NCERT Class 10 Science Chapter 1 Activities — Chemical Reactions and Equations

NCERT Class 10 Science — Chemical Reactions and Equations, Activities 1.1 to 1.11 on pages 1 to 12.

Also here. Solutions to the end-of-chapter questions for this chapter and the in-text questions are on their own pages.

Chapter 1 is built almost entirely out of its activities. Every type of reaction the chapter names is introduced by one — combination by Activity 1.4, decomposition by 1.5 to 1.8, displacement by 1.9, double displacement by 1.10, oxidation by 1.11 — so an activity here is not an optional extra at the side of the page. It is the evidence the next paragraph argues from.

The eleven activities are doing three jobs between them:

• Activities 1.1 to 1.3 establish how you know a reaction has happened at all. The chapter draws the conclusion itself on page 2: a change in state, a change in colour, the evolution of a gas, or a change in temperature.
• Activities 1.4 to 1.11 each demonstrate one type of reaction, in the order the chapter introduces them.
• Activities 1.5, 1.6 and 1.11 additionally show that heating is not a neutral act — it supplies the energy a decomposition needs.

Key insight. The observation is the point, not the procedure. Every one of these activities is engineered so that exactly one of the four indicators is unmistakable — the flame in 1.1, the colour in 1.2, the bubbles and warmth in 1.3, the heat in 1.4. If a student sees nothing, the reaction has not gone wrong so much as the set-up has, and the fix is almost always the same three things: a clean surface, a dry tube, and enough heat.

Activity 1.1

Clean a magnesium ribbon 3–4 cm long with sandpaper, hold it with tongs, burn it in a spirit lamp or burner, and collect the ash in a watch-glass. The book prints a caution with this one: it needs the teacher’s assistance, and eye protection is advisable.

What you should observe. The ribbon burns with a brilliant white flame — the chapter calls it “a dazzling white flame” — and leaves a white powder.

Why. Magnesium has a strong affinity for oxygen and burns in the oxygen of the air. The reaction releases a great deal of energy, much of it as light, which is why the flame is bright enough to be uncomfortable to look at directly — the reason for the caution.

What it sets up. This is the reaction the chapter turns into its first word-equation on page 2, and then into its first balanced equation. Everything about writing chemical equations in this chapter starts here.

What goes wrong. Almost always the sandpaper step is skipped. Magnesium carries a dull grey coating of magnesium oxide, and an uncleaned ribbon either refuses to light or burns feebly, because the coating keeps the metal underneath away from the air. This is Question 1 of the chapter’s own Exercises, and it is worth making students predict the answer before they clean the ribbon.

Activity 1.2

Take lead nitrate solution in a test tube and add potassium iodide solution to it.

What you should observe. A yellow solid appears immediately and settles — a precipitate of lead(II) iodide.

Why. Lead ions and iodide ions meet in solution and form lead iodide, which is insoluble in cold water, so it cannot stay dissolved and drops out as a solid. The potassium and nitrate ions simply stay in solution.

A note on where the answer comes from. The chapter does not state the colour where the activity appears on page 2. It comes back to it on page 12, in a box headed “Recall Activity 1.2”, which asks for the colour of the precipitate, the balanced equation, and whether this is a double displacement reaction — so the book expects the observation to have been recorded at the time. The answers are yellow, the equation above, and yes.

What goes wrong. Nothing much can, which makes this a good demonstration to do first. If the yellow looks weak, the solutions are too dilute.

Activity 1.3

Take a few zinc granules in a conical flask or test tube and add dilute hydrochloric or sulphuric acid. The book cautions to handle the acid with care. Then touch the flask.

What you should observe. Bubbles of gas form steadily on the surface of the granules, and the flask becomes noticeably warm.

Why. Zinc is above hydrogen in the reactivity series, so it displaces hydrogen from the acid. Figure 1.2 in the book labels the gas collected through the glass tube as $\mathrm{H_2}$ directly, so the identification is the chapter’s own, not an inference.

What it sets up. Taken with 1.1 and 1.2, this completes the set of indicators. The chapter draws them together immediately afterwards on page 2: change in state, change in colour, evolution of a gas, change in temperature. Activity 1.3 is the only one of the three that shows two of them at once, which is why it is worth doing last of the three.

What goes wrong. Granules that have sat in a bottle for years are coated and slow to start. Warming the flask very gently gets it going; so does scratching the granules. If the acid is too dilute the bubbling is too slow to see in a lesson.

Activity 1.4

Take a small amount of calcium oxide (quick lime) in a beaker, add water slowly, and touch the beaker.

