Chapter 1 · Chemical Reactions and Equations

1.3 Oxidation and Reduction in Everyday Life

Oxidation is gain of oxygen (or loss of hydrogen), reduction is the reverse, and the two always happen together: $\mathrm{CuO + H_{2} \rightarrow Cu + H_{2}O}$. Then see oxidation at work in rusting iron and rancid food.

Cut an apple and leave it on the plate: within minutes the white flesh turns brown. A shiny new iron nail left in the garden is orange-brown within weeks. A packet of chips opened and forgotten for a week tastes stale and smells odd. A copper vessel slowly turns dull and then green. In every one of these, a substance is reacting with the oxygen of the air. The same kind of reaction burns the fuel in a car engine and releases the energy from the food in your cells.

Chemists call a reaction in which a substance gains oxygen an oxidation, and the reverse, in which a substance loses oxygen, a reduction. The key idea of this section is that the two are partners: oxygen taken from one substance must go somewhere, so whenever something is reduced, something else is oxidised. You will learn to spot both halves in an equation, name the oxidising and reducing agents, and then use these ideas to understand two expensive everyday problems — corrosion and rancidity — and how we fight them.

Gain and Loss of Oxygen

Heat a china dish containing about $1\ \mathrm{g}$ of reddish-brown copper powder. The surface of the powder turns black: copper has combined with oxygen from the air to form copper(II) oxide: $$\mathrm{2Cu(s) + O_{2}(g) \xrightarrow{\text{heat}} 2CuO(s)}$$ Now pass hydrogen gas over the hot black copper oxide. The black solid turns reddish-brown again, and droplets of water collect on the cooler part of the tube. The oxygen has been taken away from the copper oxide by the hydrogen: $$\mathrm{CuO(s) + H_{2}(g) \xrightarrow{\text{heat}} Cu(s) + H_{2}O(l)}$$ In this second reaction copper oxide loses oxygen while hydrogen gains it.

📘 Definition 1.10 — Oxidation and reduction
At this stage we define them through oxygen and hydrogen:
  • Oxidation is the gain of oxygen or the loss of hydrogen by a substance.
  • Reduction is the loss of oxygen or the gain of hydrogen by a substance.
Oxidised — gains O$\mathrm{H_{2} \rightarrow H_{2}O}$ in $\mathrm{CuO + H_{2} \rightarrow Cu + H_{2}O}$
Reduced — loses O$\mathrm{CuO \rightarrow Cu}$ in the same reaction
Oxidised — loses H$\mathrm{H_{2}S \rightarrow S}$ in $\mathrm{H_{2}S + Cl_{2} \rightarrow 2HCl + S}$
Reduced — gains H$\mathrm{Cl_{2} \rightarrow HCl}$ in the same reaction
📐 Principle 1.3 — Oxidation and reduction go together
Oxygen (or hydrogen) cannot simply disappear or appear: if one reactant loses it, another reactant gains it. So in a reaction in which one substance is oxidised, another substance is reduced at the same time. Such a reaction is called an oxidation–reduction or redox reaction.
📘 Definition 1.11 — Oxidising agent and reducing agent
In a redox reaction, the oxidising agent is the substance that brings about oxidation — it gives oxygen (or takes hydrogen), and is itself reduced. The reducing agent is the substance that brings about reduction — it takes oxygen (or gives hydrogen), and is itself oxidised.
✏️ Example 1 — Hydrogen and hot copper oxide
For $\mathrm{CuO(s) + H_{2}(g) \xrightarrow{\text{heat}} Cu(s) + H_{2}O(l)}$, name (a) the substance oxidised, (b) the substance reduced, (c) the oxidising agent and (d) the reducing agent. (e) What colour change shows that the reaction has happened?
Follow the O The oxygen atom starts in $\mathrm{CuO}$ and ends in $\mathrm{H_{2}O}$.
(a), (b) Hydrogen gains oxygen: $\mathrm{H_{2}}$ is oxidised. Copper(II) oxide loses oxygen: $\mathrm{CuO}$ is reduced.
(c), (d) $\mathrm{CuO}$ gives oxygen to hydrogen, so it is the oxidising agent. $\mathrm{H_{2}}$ takes oxygen from copper oxide, so it is the reducing agent.
(e) The black copper(II) oxide turns reddish-brown (copper metal), and water droplets form.
Interactive 1 — Follow the Oxygen: Who Is Oxidised, Who Is Reduced?

