Chemical Reactions and Equations — Class 10 Science Solutions

This question checks the chapter’s oxygen-based definitions: a substance that gains oxygen is oxidised, and one that loses oxygen is reduced (NCERT, p. 12). In \(2\text{PbO} + \text{C} \rightarrow 2\text{Pb} + \text{CO}_2\), lead oxide (PbO, the compound) loses its oxygen to become lead, so PbO is reduced — that makes statement (d) correct.

Carbon (C) gains oxygen to form CO₂, so carbon is oxidised. Carbon dioxide is a product of that oxidation; it cannot itself be “getting oxidised”, so statement (b) is incorrect. Statement (a) is the trap: it says the element lead is reduced, but the substance undergoing reduction is the compound lead oxide, not the free element lead.

The accepted answer is (iii) — (a), (b) and (c). Note honestly: statement (c) reads as chemically true (carbon does gain oxygen), which is exactly why this MCQ is discussed so often — the official key groups (a), (b) and (c) as incorrect.

Correct answer: (iii) (a), (b) and (c).

Watch what moves: aluminium (an element) replaces iron from its compound iron(III) oxide. An element taking the place of a less reactive element inside a compound is the signature of a displacement reaction (NCERT, p. 10-11).

It is not a combination (two reactants give one product), not a decomposition (one reactant splits), and not a double displacement (two compounds exchanging ions). The reaction is also redox — aluminium gains oxygen, iron(III) oxide loses it — but displacement is the classification the options permit.

Correct answer: (d) displacement reaction.

The general rule that decides this: a metal plus a dilute acid gives a salt plus hydrogen gas. Iron is more reactive than hydrogen, so iron displaces hydrogen from hydrochloric acid — the same gas-forming chemistry you saw with zinc and dilute sulphuric acid in Activity 1.3 and Figure 1.2 (NCERT, p. 2).

The salt is iron chloride: \(\text{Fe} + 2\text{HCl} \rightarrow \text{FeCl}_2 + \text{H}_2\). Bubbles of hydrogen form and the acid slowly turns pale green as iron dissolves.

Common error: option (d) — “iron salt and water” — describes neutralisation (acid + base gives salt + water).

Metal + acid is a different reaction: it produces hydrogen gas, not water.

The correct answer is (a).

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

A balanced chemical equation has the same number of atoms of every element on the reactant (LHS) and product (RHS) sides.

Balancing is not a convention — it follows the law of conservation of mass (NCERT, p. 3): atoms are neither created nor destroyed in a chemical reaction, so the total mass of the reactants must equal the total mass of the products. If an equation is unbalanced, it implies mass has appeared or vanished, which is impossible.

We balance by putting coefficients (whole numbers) before formulae — never by changing subscripts, because a subscript change alters the substance itself.

Common error: students “fix” an imbalance by editing the formula, writing \(\text{H}_2\text{O}_4\) instead of \(4\text{H}_2\text{O}\).

The textbook warns against exactly this (NCERT, p. 4): a coefficient multiplies the whole formula; a subscript change makes a new compound.

Part (a): Hydrogen and nitrogen are both diatomic gases, so each starts as \(\text{H}_2\) and \(\text{N}_2\), and ammonia is \(\text{NH}_3\). Two nitrogens on the left need two \(\text{NH}_3\) on the right, which introduces six hydrogens — so three \(\text{H}_2\) are needed:

\[ \text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightarrow 2\text{NH}_3(\text{g}) \]

Part (b): Hydrogen sulphide is \(\text{H}_2\text{S}\).

“Burns in air” means it reacts with oxygen, \(\text{O}_2\).

Two \(\text{H}_2\text{S}\) give four hydrogens — enough for two waters — and the sulphur of two \(\text{H}_2\text{S}\) gives two \(\text{SO}_2\).

Oxygen must then supply \(2 + 4 = 6\) atoms, i.e.

three \(\text{O}_2\):

\[ 2\text{H}_2\text{S}(\text{g}) + 3\text{O}_2(\text{g}) \rightarrow 2\text{H}_2\text{O}(\text{g}) + 2\text{SO}_2(\text{g}) \]

Part (c): Barium is \(\text{Ba}^{2+}\) and sulphate is \(\text{SO}_4^{2-}\), so barium sulphate is \(\text{BaSO}_4\) with a 1:1 formula; aluminium is \(\text{Al}^{3+}\) and chloride is \(\text{Cl}^-\), giving \(\text{AlCl}_3\).