What you should observe. The mixture hisses and swells, and the beaker becomes distinctly hot — too hot to keep a hand on comfortably.

Why. Two reactants combine to give a single product, which is what makes this the chapter’s example of a combination reaction. The energy released is large because the Ca–O and O–H bonds formed in calcium hydroxide are more stable than what they replace.

What goes wrong. Add the water slowly, as the book says. Poured in all at once, the reaction can spit hot, strongly alkaline slurry out of the beaker. This is the activity on the page that most needs the teacher to do it rather than the class.

Activity 1.5

Take about 2 g of ferrous sulphate crystals in a dry boiling tube, note their colour, heat over a flame, and observe the colour afterwards. Figure 1.4 shows the correct way to hold the tube and to waft the gas gently towards the nose rather than inhaling over the mouth of the tube.

What you should observe. The green crystals lose their colour and a brown solid is left, with the sharp smell of burning sulphur.

Why. Two changes happen in succession, and the chapter is careful about the order. Ferrous sulphate crystals are $\mathrm{FeSO_4}\cdot 7\mathrm{H_2O}$; heating first drives off the water of crystallisation, which is what removes the green colour. Only then does the anhydrous salt decompose to ferric oxide, sulphur dioxide and sulphur trioxide. The odour is the $\mathrm{SO_2}$.

What it sets up. A single reactant breaking down into simpler products is the definition of a decomposition reaction, and this is the chapter’s first example.

What goes wrong. The tube must be dry. Water already in the tube condenses on the cool upper wall and makes it impossible to tell whether the crystals have lost their own water. And the odour should be wafted, never inhaled directly — sulphur dioxide is an irritant.

Activity 1.6

Take about 2 g of lead nitrate powder in a boiling tube, hold it with tongs, and heat it over a flame.

What you should observe. The white powder crackles and dense brown fumes come off.

Why. Another thermal decomposition, and a more dramatic one: one solid gives a solid and two gases. The brown colour is nitrogen dioxide specifically — neither lead oxide nor oxygen is brown.

What goes wrong. The crackling is the powder decrepitating, and it can throw solid out of the tube, which is why the book specifies tongs and a boiling tube rather than a test tube. Do it in a fume cupboard or a well-ventilated room: nitrogen dioxide is toxic. Point the mouth of the tube away from everyone.

Activity 1.7

Set up the electrolysis of water: a plastic mug with two carbon electrodes through rubber stoppers in its base, connected to a 6 V battery; fill with water, add a few drops of dilute sulphuric acid, and invert a water-filled test tube over each electrode. Switch on and leave it. Then test each gas with a burning candle — the book marks this step as one for the teacher.

What you should observe. Bubbles at both electrodes, but not in equal amounts: the tube over the cathode fills roughly twice as fast as the tube over the anode.

Why. The equation is the whole answer to the two-to-one ratio: two molecules of water give two of hydrogen and one of oxygen, so twice the volume of hydrogen is produced in the same time. The chapter treats this as important enough to ask it directly — page 10 asks why the amount of gas in one tube is double that in the other, and to name the gas.

Why the acid. Pure water conducts electricity too poorly for this to work in a lesson. The few drops of sulphuric acid supply ions and let a current flow; the acid is not consumed.

What goes wrong. Two things, both common. If the test tubes are not completely full of water when inverted, the trapped air ruins the volume comparison — which is the one measurement this activity exists to produce. And if the electrodes are anything but carbon, the anode is attacked instead of giving off oxygen, and no oxygen collects at all.

Activity 1.8

Take about 2 g of silver chloride in a china dish, note its colour, leave it in sunlight for some time, and look again.

What you should observe. The white solid darkens to grey.

Why. Light supplies the energy for this decomposition, as heat did in 1.5 and 1.6 and electricity did in 1.7. The grey is finely divided silver metal. The chlorine escapes as a gas.

What it sets up. The chapter uses this and the corresponding silver bromide reaction to make its point about black-and-white photography, and then to define endothermic reactions — reactions that absorb energy. Taken with 1.5 to 1.7, it shows that the energy can arrive as heat, light or electricity.

What goes wrong. This one needs real sunlight and patience; a classroom window on an overcast day will not do it in a single period. Set it up at the start of a lesson and look at it at the end. Silver chloride kept in a clear bottle on a sunny shelf will already be partly grey before you start, which spoils the comparison — take it from a fresh or dark-glass bottle.

Activity 1.9

Clean three iron nails with sandpaper. Put about 10 mL of copper sulphate solution in each of two test tubes, A and B. Tie two nails with thread and lower them into tube B for about 20 minutes, keeping the third nail aside. Then take the nails out and compare — both the colour of the two solutions, and the nails against the one kept aside.