Each ball is an atom. The atoms that are handed from one reactant to the other glow with an amber ring. Watch where they go, predict which reactant is oxidised and which is reduced, then let the animation finish to check. The four answers appear when the atoms have moved.

oxidised?
reduced?
oxidising agent?
reducing agent?

Oxidising and Reducing Agents at Work

Redox reactions are everywhere once you start looking. Carbon (as coke) is the cheap reducing agent that frees zinc and iron from their oxides in industry; a strong oxidising agent such as manganese dioxide is used to make chlorine from hydrochloric acid; and every fuel that burns and every cell that respires is being oxidised.

🧭 Problem-solving strategy — spotting oxidation and reduction
  1. Look for oxygen first. Find a reactant that has oxygen in it and a product that does not (or the other way round). The reactant that loses oxygen is reduced; the reactant that gains it is oxidised.
  2. If oxygen does not move, follow hydrogen. The reactant that loses hydrogen is oxidised; the reactant that gains hydrogen is reduced.
  3. Name the agents by swapping. The substance reduced is the oxidising agent; the substance oxidised is the reducing agent.
  4. Check that both halves exist. If you have found only an oxidation, look again for the matching reduction.
✏️ Example 2 — Extracting zinc with carbon
Zinc is obtained by heating zinc oxide with carbon (coke): $\mathrm{ZnO(s) + C(s) \xrightarrow{\text{heat}} Zn(s) + CO(g)}$. Check that the equation is balanced and identify the substance oxidised, the substance reduced, the oxidising agent and the reducing agent.
Balance Zn $1=1$; O $1=1$; C $1=1$. ✓ Already balanced.
Follow the O Oxygen moves from $\mathrm{ZnO}$ to carbon, forming $\mathrm{CO}$.
Halves $\mathrm{ZnO}$ loses oxygen — reduced. Carbon gains oxygen — oxidised.
Agents Oxidising agent: $\mathrm{ZnO}$. Reducing agent: $\mathrm{C}$. (This is why coke is called a reducing agent in metallurgy — Chapter 3.)
✏️ Example 3 — Making chlorine from hydrochloric acid
When manganese dioxide is warmed with concentrated hydrochloric acid, greenish-yellow chlorine gas is given off: $$\mathrm{MnO_{2}(s) + 4HCl(aq) \xrightarrow{\text{heat}} MnCl_{2}(aq) + 2H_{2}O(l) + Cl_{2}(g)}$$ (a) Verify the balance. (b) Which substance is oxidised and which is reduced? Give a reason for each.
(a) Mn $1=1$; O $2=2$; H $4=4$; Cl $4=2+2$. ✓
Oxygen $\mathrm{MnO_{2}}$ has two oxygen atoms; $\mathrm{MnCl_{2}}$ has none. Manganese dioxide loses oxygen, so $\mathrm{MnO_{2}}$ is reduced (it is the oxidising agent).
Hydrogen Oxygen does not go to $\mathrm{HCl}$ as such, so follow hydrogen: in forming chlorine, $\mathrm{HCl}$ loses hydrogen (the hydrogen ends up in water). So $\mathrm{HCl}$ is oxidised (it is the reducing agent).
Check Both halves are present: oxygen leaves $\mathrm{MnO_{2}}$ and hydrogen leaves $\mathrm{HCl}$, and they meet to form water.
✏️ Example 4 — When only hydrogen moves
When chlorine gas is bubbled into a solution of hydrogen sulphide (the gas that smells of rotten eggs), a pale yellow solid, sulphur, separates: $\mathrm{H_{2}S(aq) + Cl_{2}(g) \rightarrow 2HCl(aq) + S(s)}$. Identify the substance oxidised, the substance reduced, and the two agents.
No oxygen No substance here contains oxygen, so follow hydrogen.
H leaves $\mathrm{H_{2}S}$ loses its hydrogen and becomes $\mathrm{S}$: hydrogen sulphide is oxidised; it is the reducing agent.
H arrives $\mathrm{Cl_{2}}$ gains hydrogen and becomes $\mathrm{HCl}$: chlorine is reduced; it is the oxidising agent.
Check Balance: H $2=2$; S $1=1$; Cl $2=2$. ✓
⚠️ Common mistakes with redox

1. Mixing up the agent and what happens to it. The oxidising agent is reduced; the reducing agent is oxidised. The oxidising agent hands its oxygen to the other substance, so it is left with less oxygen — it has been reduced. The reducing agent takes that oxygen — it has been oxidised.