Three sulphates on the left need three \(\text{BaSO}_4\), which uses three \(\text{BaCl}_2\) and supplies six chlorides — exactly enough for two \(\text{AlCl}_3\):

\[ 3\text{BaCl}_2 + \text{Al}_2(\text{SO}_4)_3 \rightarrow 2\text{AlCl}_3 + 3\text{BaSO}_4 \]

Part (d): Potassium and water give potassium hydroxide \(\text{KOH}\) and hydrogen.

Hydrogen must leave as the diatomic \(\text{H}_2\), so two waters are needed to supply two hydrogens for it, and those two waters also supply the two oxygens that fix two KOH:

\[ 2\text{K} + 2\text{H}_2\text{O} \rightarrow 2\text{KOH} + \text{H}_2 \]

Common error: in part (d), students write \(\text{Na} + \text{H}_2\text{O} \rightarrow \text{NaOH} + \text{H}_2\) unbalanced, forgetting hydrogen leaves as \(\text{H}_2\) and that every atom on the right must be matched on the left.

Apply the hit-and-trial routine from pages 3-5: box every formula, count atoms, then adjust with coefficients only.

Part (a): Calcium nitrate already holds two nitrate groups, so two \(\text{HNO}_3\) supply them. That gives two extra hydrogens plus the two already on calcium hydroxide = four, needing two waters:

\[ 2\text{HNO}_3 + \text{Ca(OH)}_2 \rightarrow \text{Ca(NO}_3)_2 + 2\text{H}_2\text{O} \]

Part (b): Sodium sulphate fixes two sodiums, so two \(\text{NaOH}\) are needed.

Two waters form from the two OH groups and the two hydrogens of sulphuric acid:

\[ 2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O} \]

Part (c): Check the counts before touching anything — each element already matches on both sides (Na 1 = 1, Cl 1 = 1, Ag 1 = 1, N 1 = 1, O 3 = 3).

This equation is already balanced; the mark is earned by saying so and proving it with the count:

\[ \text{NaCl} + \text{AgNO}_3 \rightarrow \text{AgCl} + \text{NaNO}_3 \]

Part (d): Barium sulphate takes the only barium and the only sulphate, leaving two chlorides and two hydrogens free to pair as \(\text{HCl}\):

\[ \text{BaCl}_2 + \text{H}_2\text{SO}_4 \rightarrow \text{BaSO}_4 + 2\text{HCl} \]
For part (a), the final atom check looks like this — the same table the textbook builds for \(\text{Zn} + \text{H}_2\text{SO}_4\) (NCERT, p. 3):

Common error: skipping the final count.

Part (c) shows why the check matters — an equation can look unfinished when it is actually balanced, and the count is the evidence.

Part (a): Calcium is \(\text{Ca}^{2+}\) and hydroxide is \(\text{OH}^-\), so calcium hydroxide is \(\text{Ca(OH)}_2\); carbonate is \(\text{CO}_3^{2-}\), so calcium carbonate is \(\text{CaCO}_3\). One carbon dioxide and one water complete the balance:

\[ \text{Ca(OH)}_2 + \text{CO}_2 \rightarrow \text{CaCO}_3 + \text{H}_2\text{O} \]
This is the whitewashing finish reaction (NCERT, p. 7) — the slow formation of the shiny calcium carbonate layer on walls.

Part (b): Zinc is more reactive than silver, so it displaces silver from silver nitrate. Each silver nitrate supplies one silver, but zinc nitrate \(\text{Zn(NO}_3)_2\) needs two nitrate groups, so two \(\text{AgNO}_3\) are needed:

\[ \text{Zn} + 2\text{AgNO}_3 \rightarrow \text{Zn(NO}_3)_2 + 2\text{Ag} \]

Part (c): Aluminium forms \(\text{Al}^{3+}\) and copper chloride is \(\text{CuCl}_2\), so aluminium chloride is \(\text{AlCl}_3\).

