Std 10 Science Part 1 · Chapter 3

3. Chemical Reactions and Equations

Complete solutions to the Exercise, with balanced equations and stepwise balancing. Maharashtra State Board (Balbharati), Std 10 Science and Technology Part 1.

Advertisement

Exercise

1Choose the correct option from the bracket and explain the statement giving reason. (Oxidation, displacement, electrolysis, reduction, zinc, copper, double displacement, decomposition)

(a) To prevent rusting, a layer of …… metal is applied on iron sheets.

To prevent rusting, a layer of zinc metal is applied on iron sheets.

Reason: Zinc is more reactive than iron. The zinc layer keeps air and moisture away from the iron, so the iron does not rust. This process is called galvanizing.

(b) The conversion of ferrous sulphate to ferric sulphate is …… reaction.

The conversion of ferrous sulphate to ferric sulphate is oxidation reaction.

Reason: In this change, Fe2+ → Fe3+ + e−. The ferrous ion loses one electron and its positive charge increases by one unit. Losing electrons is oxidation.

(c) When electric current is passed through acidulated water …… of water takes place.

When electric current is passed through acidulated water electrolysis of water takes place.

Reason: Water decomposes into hydrogen and oxygen by means of electrical energy. Decomposition by electricity is called electrolysis.

2H2O(l) electric current→ 2H2↑ + O2↑

(d) Addition of an aqueous solution of ZnSO4 to an aqueous solution of BaCl2 forms a white precipitate is an example of …… reaction.

Addition of an aqueous solution of ZnSO4 to an aqueous solution of BaCl2 forms a white precipitate is an example of double displacement reaction.

Reason: The ions of the two reactants are exchanged, and a white precipitate of barium sulphate is formed.

BaCl2(aq) + ZnSO4(aq) → BaSO4↓ (white) + ZnCl2(aq)
2Write answers to the following.

(a) What is the reaction called when oxidation and reduction take place simultaneously? Explain with one example.

A reaction in which oxidation and reduction take place simultaneously is called a redox reaction.

Example: When hydrogen gas is passed over black copper oxide, reddish copper is formed.

CuO + H2 → Cu + H2O
  • CuO loses oxygen, so copper oxide is reduced.
  • H2 gains oxygen to form water, so hydrogen is oxidised.
  • H2 is the reductant and CuO is the oxidant.

Redox reaction = Reduction + Oxidation.

(b) How can the rate of the chemical reaction, namely, decomposition of hydrogen peroxide be increased?

2H2O2(l) → 2H2O(l) + O2↑

Hydrogen peroxide decomposes slowly at room temperature. The rate can be increased by:

  1. Adding a catalyst: adding manganese dioxide (MnO2) powder makes it decompose rapidly.
  2. Increasing the temperature: heating it increases the rate.
  3. Increasing the concentration: concentrated hydrogen peroxide decomposes faster.

(c) Explain the term reactant and product giving examples.

  • Reactants: The substances which take part in a chemical reaction are called reactants.
  • Products: The new substances formed as a result of a chemical reaction are called products.
C + O2 → CO2
CuSO4(aq) + Zn(s) → ZnSO4(aq) + Cu(s)

In the first reaction, coal (carbon) and oxygen are reactants, and carbon dioxide is the product. In the second, copper sulphate and zinc are reactants, and zinc sulphate and copper are products.

(d) Explain the types of reaction with reference to oxygen and hydrogen. Illustrate with examples.

1. Oxidation: The reaction in which a reactant combines with oxygen or loses hydrogen is called oxidation.

2Mg + O2 → 2MgO (Mg combines with oxygen)
MgH2 → Mg + H2↑ (MgH2 loses hydrogen)

2. Reduction: The reaction in which a reactant gains hydrogen or loses oxygen is called reduction.

CuO + H2 → Cu + H2O (CuO loses oxygen)
Vegetable oil + H2 60 °C, Ni→ Vanaspati ghee (oil gains hydrogen)

3. Redox: Oxidation and reduction occur together. In CuO + H2 → Cu + H2O, CuO is reduced and H2 is oxidised.

(e) Explain the similarity and difference in two events, namely adding NaOH to water and adding CaO to water.

Similarity: In both events heat is given out, so the temperature rises. Both are exothermic. Both form a basic solution.

Difference:

PointNaOH added to waterCaO added to water
Type of changePhysical change (dissolution)Chemical change (combination reaction)
New substanceNo new substance is formed; NaOH only dissolvesA new substance, calcium hydroxide, is formed
NameExothermic processExothermic reaction
CaO(s) + H2O(l) → Ca(OH)2(aq) + heat

Advertisement

3Explain the following terms with examples.