2. Thinking oxidation needs oxygen gas. Hydrogen sulphide is oxidised by chlorine, with no oxygen in sight, because it loses hydrogen.

3. Naming an element instead of the substance. In $\mathrm{CuO + H_{2} \rightarrow Cu + H_{2}O}$ the substance reduced is copper(II) oxide, $\mathrm{CuO}$, not “copper”.

4. Forgetting that burning is oxidation. Combustion of a fuel such as methane, $\mathrm{CH_{4} + 2O_{2} \rightarrow CO_{2} + 2H_{2}O}$, is an oxidation: carbon gains oxygen. Respiration is the slow oxidation of glucose.

🚀 Beyond the board syllabus (JEE Main · NEET · JEE Advanced) — oxidation as loss of electrons

In JEE Main, NEET and JEE Advanced (as part of Class 11 redox reactions); not in the Class 10 CBSE syllabus.

The oxygen-and-hydrogen definitions are where chemistry started, but chemists now use a more general one: oxidation is loss of electrons, reduction is gain of electrons (memory aid: OIL RIG — Oxidation Is Loss, Reduction Is Gain). When magnesium burns, each magnesium atom loses two electrons, $\mathrm{Mg \rightarrow Mg^{2+} + 2e^{-}}$, and each oxygen atom gains two, $\mathrm{O + 2e^{-} \rightarrow O^{2-}}$, forming the ions of magnesium oxide. With this definition even the displacement of Section 1.2 is a redox reaction: in $\mathrm{Fe + Cu^{2+} \rightarrow Fe^{2+} + Cu}$, iron loses two electrons (oxidised) and the copper ion gains them (reduced). You will study this fully in Class 11 Chemistry, Ch. 7 — Redox Reactions; for the board exam, use the oxygen/hydrogen definitions.

Corrosion

Iron railings, car bodies, ships, bridges and water pipes are all slowly eaten away by the air around them. New iron is grey and shiny; left outdoors it becomes covered in a flaky, reddish-brown powder, rust. Silver jewellery and spoons turn black, and copper vessels and statues lose their shine and become coated with a green layer. All of these are examples of corrosion.

📘 Definition 1.12 — Corrosion and rusting
Corrosion is the slow eating away of a metal when it is attacked by substances around it, such as moisture, oxygen, acids or sulphur compounds in the air. The corrosion of iron is called rusting; rust is hydrated iron(III) oxide, $\mathrm{Fe_{2}O_{3}\cdot xH_{2}O}$, where $x$ (the amount of water) varies.
Iron → rustreddish-brown, flaky; needs both oxygen (air) and water
Silver → black coatingsilver sulphide, from sulphur compounds in the air
Copper → green coatingbasic copper carbonate, from moist air containing carbon dioxide

Rusting is an oxidation: iron gains oxygen. Water is needed too, and is built into the rust. An overall equation is $$\mathrm{4Fe(s) + 3O_{2}(g) + 2xH_{2}O(l) \rightarrow 2Fe_{2}O_{3}\cdot xH_{2}O(s)}$$ Rust does not stick firmly to the iron: it flakes off and exposes fresh metal underneath, so rusting continues until the object is eaten through. This is why rust is so destructive, while aluminium, which forms a thin, firmly attached oxide coat, is protected by its own corrosion.

A classic experiment shows exactly what iron needs in order to rust. Three clean iron nails are placed in three test tubes:

After a few days only the nail in tube A has rusted. Rusting needs both oxygen and water; take either away and it stops. Salt dissolved in the water (sea spray, salt spread on icy roads in cold countries) makes rusting much faster.

✏️ Example 5 — Reading the rusting experiment
(a) Why is the water in tube B boiled, and why is the oil layer added? (b) What is the job of the anhydrous calcium chloride in tube C? (c) A student finds a little rust in tube B after a week. Suggest one reason. (d) Check that the rusting equation above is balanced when $x=3$.
(a) Boiling drives out the air (oxygen) dissolved in water; the oil layer stops air from dissolving back in. So the nail has water but no oxygen.
(b) It is a drying agent: it absorbs water vapour, so the air around the nail is dry.
(c) Some air was left in the water (it was not boiled long enough, or air dissolved before the oil was added), or the oil layer did not cover the surface completely.
(d) With $x=3$: $\mathrm{4Fe + 3O_{2} + 6H_{2}O \rightarrow 2Fe_{2}O_{3}\cdot 3H_{2}O}$. Fe: $4=2\times2$; O: $6+6=12$ and $2\times(3+3)=12$; H: $12=2\times3\times2=12$. ✓ (It balances for any $x$.)
Interactive 2 — Rust Race: What Does Iron Need to Rust?