Two aluminiums give six chlorides, matching three \(\text{CuCl}_2\):

\[ 2\text{Al} + 3\text{CuCl}_2 \rightarrow 2\text{AlCl}_3 + 3\text{Cu} \]

Part (d): This is the ion-exchange pattern of Activity 1.10 (NCERT, p. 11): barium and sulphate form the insoluble \(\text{BaSO}_4\), while potassium and chloride pair as \(\text{KCl}\).

Two potassium chlorides are needed to use both chlorides of barium chloride:

\[ \text{BaCl}_2 + \text{K}_2\text{SO}_4 \rightarrow \text{BaSO}_4 + 2\text{KCl} \]

Common error: forgetting the brackets in \(\text{Ca(OH)}_2\) and \(\text{Zn(NO}_3)_2\).

The subscript outside the bracket multiplies the whole group — write \(2\text{AgNO}_3\), never \(\text{Ag}_2\text{NO}_3\).

Part (a): Potassium is \(\text{K}^+\) and bromide is \(\text{Br}^-\), so potassium bromide is \(\text{KBr}\); barium iodide is \(\text{BaI}_2\) (Ba²⁺ with two I⁻). Barium bromide is written \(\text{BaBr}_2\), and the (s) tells you it precipitates.

Exchange the ions — two \(\text{KBr}\) supply the two bromides \(\text{BaBr}_2\) needs, and \(\text{BaI}_2\) supplies two iodides for two KI:

\[ 2\text{KBr(aq)} + \text{BaI}_2\text{(aq)} \rightarrow 2\text{KI(aq)} + \text{BaBr}_2\text{(s)} \]
Two compounds exchange ions, so this is a double displacement reaction. The (s) makes it also a precipitation reaction, but the question asks for the type, and the ion exchange is the defining feature.

Part (b): One reactant (\(\text{ZnCO}_3\)) splits into two simpler products on heating:

\[ \text{ZnCO}_3\text{(s)} \xrightarrow{\text{Heat}} \text{ZnO(s)} + \text{CO}_2\text{(g)} \]
Type: decomposition (thermal, since heat drives it).

Part (c): Two elements combine into a single product. Both are diatomic gases, and one molecule of each makes two molecules of hydrogen chloride:

\[ \text{H}_2\text{(g)} + \text{Cl}_2\text{(g)} \rightarrow 2\text{HCl(g)} \]
Type: combination.

Part (d): Magnesium displaces hydrogen from the acid because magnesium is more reactive than hydrogen. One magnesium gives one \(\text{MgCl}_2\), needing two chlorides — hence two \(\text{HCl}\) — and the freed hydrogens pair as \(\text{H}_2\):

\[ \text{Mg(s)} + 2\text{HCl(aq)} \rightarrow \text{MgCl}_2\text{(aq)} + \text{H}_2\text{(g)} \]
Type: displacement.

Common error in (a): answering “precipitation” instead of “double displacement”.

Precipitation describes what happens (a solid falls out); the reaction type is the ion exchange.

The dividing question is simple: does the reaction release heat to its surroundings or absorb it? Exothermic reactions release heat along with the products (NCERT, p. 7) — the mixture warms up. Examples: burning of natural gas (\(\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} + \text{heat}\)), respiration, and the decomposition of vegetable matter into compost.

Endothermic reactions absorb energy (NCERT, p. 9-10) — the mixture cools. Examples: thermal decomposition of limestone (\(\text{CaCO}_3 \xrightarrow{\text{Heat}} \text{CaO} + \text{CO}_2\)), electrolysis of water, and the barium hydroxide + ammonium chloride group activity on page 10, where the test tube turns cold in your palm.

Common error: labelling electrolysis exothermic because “energy is involved”.

Energy supplied to break the reactant apart is the sign of an endothermic reaction.

