double displacement reaction.

(a) Oxidation:

Definition: Oxidation is the process in which a substance gains oxygen (or loses hydrogen).

Example 1: Burning of magnesium in oxygen:
$$2\mathrm{Mg}(\mathrm{s}) + \mathrm{O}_2(\mathrm{g}) \rightarrow 2\mathrm{MgO}(\mathrm{s})$$
Magnesium gains oxygen → Magnesium is oxidised.

Example 2: Oxidation of carbon:
$$2\mathrm{C}(\mathrm{s}) + \mathrm{O}_2(\mathrm{g}) \rightarrow 2\mathrm{CO}(\mathrm{g})$$
Carbon gains oxygen → Carbon is oxidised.

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(b) Reduction:

Definition: Reduction is the process in which a substance loses oxygen (or gains hydrogen).

Example 1: Reduction of copper oxide by hydrogen:
$$\mathrm{CuO}(\mathrm{s}) + \mathrm{H}_2(\mathrm{g}) \rightarrow \mathrm{Cu}(\mathrm{s}) + \mathrm{H}_2\mathrm{O}(\mathrm{l})$$
CuO loses oxygen → CuO (copper oxide) is reduced to copper.

Example 2: Reduction of zinc oxide by carbon:
$$\mathrm{ZnO}(\mathrm{s}) + \mathrm{C}(\mathrm{s}) \rightarrow \mathrm{Zn}(\mathrm{s}) + \mathrm{CO}(\mathrm{g})$$
ZnO loses oxygen → ZnO is reduced to zinc.

Given: A shiny brown coloured element 'X' turns black on heating in air.

Identification: The shiny brown coloured element is Copper (Cu).

Reaction: When copper is heated in air (oxygen), it reacts with oxygen to form copper(II) oxide, which is black in colour.

$$2\mathrm{Cu}(\mathrm{s}) + \mathrm{O}_2(\mathrm{g}) \xrightarrow{\text{Heat}} 2\mathrm{CuO}(\mathrm{s})$$

Answer:
- Element 'X' = Copper (Cu)
- Black coloured compound formed = Copper(II) oxide (CuO)

This is an oxidation reaction as copper gains oxygen.

Reason:

Iron articles are painted to prevent corrosion (rusting).

Explanation: When iron is exposed to moisture and oxygen in the air, it undergoes a chemical reaction to form hydrated iron(III) oxide, commonly known as rust ($\mathrm{Fe}_2\mathrm{O}_3 \cdot x\mathrm{H}_2\mathrm{O}$), which is reddish-brown in colour.

When paint is applied on iron articles, it forms a protective layer that prevents iron from coming in contact with moisture and oxygen. This stops the process of rusting/corrosion.

Thus, painting is a method of preventing corrosion and increasing the life of iron articles.

Reason:

Oil and fat containing food items are flushed with nitrogen gas to prevent oxidation (rancidity).

Explanation: When fats and oils in food are exposed to oxygen in the air, they get oxidised and become rancid — their smell and taste change, making the food unfit for consumption. This process is called rancidity.

Nitrogen is a chemically inert (unreactive) gas. When bags of chips or other fat-containing food items are flushed with nitrogen, the oxygen inside the packet is replaced by nitrogen. Since there is no oxygen, the fats and oils cannot get oxidised.

This helps in preserving the food for a longer time and prevents it from becoming rancid.

(a) Corrosion:

Definition: The process by which metals are slowly eaten away (attacked) by substances such as moisture, acids, or other chemicals present in the environment is called corrosion.

Explanation: When a metal surface comes in contact with air, moisture, or chemicals, it undergoes oxidation and forms an undesirable compound on its surface.

Example: Rusting of iron — when iron is exposed to moist air, it reacts with oxygen and water to form a reddish-brown layer of hydrated iron(III) oxide (rust).
$$4\mathrm{Fe}(\mathrm{s}) + 3\mathrm{O}_2(\mathrm{g}) + x\mathrm{H}_2\mathrm{O}(\mathrm{l}) \rightarrow 2\mathrm{Fe}_2\mathrm{O}_3 \cdot x\mathrm{H}_2\mathrm{O}(\mathrm{s})$$

Other examples: green coating on copper, black coating on silver.

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(b) Rancidity:

Definition: The process by which fats and oils in food are oxidised on exposure to air, resulting in a change in their smell and taste (making the food unfit for consumption), is called rancidity.

Explanation: When food items containing fats and oils are left open in air for a long time, the oxygen in air oxidises the fats and oils. This produces new compounds with an unpleasant smell and taste.

Example: Butter left open in air for a long time becomes rancid — it develops a bad smell and taste due to oxidation of the fats present in it.

Prevention: Rancidity can be prevented by:
- Adding antioxidants to food.
- Storing food in airtight containers.
- Flushing food packets with inert gases like nitrogen.