Five iron nails, five conditions. Press ▶ Play to let the days pass, or drag the slider. Watch which nails grow rust, and how fast. Change tube E to compare ways of protecting iron. The amounts of rust are illustrative, not measurements.

day—
A: air + water—
D: air + salt water—
E: protected nail—

Why it matters, and what we do about it. Corrosion damages car bodies, bridges, iron railings, ships and every structure made of metal, especially iron; replacing and protecting corroded metal costs every country a large share of its income each year. The simplest defences keep air and water away from the iron: painting, oiling or greasing, and coating it with a more resistant metal such as zinc (galvanising). Mixing iron with other metals to make an alloy such as stainless steel also stops rusting. You will study these methods in Section 3.3.

Rancidity

Corrosion is oxidation of metals. Food that contains fats and oils is oxidised too. Butter left out of the fridge, cooking oil kept for months in a clear bottle on a sunny shelf, or chips and namkeen in an opened packet slowly develop a bad smell and an unpleasant taste. The fats and oils have reacted with oxygen from the air.

📘 Definition 1.13 — Rancidity and antioxidants
Rancidity is the change in smell and taste of food containing fats and oils when they are oxidised by the oxygen of the air; the food is then said to be rancid. An antioxidant is a substance added to food to slow down this oxidation.

Heat and light speed up the oxidation, so rancidity can be slowed down by:

✏️ Example 6 — Fresh chips, stale chips
(a) Why do manufacturers fill chip packets with nitrogen rather than air? (b) A packet of chips is opened and left with its top folded loosely. Why do the chips taste stale after a week, while an unopened packet with the same date tastes fresh? (c) Suggest two ways a family can keep a large tin of cooking oil from turning rancid.
(a) Nitrogen is unreactive and contains no oxygen, so the fats and oils in the chips cannot be oxidised; the chips do not turn rancid. (The gas also stops the chips from being crushed.)
(b) Once the packet is opened, air replaces the nitrogen. Oxygen slowly oxidises the oil in the chips — the chips become rancid. The sealed packet still contains only nitrogen.
(c) Any two of: keep the lid tightly closed (airtight); store it in a cool, dark place away from sunlight and the stove; pour small amounts into a smaller bottle for daily use so the large tin is opened less often.