Respiration releases energy, and releasing energy is exactly what makes a reaction exothermic. Follow the food chain: carbohydrates from rice, potatoes and bread are digested down to glucose (\(\text{C}_6\text{H}_{12}\text{O}_6\)); glucose then combines with oxygen in the body’s cells (NCERT, p. 7). The equation on page 7 shows the product list ends in energy:

\[ \text{C}_6\text{H}_{12}\text{O}_6\text{(aq)} + 6\text{O}_2\text{(aq)} \rightarrow 6\text{CO}_2\text{(aq)} + 6\text{H}_2\text{O(l)} + \text{energy} \]
That energy keeps us alive — it powers movement, growth and repair. Because the reaction supplies energy out, not in, it is exothermic, not endothermic.

Common error: quoting the equation without saying what “exothermic” means here — state that energy is a product of the reaction.

The two definitions are mirror images. A combination reaction builds one product from two or more reactants; a decomposition reaction breaks one reactant into two or more simpler products (NCERT, p. 6-8). What combination does forward, decomposition undoes — which is why the textbook calls them opposite. One pair that shows this clearly:

\[ \text{Combination:}\quad \text{CaO} + \text{H}_2\text{O} \rightarrow \text{Ca(OH)}_2 \]
\[ \text{Decomposition:}\quad \text{CaCO}_3 \xrightarrow{\text{Heat}} \text{CaO} + \text{CO}_2 \]

Common error: writing a definition without any equation.

The question explicitly demands equations for both types, and each equation must show the direction: two substances collapsing into one, and one substance splitting into two.

Decomposition needs an energy input, and the source gives the name: heat → thermal decomposition, light → photolytic, electricity → electrolysis (NCERT, p. 8-9). Pick one reaction per energy source and write the source above the arrow — the question scores that detail.

Heat: green ferrous sulphate crystals turn brown, giving ferric oxide, sulphur dioxide and sulphur trioxide:

\[ 2\text{FeSO}_4\text{(s)} \xrightarrow{\text{Heat}} \text{Fe}_2\text{O}_3\text{(s)} + \text{SO}_2\text{(g)} + \text{SO}_3\text{(g)} \]

Heat (second option): lead nitrate gives brown fumes of nitrogen dioxide:

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

Light: white silver chloride turns grey in sunlight, splitting into silver and chlorine — the reaction behind black and white photography:

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

Electricity: electrolysis of water splits it into hydrogen and oxygen (see Figure 1.6 in the intext section below):

\[ 2\text{H}_2\text{O(l)} \xrightarrow{\text{Electricity}} 2\text{H}_2\text{(g)} + \text{O}_2\text{(g)} \]

Common error: forgetting the energy source above the arrow — Q12 explicitly tests the three labels heat, light and electricity, and the source is part of the answer.

Count the actors. In a displacement reaction, one element pushes a less reactive element out of its compound — so the pattern is element + compound → new element + new compound (NCERT, p. 10-11). Zinc displaces copper from copper sulphate:

\[ \text{Zn(s)} + \text{CuSO}_4\text{(aq)} \rightarrow \text{ZnSO}_4\text{(aq)} + \text{Cu(s)} \]
In a double displacement reaction, two compounds exchange ions — the positive ion of one pairs with the negative ion of the other (NCERT, p. 11). Sodium sulphate and barium chloride swap partners:

\[ \text{Na}_2\text{SO}_4\text{(aq)} + \text{BaCl}_2\text{(aq)} \rightarrow \text{BaSO}_4\text{(s)} + 2\text{NaCl(aq)} \]

Common error: answering definitions only.

The question says “write equations” — each contrast must pair its definition with a balanced equation, or marks are lost.

Copper is more reactive than silver, so copper atoms displace silver from its nitrate solution — silver drops out of solution as the metal while copper goes in as copper(II) nitrate (NCERT, p. 10-11 displacement pattern). Each \(\text{Cu}^{2+}\) ion needs two nitrate groups, so two \(\text{AgNO}_3\) are consumed:

\[ \text{Cu(s)} + 2\text{AgNO}_3\text{(aq)} \rightarrow \text{Cu(NO}_3)_2\text{(aq)} + 2\text{Ag(s)} \]
This is also a redox reaction — copper gains oxygen (from nitrate) and silver(I) loses it — but the question names the driving idea: displacement of a less reactive metal by a more reactive one.

Common error: writing \(\text{Cu} + \text{AgNO}_3 \rightarrow \text{CuNO}_3 + \text{Ag}\).