What you should observe. The nails come out coated with a brown deposit, and the blue of the solution in B is paler than in A.

Why. Iron is more reactive than copper, so it displaces copper from copper sulphate solution. The brown coating is copper metal deposited on the nail. The blue fades because the blue colour belongs to copper(II) ions in solution, and they are being removed; what replaces them is iron(II) sulphate, which is a pale green.

What it sets up. This is the chapter’s definition of a displacement reaction: one element removing another from its compound.

What goes wrong. The nails must be cleaned, for the same reason the magnesium ribbon must be — rust is iron oxide, and it stops the iron beneath from meeting the solution. The untouched third nail matters more than it looks: without it there is nothing to compare the coated nails against, and students routinely forget to keep it. Twenty minutes is a genuine minimum; at five minutes the colour change is arguable.

Activity 1.10

Take about 3 mL of sodium sulphate solution in one test tube and about 3 mL of barium chloride solution in another, and mix them.

What you should observe. A white solid forms at once and clouds the mixture.

Why. The chapter explains it in terms of ions: the white precipitate of $\mathrm{BaSO_4}$ is formed by $\mathrm{SO_4^{2-}}$ meeting $\mathrm{Ba^{2+}}$, while the other product, sodium chloride, stays dissolved. Because the two reactants have exchanged partners, this is a double displacement reaction, and because it produces a precipitate it is also a precipitation reaction.

What goes wrong. Very little — which is why it is the chapter’s chosen illustration. If nothing appears, one of the two bottles is not what its label says.

Activity 1.11

Heat a china dish containing about 1 g of copper powder.

What you should observe. The bright reddish-brown powder dulls and its surface turns black.

Why. Oxygen from the air is added to copper. In the chapter’s terms, the copper has been oxidised — it has gained oxygen.

What it sets up. The chapter immediately turns the experiment round: pass hydrogen over the hot black copper oxide and the surface goes brown again as copper is recovered.

$$\mathrm{CuO}(s) + \mathrm{H_2}(g) \xrightarrow{\ \text{heat}\ } \mathrm{Cu}(s) + \mathrm{H_2O}(l)$$

Here the copper oxide loses oxygen and is reduced, while the hydrogen gains it and is oxidised — the two happening together in one reaction, which is what the chapter means by a redox reaction.

What goes wrong. Only the surface reacts, so stirring the powder part-way through makes the change much more convincing. If the dish is not hot enough the black coating is patchy and easily mistaken for soot from the flame — a reason to heat from below with a clean blue flame rather than a smoky one.

Common mistakes

• Treating Activity 1.1’s sandpaper step as tidiness. It is the experiment. The oxide coating is what the cleaning removes, and the chapter’s own Exercise question asks why — students who have cleaned the ribbon without being asked to think about it have nothing to say.
• Recording “a gas is given off” in Activity 1.3 and stopping there. The flask also gets warm. This activity shows two of the four indicators at once, and the temperature change is the one that gets missed because nobody touches the flask.
• Reading Activity 1.5 as a single change. The loss of green and the decomposition to ferric oxide are two separate steps — water of crystallisation first, then the salt itself. A student who says “ferrous sulphate turns brown because it decomposes” has skipped the step the colour actually belongs to.
• Getting Activity 1.7’s ratio the wrong way round. Hydrogen is the doubled one, at the cathode. The equation settles it — $2\mathrm{H_2O}$ gives $2\mathrm{H_2}$ and one $\mathrm{O_2}$ — but under exam pressure the tube with more gas is often labelled oxygen because oxygen feels like the bigger molecule.
• Calling Activity 1.9 a double displacement. One element displaces another; nothing swaps partners. Activity 1.10 is the double displacement, and the two sit close enough together in the chapter to be confused.
• Describing Activity 1.11 as the copper “burning”. It is not a flame reaction. Oxygen is adding to the surface of the solid, which is the whole reason the chapter can use it to define oxidation in terms of gain of oxygen.

Practise next

• The end-of-chapter Exercises for this chapter, where Question 1 asks exactly why the magnesium ribbon of Activity 1.1 must be cleaned, and Questions 5 to 8 ask for the balanced equations behind these activities.
• The in-text questions of this chapter, which include the chapter’s own “Recall Activity 1.2” box asking for the colour of the precipitate and whether it is a double displacement reaction.
• Chapter 3, Metals and Non-metals, where the displacement of Activity 1.9 becomes the reactivity series in full, and the oxidation of Activity 1.11 becomes the corrosion and extraction of metals.