(a) Endothermic reaction

A chemical reaction in which heat is absorbed from the surroundings, or has to be supplied continuously from outside, is called an endothermic reaction.

CaCO3(s) + heat → CaO(s) + CO2(g)

(b) Combination reaction

When two or more reactants combine in a reaction to form a single product, it is called a combination reaction.

2Mg + O2 → 2MgO
NH3(g) + HCl(g) → NH4Cl(s)

(c) Balanced equation

A chemical equation in which the number of atoms of each element in the reactants is the same as in the products is called a balanced equation. It follows the law of conservation of mass.

2NaOH + H2SO4 → Na2SO4 + 2H2O
Na: 2 = 2, S: 1 = 1, O: 6 = 6, H: 4 = 4

(d) Displacement reaction

The reaction in which the place of the ion of a less reactive element in a compound is taken by a more reactive element by formation of its own ions is called a displacement reaction.

CuSO4(aq) + Zn(s) → ZnSO4(aq) + Cu(s)

Zinc is more reactive than copper, so it displaces copper from copper sulphate.

4Give scientific reasons.

(a) When the gas formed on heating limestone is passed through freshly prepared lime water, the lime water turns milky.

  1. On heating, limestone (calcium carbonate) decomposes and gives carbon dioxide gas.
  2. When this CO2 is passed through lime water, it forms calcium carbonate, which is insoluble in water.
  3. This white insoluble calcium carbonate makes the lime water milky.
CaCO3(s) Δ→ CaO(s) + CO2↑
Ca(OH)2(aq) + CO2(g) → CaCO3↓ (white) + H2O(l)

(b) It takes time for pieces of Shahabad tile to disappear in HCl, but its powder disappears rapidly.

  1. Shahabad tile contains calcium carbonate, which reacts with HCl and gives CO2.
  2. The rate of a reaction depends on the size of the particles of the reactants. Smaller the particles, higher the rate.
  3. Powder has very small particles, so a larger surface comes in contact with the acid. Hence the powder reacts and disappears rapidly, while the pieces take time.
CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2↑

(c) While preparing dilute sulphuric acid from concentrated sulphuric acid in the laboratory, the concentrated sulphuric acid is added slowly to water with constant stirring.

  1. Dilution of concentrated sulphuric acid with water is a highly exothermic process. A very large amount of heat is liberated.
  2. If water is poured into the acid, the water evaporates instantly and the acid may splash out, causing an accident.
  3. When a small quantity of acid is added to water at a time with stirring, only a small amount of heat is liberated at a time, and it spreads evenly. So the dilution is safe.

(d) It is recommended to use air tight container for storing oil for long time.

  1. When oil is left in contact with air for a long time, it undergoes air oxidation.
  2. It becomes rancid: it gets a foul smell and its taste changes.
  3. An air tight container keeps air (oxygen) away from the oil, so oxidation slows down and rancidity is prevented.
5Observe the following picture and write down the chemical reaction with explanation.
Collected rustFe₂O₃·H₂OO₂Fe²⁺ (aq)e⁻Water dropCathodeAnodeO₂ + 4H⁺ + 4e⁻ → 2H₂OFe → Fe²⁺ + 2e⁻Iron

Rusting of iron: iron dissolves at the anode, oxygen is reduced at the cathode, and rust collects on the surface.

The picture shows the rusting of iron. Rusting is an electrochemical reaction. Different regions of the iron surface under the water drop act as the anode and the cathode.

1. At the anode: iron is oxidised to Fe2+ ions.

Fe(s) → Fe2+(aq) + 2e−

2. At the cathode: the electrons travel through the iron to the cathode, where oxygen is reduced.

O2(g) + 4H+(aq) + 4e− → 2H2O(l)
or O2 + 2H2O + 4e− → 4OH−

3. Formation of rust: Fe2+ ions move through the water, are oxidised further to Fe3+ ions, and form a reddish hydrated oxide. This is rust, which collects on the surface.

2Fe3+(aq) + 4H2O(l) → Fe2O3·H2O(s) + 6H+(aq)

The formula of rust is Fe2O3·xH2O. Both air and water are necessary for rusting, and salt makes it faster.

6Identify from the following reactions the reactants that undergo oxidation and reduction.

(a) Fe + S → FeS

Oxidation: Fe — iron loses 2 electrons (Fe → Fe2+ + 2e−).
Reduction: S — sulphur gains 2 electrons (S + 2e− → S2−).

(b) 2Ag2O → 4Ag + O2↑

Reduction: Ag2O — it loses oxygen and forms silver (Ag+ + e− → Ag).
Oxidation: the oxide ions of the same Ag2O lose electrons and form oxygen gas (2O2− → O2 + 4e−).