Exercises

I
Oxidation, reduction and the agents. In parts 3–9, name the substance oxidised, the substance reduced, the oxidising agent and the reducing agent.
  • 1 Define oxidation and reduction in terms of (a) oxygen and (b) hydrogen.
  • 2 Copper powder heated in air turns black. Write the equation and name the substance that is oxidised. What is the black substance?
  • 3 $\mathrm{CuO(s) + H_{2}(g) \rightarrow Cu(s) + H_{2}O(l)}$
  • 4 $\mathrm{ZnO(s) + C(s) \rightarrow Zn(s) + CO(g)}$
  • 5 $\mathrm{Fe_{2}O_{3}(s) + 3CO(g) \rightarrow 2Fe(l) + 3CO_{2}(g)}$ (in a blast furnace)
  • 6 $\mathrm{H_{2}S(g) + Cl_{2}(g) \rightarrow 2HCl(g) + S(s)}$
  • 7 $$\mathrm{3MnO_{2}(s) + 4Al(s) \rightarrow 3Mn(l) + 2Al_{2}O_{3}(s)}$$ (first check that it is balanced)
  • 8 $$\mathrm{MnO_{2}(s) + 4HCl(aq) \rightarrow MnCl_{2}(aq) + 2H_{2}O(l) + Cl_{2}(g)}$$
  • 9 $\mathrm{2Mg(s) + CO_{2}(g) \rightarrow 2MgO(s) + C(s)}$ — burning magnesium keeps burning even in carbon dioxide, leaving black specks of carbon.
  • 10 Methane burns: $\mathrm{CH_{4}(g) + 2O_{2}(g) \rightarrow CO_{2}(g) + 2H_{2}O(g)}$. Show that methane is oxidised in two ways, and name the substance reduced.
  • 11 Challenging. When ammonia gas is passed over hot copper(II) oxide, copper, nitrogen and water are formed. Balance $\mathrm{CuO + NH_{3} \rightarrow Cu + N_{2} + H_{2}O}$ and identify the substance oxidised and the substance reduced.
  • 12 Explain, in your own words, why there can never be an oxidation without a reduction in the same reaction.
1 Oxidation: gain of oxygen or loss of hydrogen. Reduction: loss of oxygen or gain of hydrogen.
2 $\mathrm{2Cu(s) + O_{2}(g) \xrightarrow{\text{heat}} 2CuO(s)}$. Copper is oxidised (it gains oxygen); the black substance is copper(II) oxide.
3 Oxidised: $\mathrm{H_{2}}$; reduced: $\mathrm{CuO}$; oxidising agent: $\mathrm{CuO}$; reducing agent: $\mathrm{H_{2}}$.
4 Oxidised: $\mathrm{C}$ (gains O); reduced: $\mathrm{ZnO}$ (loses O); oxidising agent: $\mathrm{ZnO}$; reducing agent: $\mathrm{C}$.
5 Oxidised: $\mathrm{CO}$ (becomes $\mathrm{CO_{2}}$); reduced: $\mathrm{Fe_{2}O_{3}}$; oxidising agent: $\mathrm{Fe_{2}O_{3}}$; reducing agent: $\mathrm{CO}$.
6 Oxidised: $\mathrm{H_{2}S}$ (loses H); reduced: $\mathrm{Cl_{2}}$ (gains H); oxidising agent: $\mathrm{Cl_{2}}$; reducing agent: $\mathrm{H_{2}S}$.
7 Balance: Mn $3=3$; O $6=2\times3$; Al $4=2\times2$. ✓ Oxidised: $\mathrm{Al}$ (gains O); reduced: $\mathrm{MnO_{2}}$; oxidising agent: $\mathrm{MnO_{2}}$; reducing agent: $\mathrm{Al}$.
8 Oxidised: $\mathrm{HCl}$ (loses H); reduced: $\mathrm{MnO_{2}}$ (loses O); oxidising agent: $\mathrm{MnO_{2}}$; reducing agent: $\mathrm{HCl}$.
9 Oxidised: $\mathrm{Mg}$ (gains O); reduced: $\mathrm{CO_{2}}$ (loses all its O, leaving carbon); oxidising agent: $\mathrm{CO_{2}}$; reducing agent: $\mathrm{Mg}$. Balance: Mg $2=2$; C $1=1$; O $2=2$. (This is why a carbon dioxide extinguisher must not be used on a magnesium fire.)
10 Carbon gains oxygen ($\mathrm{CH_{4} \rightarrow CO_{2}}$) and methane loses hydrogen (it goes into water): both mean oxidation. Oxygen gains hydrogen to form water, so $\mathrm{O_{2}}$ is reduced.
11 N: $\mathrm{2NH_{3}}$ for one $\mathrm{N_{2}}$; H: 6, so $\mathrm{3H_{2}O}$; O: 3, so $\mathrm{3CuO}$; Cu: $\mathrm{3Cu}$. $$\mathrm{3CuO(s) + 2NH_{3}(g) \xrightarrow{\text{heat}} 3Cu(s) + N_{2}(g) + 3H_{2}O(g)}$$ $\mathrm{NH_{3}}$ loses hydrogen — oxidised; $\mathrm{CuO}$ loses oxygen — reduced.
12 Atoms are conserved: oxygen (or hydrogen) removed from one substance does not vanish, it must end up in another substance. The substance that loses oxygen is reduced and the one that gains it is oxidised, so both happen together.
II
Corrosion.
  • 1 What is corrosion? Name the substance formed, and its colour, when (a) iron, (b) silver and (c) copper corrode.
  • 2 Name the two substances that must both be present for iron to rust. Describe briefly an experiment that proves it.
  • 3 Show by counting atoms that $\mathrm{4Fe + 3O_{2} + 2xH_{2}O \rightarrow 2Fe_{2}O_{3}\cdot xH_{2}O}$ is balanced for $x=1$ and for $x=2$.
  • 4 Why does an iron gate in a coastal city such as Chennai or Mumbai rust much faster than one in dry Jaipur?
  • 5 Why are iron railings painted, and why must even a small scratch in the paint be repainted quickly?
  • 6 Is rusting an oxidation or a reduction? Explain using the definition.
  • 7 A clean iron nail is weighed, left in moist air until it is covered with rust, and weighed again with all of the rust still on it. Is the new mass greater or smaller? Explain using the law of conservation of mass.
  • 8 Applied. Taking atomic masses $\mathrm{Fe}=56\ \mathrm{u}$ and $\mathrm{O}=16\ \mathrm{u}$ and ignoring the water in rust, $2$ atoms of iron ($112\ \mathrm{u}$) become one unit of $\mathrm{Fe_{2}O_{3}}$. What mass of $\mathrm{Fe_{2}O_{3}}$ forms when $1.12\ \mathrm{g}$ of iron rusts completely, and what mass of oxygen is taken from the air?