Copper(II) nitrate is \(\text{Cu(NO}_3)_2\), which forces the coefficient 2 on \(\text{AgNO}_3\) and balances the nitrates.

A precipitation reaction is one in which two solutions react to form an insoluble solid, called a precipitate (NCERT, p. 11). The solid does not stay dissolved — it falls out of the mixture, which is why the reaction is visible. The chapter’s own example is sodium sulphate mixed with barium chloride, where the white \(\text{BaSO}_4\) settles out (Activity 1.10):

\[ \text{Na}_2\text{SO}_4\text{(aq)} + \text{BaCl}_2\text{(aq)} \rightarrow \text{BaSO}_4\text{(s)} + 2\text{NaCl(aq)} \]
A second example from Activity 1.2: lead(II) nitrate and potassium iodide give a yellow precipitate of lead(II) iodide:

\[ \text{Pb(NO}_3)_2\text{(aq)} + 2\text{KI(aq)} \rightarrow \text{PbI}_2\text{(s)} + 2\text{KNO}_3\text{(aq)} \]

Common error: calling every double displacement reaction a precipitation reaction.

Precipitation is the outcome (an insoluble salt forms); double displacement is the mechanism.

Both examples above happen to be both — but the label must match what the question asks.

The chapter defines these purely by oxygen (and hydrogen) movement: oxidation is the gain of oxygen or loss of hydrogen; reduction is the loss of oxygen or gain of hydrogen (NCERT, p. 12). The gain/loss always happens to a substance in the reaction — name the substance, not just the process.

Part (a) Oxidation (gain of oxygen): copper gains oxygen to become copper(II) oxide on heating (Activity 1.11):

\[ 2\text{Cu} + \text{O}_2 \xrightarrow{\text{Heat}} 2\text{CuO} \]
\[ 4\text{Na} + \text{O}_2 \rightarrow 2\text{Na}_2\text{O} \]
Part (b) Reduction (loss of oxygen): copper(II) oxide loses oxygen when hydrogen is passed over it, giving back copper and water (reaction 1.29):

\[ \text{CuO} + \text{H}_2 \xrightarrow{\text{Heat}} \text{Cu} + \text{H}_2\text{O} \]
\[ \text{ZnO} + \text{C} \rightarrow \text{Zn} + \text{CO} \]

Common error: saying “copper is reduced” in the first equation.

The precise answer is that copper(II) oxide is reduced — the compound loses oxygen; the free element copper is its product.

Shiny brown and metallic is the giveaway: the element is copper. On heating in air, copper combines with oxygen to form a black coating of copper(II) oxide, CuO (NCERT, p. 12, Activity 1.11):

\[ 2\text{Cu} + \text{O}_2 \xrightarrow{\text{Heat}} 2\text{CuO} \]

Notice the reverse is possible too: passing hydrogen over the hot black oxide turns it brown again as CuO loses oxygen back to copper — reaction 1.29 on page 12. Common error: naming the compound “copper oxide” without saying copper(II) oxide or giving the formula CuO.

Iron rusts because air and moisture reach its surface and oxidise it. Paint is a physical barrier: it seals the iron away from contact with oxygen and water, so the oxidation that causes rusting cannot start (NCERT, p. 13).

The chapter stresses how serious this is — corrosion damages car bodies, bridges, iron railings and ships, and replacing the damaged iron costs enormous sums every year. The same logic explains why you see paint, oil, grease or galvanising on iron objects exposed to weather.

Common error: saying paint “stops oxidation” without naming the barrier idea — full marks needs the mechanism: paint keeps air and moisture away from the iron surface.

Fats and oils react with oxygen from the air and turn rancid — their smell and taste change (NCERT, p. 13). Flushing the packet with nitrogen pushes the oxygen out and fills the space with an unreactive gas, so the oil cannot be oxidised.

This is why chips manufacturers fill bags with nitrogen before sealing; the same idea sits behind airtight containers and added antioxidants, which slow oxidation.

Common error: saying nitrogen “protects” without stating why nitrogen is chosen.

The exam point is that nitrogen is unreactive — it occupies the space oxygen would otherwise fill.