(c) 2Mg + O2 → 2MgO

Oxidation: Mg — magnesium combines with oxygen.
Reduction: O2 — oxygen gains the electrons lost by magnesium.

(d) NiO + H2 → Ni + H2O

Reduction: NiO — nickel oxide loses oxygen.
Oxidation: H2 — hydrogen gains oxygen and forms water.

7Balance the following equation stepwise.

(a) H2S2O7(l) + H2O(l) → H2SO4(l)

Step 1: Count atoms. Left: H = 4, S = 2, O = 8. Right: H = 2, S = 1, O = 4
Step 2: Balance S by putting 2 before H2SO4. Right: H = 4, S = 2, O = 8
Step 3: All atoms are now equal on both sides.
H2S2O7(l) + H2O(l) → 2H2SO4(l)

(b) SO2(g) + H2S(aq) → S(s) + H2O(l)

Step 1: Count atoms. Left: S = 2, O = 2, H = 2. Right: S = 1, O = 1, H = 2
Step 2: Balance O by putting 2 before H2O. Right: O = 2, H = 4
Step 3: Balance H by putting 2 before H2S. Left: H = 4, S = 3
Step 4: Balance S by putting 3 before S. Right: S = 3
SO2(g) + 2H2S(aq) → 3S(s) + 2H2O(l)

(c) Ag(s) + HCl(aq) → AgCl↓ + H2↑

Step 1: Count atoms. Left: Ag = 1, H = 1, Cl = 1. Right: Ag = 1, Cl = 1, H = 2
Step 2: Balance H by putting 2 before HCl. Left: H = 2, Cl = 2
Step 3: Balance Cl by putting 2 before AgCl. Right: Ag = 2, Cl = 2
Step 4: Balance Ag by putting 2 before Ag. Left: Ag = 2
2Ag(s) + 2HCl(aq) → 2AgCl↓ + H2↑

This equation is given only for practice in balancing. In practice, silver is less reactive than hydrogen, so it does not react with dilute hydrochloric acid.

(d) NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + H2O(l)

Step 1: Count atoms. Left: Na = 1, O = 5, H = 3, S = 1. Right: Na = 2, O = 5, H = 2, S = 1
Step 2: Balance Na by putting 2 before NaOH. Left: Na = 2, O = 6, H = 4
Step 3: Balance H by putting 2 before H2O. Right: H = 4, O = 6
Step 4: All atoms are now equal on both sides.
2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2H2O(l)

Advertisement

8Identify the endothermic and exothermic reaction.

(a) HCl + NaOH → NaCl + H2O + heat

Exothermic reaction — heat is given out.

(b) 2KClO3(s) Δ→ 2KCl(s) + 3O2↑

Endothermic reaction — heat has to be supplied (Δ).

(c) CaO + H2O → Ca(OH)2 + heat

Exothermic reaction — heat is given out.

(d) CaCO3(s) Δ→ CaO(s) + CO2↑

Endothermic reaction — heat has to be supplied (Δ).

9Match the column in the following table.
ReactantsProductsType of chemical reaction
BaCl2(aq) + ZnSO4(aq)BaSO4↓ + ZnCl2(aq)Double displacement
2AgCl(s)2Ag(s) + Cl2(g)Decomposition
CuSO4(aq) + Fe(s)FeSO4(aq) + Cu(s)Displacement
H2O(l) + CO2(g)H2CO3(aq)Combination
★Project: Prepare aqueous solutions of various solid salts available in the laboratory. Observe what happens when aqueous solution of sodium hydroxide is added to these. Prepare a chart of double displacement reactions based on these observations.

Sodium hydroxide exchanges ions with many salt solutions and forms an insoluble metal hydroxide precipitate. A sample chart:

Salt solutionReaction with NaOHObservation
Copper sulphateCuSO4 + 2NaOH → Cu(OH)2↓ + Na2SO4Blue precipitate
Ferrous sulphateFeSO4 + 2NaOH → Fe(OH)2↓ + Na2SO4Dirty green precipitate
Ferric chlorideFeCl3 + 3NaOH → Fe(OH)3↓ + 3NaClReddish brown precipitate
Zinc sulphateZnSO4 + 2NaOH → Zn(OH)2↓ + Na2SO4White precipitate
Magnesium chlorideMgCl2 + 2NaOH → Mg(OH)2↓ + 2NaClWhite precipitate
Aluminium chlorideAlCl3 + 3NaOH → Al(OH)3↓ + 3NaClWhite gelatinous precipitate

Sodium hydroxide is corrosive. Do this activity only in the presence of your teacher, using dilute solutions.