1 The slow eating away of a metal by substances around it (moisture, oxygen, acids, sulphur compounds). (a) Rust, hydrated iron(III) oxide — reddish-brown; (b) silver sulphide — black; (c) basic copper carbonate — green.
2 Oxygen (air) and water. Three nails: one in tap water open to air (rusts); one in boiled water under a layer of oil (no air — does not rust); one in dry air with anhydrous calcium chloride (no water — does not rust).
3 $x=1$: Fe $4=4$; O $6+2=8$ and $2\times(3+1)=8$; H $4=4$. $x=2$: Fe $4=4$; O $6+4=10$ and $2\times(3+2)=10$; H $8=8$. ✓
4 Coastal air is more humid (more water), and it carries salt from sea spray; dissolved salt speeds up rusting a great deal.
5 Paint keeps air and water away from the iron. At a scratch the iron is exposed again and rust starts there and spreads under the paint, so it must be repainted.
6 Oxidation: iron gains oxygen to form iron(III) oxide.
7 Greater. Oxygen (and water) from the air have combined with the iron. Mass is conserved for the whole system — the gain of the nail equals the mass of oxygen and water taken from the air.
8 Mass of $\mathrm{Fe_{2}O_{3}}$ $=2\times56+3\times16=160\ \mathrm{u}$ for every $112\ \mathrm{u}$ of iron. So $1.12\ \mathrm{g}$ of iron gives $1.12\times\tfrac{160}{112}=1.60\ \mathrm{g}$ of $\mathrm{Fe_{2}O_{3}}$, taking $1.60-1.12=0.48\ \mathrm{g}$ of oxygen from the air.
III
Rancidity.
  • 1 What is rancidity? Which kind of reaction causes it?
  • 2 Why are packets of chips flushed with nitrogen gas before they are sealed?
  • 3 Give two ways of slowing down rancidity at home.
  • 4 What is an antioxidant? Name one natural antioxidant.
  • 5 Why is it better to store cooking oil in a dark or opaque bottle, away from the window?
  • 6 Butter kept in the refrigerator stays fresh for weeks, but on a kitchen shelf in summer it tastes “off” in a few days. Explain.
1 The change in smell and taste of food containing fats and oils when they are oxidised by air. It is caused by oxidation.
2 Nitrogen is unreactive and keeps oxygen out, so the oils in the chips are not oxidised and do not turn rancid.
3 Any two: store in airtight containers; keep in the refrigerator; keep away from light; use food with added antioxidants.
4 A substance that slows down oxidation, added to food containing fats and oils. Vitamin E is a natural antioxidant.
5 Light (and the heat of sunlight) speeds up the oxidation of oils; a dark bottle in a cool place slows it down.
6 Oxidation of the fat in butter is much slower at low temperature, so it turns rancid slowly in the refrigerator but quickly in a warm kitchen.
IV
Mixed problems.
  • 1 Say whether each change is an oxidation or a reduction: (a) $\mathrm{Cu \rightarrow CuO}$; (b) $\mathrm{CuO \rightarrow Cu}$; (c) $\mathrm{Cl_{2} \rightarrow HCl}$; (d) $\mathrm{H_{2}S \rightarrow S}$.
  • 2 In a blast furnace, iron oxide is turned into iron by carbon monoxide. What is the role of carbon monoxide? Write the equation.
  • 3 Respiration is described as the “oxidation of glucose”. Use the equation for respiration to explain why.
  • 4 A student writes: “In a redox reaction, the oxidising agent is the substance that gets oxidised.” Correct the statement.
  • 5 Applied. Old copper vessels with a green coating are traditionally cleaned with tamarind or lemon juice. What is the green coating, and why does an acidic juice remove it?
  • 6 Challenging. Balance $\mathrm{Fe_{3}O_{4} + H_{2} \rightarrow Fe + H_{2}O}$ (the reverse of the iron–steam reaction of Section 1.1) and name the substance oxidised and the substance reduced.
1 (a) oxidation (gain of O); (b) reduction (loss of O); (c) reduction (gain of H); (d) oxidation (loss of H).
2 It is the reducing agent: it takes oxygen from iron oxide. $\mathrm{Fe_{2}O_{3}(s) + 3CO(g) \rightarrow 2Fe(l) + 3CO_{2}(g)}$.
3 $$\mathrm{C_{6}H_{12}O_{6}(aq) + 6O_{2}(aq) \rightarrow 6CO_{2}(aq) + 6H_{2}O(l)}$$ The carbon of glucose gains oxygen (forming carbon dioxide) and glucose loses its hydrogen (to water), so glucose is oxidised.
4 “The oxidising agent is the substance that gets reduced (it causes the other substance to be oxidised).”
5 It is basic copper carbonate, formed by corrosion of copper in moist air. Being basic (a carbonate and a hydroxide), it reacts with and dissolves in acids, such as the tartaric acid in tamarind and the citric acid in lemon (Chapter 2), exposing the shiny copper again.
6 O: 4, so $\mathrm{4H_{2}O}$ and $\mathrm{4H_{2}}$; Fe: $\mathrm{3Fe}$. $\mathrm{Fe_{3}O_{4}(s) + 4H_{2}(g) \rightarrow 3Fe(s) + 4H_{2}O(g)}$. $\mathrm{H_{2}}$ is oxidised; $\mathrm{Fe_{3}O_{4}}$ is reduced.
🎯 Exam corner BoardBoard · A–RBoard · Case-basedJEE Main / NEET