Part (a) Corrosion: when a metal is attacked by substances around it — moisture, air, acids — it corrodes (NCERT, p. 13). Rusting of iron is the classic example: iron reacts with oxygen and moisture to form a reddish-brown layer of rust. The same process appears as the black coating on silver and the green coating on copper.

It is a serious economic problem because it damages car bodies, bridges, railings and ships.

Part (b) Rancidity: the oxidation of fats and oils, which changes their smell and taste (NCERT, p. 13). Food left too long tastes stale because the fat has oxidised. It is prevented by adding antioxidants, keeping food in airtight containers, and flushing packets with nitrogen — the reason chips bags are filled with nitrogen.

Common error: defining both terms with no example.

The question says “with one example each” — a definition alone loses the example mark.

Also, link both back to oxidation: corrosion is oxidation of a metal, rancidity is oxidation of a fat.

Magnesium is reactive enough to combine with oxygen in ordinary air, so a ribbon left on a shelf is already coated with a layer of magnesium oxide. That oxide is stable and does not burn — it sits between the flame and the metal underneath.

Rubbing with sandpaper scrapes the coating off and exposes fresh magnesium, so the metal can actually burn with the dazzling white flame of Activity 1.1 (NCERT, p. 1). Cleaning is not tidiness; it removes the barrier that would stop the reaction you want to observe.

Common error: answering “to make it pure”.

The ribbon is not impure — it is already oxidised on the surface, and sandpaper removes that oxide layer to expose fresh metal.

Part (i): Hydrogen and chlorine are both diatomic gases, so each starts as \(\text{H}_2\) and \(\text{Cl}_2\). They combine in a 1:1 ratio to give two molecules of hydrogen chloride:

\[ \text{H}_2 + \text{Cl}_2 \rightarrow 2\text{HCl} \]

Part (ii): Barium is \(\text{Ba}^{2+}\) and sulphate is \(\text{SO}_4^{2-}\), so barium sulphate is \(\text{BaSO}_4\); aluminium is \(\text{Al}^{3+}\) and chloride is \(\text{Cl}^-\), so aluminium chloride is \(\text{AlCl}_3\) — the valency swap gives these formulae.

Three sulphates on the left need three \(\text{BaSO}_4\), which takes three \(\text{BaCl}_2\) and supplies six chlorides — enough for two \(\text{AlCl}_3\):

\[ 3\text{BaCl}_2 + \text{Al}_2(\text{SO}_4)_3 \rightarrow 3\text{BaSO}_4 + 2\text{AlCl}_3 \]

Part (iii): Sodium and water give sodium hydroxide, but hydrogen leaves as the diatomic \(\text{H}_2\).

Two waters supply the two oxygens for two NaOH and the two hydrogens for one \(\text{H}_2\):

\[ 2\text{Na} + 2\text{H}_2\text{O} \rightarrow 2\text{NaOH} + \text{H}_2 \]

Common error: in part (iii), leaving \(\text{Na} + \text{H}_2\text{O} \rightarrow \text{NaOH} + \text{H}_2\) unbalanced — hydrogen cannot appear as a single atom; it must be \(\text{H}_2\), and that forces the 2:2 coefficients.

Part (i): “Solutions in water” is the signal for (aq). Barium sulphate is called insoluble, so it is (s), the precipitate of Activity 1.10. Sodium chloride stays dissolved as (aq):

\[ \text{BaCl}_2\text{(aq)} + \text{Na}_2\text{SO}_4\text{(aq)} \rightarrow \text{BaSO}_4\text{(s)} + 2\text{NaCl(aq)} \]

Part (ii): Sodium hydroxide solution and hydrochloric acid solution are both (aq); the products are sodium chloride solution, still (aq), and water, (l). This is a neutralisation reaction:

\[ \text{NaOH(aq)} + \text{HCl(aq)} \rightarrow \text{NaCl(aq)} + \text{H}_2\text{O(l)} \]

Common error: leaving state symbols off.

The question explicitly demands them — (aq) on everything dissolved, (s) on the precipitate, (l) on water.

Part (i): X is calcium hydroxide, commonly called slaked lime, with formula \(\text{Ca(OH)}_2\). The “Do You Know?” box on page 7 states that a solution of slaked lime is used for whitewashing walls.