E1. Board · 1 mark In the reaction $\mathrm{CuO + H_{2} \xrightarrow{\text{heat}} Cu + H_{2}O}$, the substance oxidised is (A) $\mathrm{CuO}$   (B) $\mathrm{H_{2}}$   (C) $\mathrm{Cu}$   (D) $\mathrm{H_{2}O}$

Idea Oxidised = gains oxygen. Hydrogen gains the oxygen lost by copper oxide. (C) and (D) are products, not substances that react.
Answer (B).

E2. Board · A–R Assertion (A): Packets of potato chips are flushed with nitrogen before sealing. Reason (R): Nitrogen is an unreactive gas that keeps oxygen away from the oils in the chips and so prevents rancidity. Choose: (A) Both A and R are true and R is the correct explanation of A. (B) Both A and R are true but R is not the correct explanation of A. (C) A is true but R is false. (D) A is false but R is true.

Idea Both are true, and R gives exactly the reason for A: no oxygen, no oxidation of fats.
Answer (A).

E3. Board · 3 marks Three test tubes each contain a clean iron nail: P — in tap water, open to the air; Q — in boiled distilled water covered with a layer of oil; R — in dry air with anhydrous calcium chloride, corked. (a) In which tube does the nail rust? (b) Give the purpose of the oil in Q and of the calcium chloride in R. (c) What do you conclude?

(a) Only in P, where both air and water are present.
(b) The oil stops air from dissolving in the boiled water (no oxygen in Q); calcium chloride absorbs moisture (no water in R).
Answer (c) Iron rusts only when both oxygen (air) and water are present.

E4. Board · Case-based A town by the sea has a steel footbridge over a railway line. The engineers report that the bridge needs repainting every three years, while an identical bridge in an inland desert town lasts more than ten years between coats. Where the paint has chipped, flaky reddish-brown patches have appeared and are spreading under the paint. (i) Name the process and the reddish-brown substance, and write a balanced equation for its formation. (2 marks) (ii) Give two reasons why the seaside bridge suffers more. (1 mark) (iii) Why does the damage spread under the paint from a chip? (1 mark)

(i) Rusting (corrosion of iron); rust is hydrated iron(III) oxide, $\mathrm{Fe_{2}O_{3}\cdot xH_{2}O}$: $$\mathrm{4Fe(s) + 3O_{2}(g) + 2xH_{2}O(l) \rightarrow 2Fe_{2}O_{3}\cdot xH_{2}O(s)}$$
(ii) Sea air is more humid (more water), and salt from sea spray speeds up rusting.
Answer (iii) At the chip, iron is exposed to air and water; rust is flaky and does not protect the metal, so it keeps forming and creeps under the edges of the paint.

E5. JEE Main / NEET Foundation. In $\mathrm{Cr_{2}O_{3} + 2Al \rightarrow Al_{2}O_{3} + 2Cr}$, the substance reduced and the reducing agent are, respectively, (A) $\mathrm{Cr_{2}O_{3}}$, $\mathrm{Al}$   (B) $\mathrm{Al}$, $\mathrm{Cr_{2}O_{3}}$   (C) $\mathrm{Al}$, $\mathrm{Al}$   (D) $\mathrm{Cr_{2}O_{3}}$, $\mathrm{Cr_{2}O_{3}}$

Idea Chromium(III) oxide loses oxygen — it is reduced. Aluminium takes that oxygen — it is oxidised, so it is the reducing agent. The substance reduced and the reducing agent are never the same substance.
Answer (A).

E6. Board · 2 marks Volcanic gases contain both hydrogen sulphide and sulphur dioxide, which react to deposit yellow sulphur around volcanic vents: $\mathrm{2H_{2}S + SO_{2} \rightarrow 3S + 2H_{2}O}$. Identify the substance oxidised and the substance reduced, giving a reason for each.

Idea Follow hydrogen and oxygen separately: H leaves $\mathrm{H_{2}S}$; O leaves $\mathrm{SO_{2}}$; they meet in water.
Answer $\mathrm{H_{2}S}$ is oxidised (it loses hydrogen); $\mathrm{SO_{2}}$ is reduced (it loses oxygen).
🔬 Research–Teaching NexusResearch-tutored · optional, about a week of short observations

Research question. Rusting uses up oxygen. Can you measure what fraction of the air is oxygen using nothing more than a pad of steel wool and a glass of water?

Why it matters. Corrosion is one of the most expensive chemical reactions on Earth: a widely quoted international study (NACE, 2016) estimated its global cost at about US$2.5 trillion a year, roughly 3% of the world's economic output — bridges, pipelines, ships and power stations all suffer. Understanding exactly what rusting consumes is the first step in preventing it. The same idea lets you measure the composition of the air without any instruments.

What to do.

  1. Wet a small pad of steel wool (iron) with water — or rinse it in vinegar first to remove its protective coating — and push it to the closed end of a narrow glass tube or measuring cylinder.
  2. Stand the tube upside down in a dish of water and mark the starting water level. Measure the length (or volume) of the trapped air column.
  3. Record the water level every day for a week. Note when the steel wool turns brown.
  4. When the level stops changing, calculate the fraction of the air that was used up: (rise in water level) ÷ (initial length of air column). Express it as a percentage.
  5. Set up a control with dry steel wool and a drying agent, and compare. Explain the difference using the rusting equation.

Tools: Steel wool, a test tube or measuring cylinder, a dish of water, a ruler or marker; optionally vinegar.

Hand in: A table of daily readings, a simple graph of water rise against time, your percentage of oxygen, and two sources of error.

Going further. Repeat with salt water on the steel wool. Does the water rise faster? Does it rise further? Use your answer to explain why ships and coastal bridges need extra protection.

Instructor note: the water rises by about one-fifth of the air column (dry air is about 21% oxygen by volume), e.g. a $50\ \mathrm{mL}$ air column shrinks by about $10\ \mathrm{mL}$, giving $10\div50=20\%$. The dry control shows little or no rise, because rusting needs water. Salt water speeds up the rise but the final rise is the same, since the amount of oxygen is fixed. Common errors: not enough steel wool, temperature changes, reading the level without equalising the water levels inside and outside.

🧪 Self-check quiz

0/5 correct

Q1. In terms of oxygen and hydrogen, oxidation is

Q2. In $\mathrm{ZnO + C \rightarrow Zn + CO}$, the reducing agent is

Q3. For an iron nail to rust, it must be exposed to

Q4. Which of these does not help to prevent rancidity?

Q5. In $\mathrm{H_{2}S + Cl_{2} \rightarrow 2HCl + S}$, the substance reduced is

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