Part (ii): Here is a wording quirk you must not fall for. The question says “reaction of X with water”, but X (slaked lime) is itself the product of that reaction. The accepted answer gives the slaking equation — quick lime (calcium oxide) reacting with water to form the slaked lime solution, releasing heat (reaction 1.13, p. 6):

\[ \text{CaO(s)} + \text{H}_2\text{O(l)} \rightarrow \text{Ca(OH)}_2\text{(aq)} + \text{Heat} \]

Common error: answering part (ii) with \(\text{Ca(OH)}_2 + \text{CO}_2 \rightarrow \text{CaCO}_3 + \text{H}_2\text{O}\).

That is the whitewashing finish reaction (page 7), not the reaction of X with water — it belongs in the FAQ about why fresh whitewash shines.

Work from the formula, not from memory. Electrolysis of water is the decomposition of water by electricity (NCERT, p. 9):

\[ 2\text{H}_2\text{O(l)} \xrightarrow{\text{Electricity}} 2\text{H}_2\text{(g)} + \text{O}_2\text{(g)} \]
Each water molecule has two hydrogen atoms and one oxygen atom. So two molecules of water yield two molecules of \(\text{H}_2\) (four hydrogen atoms) but only one molecule of \(\text{O}_2\) (two oxygen atoms).

At the same pressure, gas volume is proportional to the number of molecules — so twice as many hydrogen molecules means twice the volume.

The doubled gas is hydrogen, collected at the cathode (the negative electrode).

Common error: naming oxygen as the doubled gas by misreading \(2\text{H}_2 + \text{O}_2\).

Hydrogen carries the coefficient 2 — that coefficient is exactly why its volume is double.

Iron is more reactive than copper, so iron displaces copper from copper sulphate solution (NCERT, p. 10-11, Activity 1.9). The blue copper sulphate solution fades because the \(\text{Cu}^{2+}\) ions leave the solution as copper metal, depositing a brownish coat on the nail. What replaces them is iron(II) sulphate, whose solution is pale green:

\[ \text{Fe(s)} + \text{CuSO}_4\text{(aq)} \rightarrow \text{FeSO}_4\text{(aq)} + \text{Cu(s)} \]

Common error: saying the solution “loses its colour” without naming the new colour.

Full answer: blue fades to pale green (iron sulphate) and the nail gains a brownish deposit of copper.

In a double displacement reaction, two compounds exchange their ions — the positive ion of one pairs with the negative ion of the other (NCERT, p. 11). Silver nitrate and sodium chloride are a clean example: silver (\(\text{Ag}^+\)) pairs with chloride (\(\text{Cl}^-\)) to give insoluble silver chloride, while sodium (\(\text{Na}^+\)) pairs with nitrate (\(\text{NO}_3^-\)) to give sodium nitrate:

\[ \text{AgNO}_3\text{(aq)} + \text{NaCl(aq)} \rightarrow \text{AgCl(s)} + \text{NaNO}_3\text{(aq)} \]

Common error: giving Activity 1.10’s own reaction (\(\text{Na}_2\text{SO}_4 + \text{BaCl}_2\)) — the question explicitly rules that one out.

Offer a different ion exchange, like the lead nitrate + potassium iodide example from Activity 1.2.

Apply the oxygen-based rule from page 12: gain of oxygen = oxidised; loss of oxygen = reduced.

Part (i): Sodium gains oxygen (it becomes \(\text{Na}_2\text{O}\)), so sodium is oxidised. The oxygen gas that provides the oxygen is the substance that gets reduced — \(\text{O}_2\) is reduced.

Part (ii): Copper(II) oxide loses oxygen (it becomes copper metal), so \(\text{CuO}\) is reduced. Hydrogen gains oxygen (it becomes water), so \(\text{H}_2\) is oxidised. This is reaction 1.29, the reverse of copper oxidation.

Common error: in part (i), students say “nothing is reduced” because oxygen is a reactant.

But oxidation and reduction always happen together — the substance that supplies the oxygen is the one reduced.

And in part (ii), write “copper(II) oxide is reduced”, not “copper is reduced” — copper metal is the product, not the substance losing oxygen.