Redox Reactions | Teaching Notes | Chemistry | Form 2 | CBC

Topic: Redox Reactions

Subtopic: Oxidation and Reduction

INTRODUCTION TO REDOX REACTIONS In everyday life, we see many chemical changes around us. When an iron nail rusts, when charcoal burns to provide heat, or when we use a dry cell torch battery, a special type of chemical reaction is taking place. These reactions are called redox reactions. The word redox comes from two words: reduction and oxidation. In this topic, we will explore what these terms mean, how to identify them using different chemical models, and how to test for them in our school laboratory. ANALYSING OXIDATION AND REDUCTION IN TERMS OF: (A) OXYGEN/HYDROGEN EXCHANGE, (B) ELECTRON TRANSFER, (C) OXIDATION STATE CHANGES To understand redox reactions, scientists use three different definitions or models. Each model is useful depending on the chemical reaction we are analysing. (A) OXYGEN AND HYDROGEN EXCHANGE MODEL This is the simplest and oldest definition of oxidation and reduction. It is based on whether a substance gains or loses oxygen or hydrogen during a chemical reaction. Oxidation is defined as: 1. The gain of oxygen by a substance. 2. The loss of hydrogen from a substance. Reduction is defined as: 1. The loss of oxygen from a substance. 2. The gain of hydrogen by a substance. Let us look at two clear examples to understand this exchange. Example 1: Gain and Loss of Oxygen When black copper(II) oxide is heated with hydrogen gas, pinkish-brown copper metal and water are formed. Word equation: Copper(II) oxide + Hydrogen → Copper + Water Chemical equation: CuO(s) + H₂(g) → Cu(s) + H₂O(l) In this reaction: • CuO loses oxygen to become Cu. Since it has lost oxygen, copper(II) oxide has undergone reduction. • H₂ gains oxygen to become H₂O. Since it has gained oxygen, hydrogen gas has undergone oxidation. Example 2: Gain and Loss of Hydrogen When hydrogen sulphide gas reacts with chlorine gas, yellow sulphur and hydrogen chloride gas are formed. Word equation: Hydrogen sulphide + Chlorine → Sulphur + Hydrogen chloride Chemical equation: H₂S(g) + Cl₂(g) → S(s) + 2HCl(g) In this reaction: • H₂S loses hydrogen to become S. The loss of hydrogen means hydrogen sulphide has undergone oxidation. • Cl₂ gains hydrogen to become HCl. The gain of hydrogen means chlorine gas has undergone reduction.
(B) ELECTRON TRANSFER MODEL While the oxygen-hydrogen model is easy to see, many redox reactions do not involve oxygen or hydrogen at all. To solve this, scientists developed a broader model based on the movement of electrons. Oxidation is defined as the loss of electrons by an atom, molecule, or ion. Reduction is defined as the gain of electrons by an atom, molecule, or ion. An easy way to remember this model is by using the word: OIL RIG • OIL: Oxidation Is Loss of electrons. • RIG: Reduction Is Gain of electrons. Let us look at a reaction between a metal and a non-metal to see this model in action. Example: Reaction between Sodium and Chlorine When sodium metal reacts with chlorine gas, table salt (sodium chloride) is formed. Chemical equation: 2Na(s) + Cl₂(g) → 2NaCl(s) Sodium chloride is an ionic compound containing Na⁺ ions and Cl^- ions. We can write separate "half-equations" to show what happens to the electrons of each element: 1. Sodium half-equation (Oxidation): Each sodium atom loses one electron to form a stable sodium ion. Na → Na⁺ + e^- Since sodium has lost an electron, it has undergone oxidation. 2. Chlorine half-equation (Reduction): Each chlorine molecule gains two electrons to form two chloride ions. Cl₂ + 2e^- → 2Cl^- Since chlorine has gained electrons, it has undergone reduction.
(C) OXIDATION STATE CHANGES MODEL The most advanced and reliable way to define redox reactions is by using oxidation states (sometimes called oxidation numbers). An oxidation state is a positive or negative number assigned to an element in a chemical species. It represents the charge an atom would have if all bonds were completely ionic. To use this model, we must know the definitions: Oxidation is an increase in the oxidation state of an element. Reduction is a decrease in the oxidation state of an element.

Ethylene reacts with HBr to give 1-bromoethane. oxidation state of one carbon goes from -2 to -3% 3B other goes from -2 to -1. this is not a redox reaction

Image: Unknown author Unknown author · CC BY · Wikimedia Commons

Simple Rules for Assigning Oxidation States: 1. The oxidation state of any uncombined element (e.g., Mg, O₂, H₂, Fe, Na) is always 0. 2. The oxidation state of a simple, single-atom ion is equal to its ionic charge (e.g., Na⁺ is +1, Mg²⁺ is +2, Cl^- is -1, O^2- is -2). 3. The oxidation state of oxygen in almost all its compounds is -2 (except in peroxides where it is -1). 4. The oxidation state of hydrogen in almost all its compounds is +1 (except in metal hydrides where it is -1). 5. The sum of all oxidation states in a neutral compound is always equal to 0. 6. The sum of all oxidation states in a polyatomic ion is equal to the charge on that ion. Let us practice calculating an oxidation state. Worked Example: Find the oxidation state of sulphur (S) in sulphuric acid, H₂SO₄. Step 1: Identify known oxidation states: Hydrogen (H) = +1 Oxygen (O) = -2 Let the oxidation state of Sulphur be x. Step 2: Set up the algebraic equation based on Rule 5 (the sum equals 0): 2(Oxidation state of H) + (Oxidation state of S) + 4(Oxidation state of O) = 0 2(+1) + x + 4(-2) = 0 Step 3: Solve for x: +2 + x - 8 = 0 x - 6 = 0 x = +6 Therefore, the oxidation state of sulphur in H₂SO₄ is +6. Example of Redox using Oxidation States: Consider the reaction of carbon with oxygen to form carbon dioxide: C + O₂ → CO₂ • The oxidation state of pure C on the reactant side is 0. • The oxidation state of pure O₂ on the reactant side is 0. • In the product CO₂, oxygen has an oxidation state of -2. • To find carbon in CO₂: x + 2(-2) = 0 → x = +4. • Carbon changed from 0 to +4 (increase in oxidation state = oxidation). • Oxygen changed from 0 to -2 (decrease in oxidation state = reduction).
EXAMINING A REDOX REACTION (REACTION INVOLVING BOTH OXIDATION AND REDUCTION) A very important rule in chemistry is that oxidation and reduction must always occur at the same time. They are like two sides of a coin. If one chemical species loses electrons or oxygen, another chemical species must be right there to gain those electrons or oxygen. You cannot have oxidation without reduction, and you cannot have reduction without oxidation. Let us examine a complete, classic redox reaction in detail: the reaction between heated copper(II) oxide and hydrogen gas.

Cuprous oxide 2

Image: Chemicalinterest · Public Domain · Wikimedia Commons

When we pass dry hydrogen gas over heated black copper(II) oxide powder inside a glass tube, we observe a distinct colour change. The black powder slowly turns into a shiny pinkish-brown solid, and clear, colourless droplets of liquid collect at the cool end of the tube. Let us examine the chemical equation for this reaction: CuO(s) + H₂(g) → Cu(s) + H₂O(l) By examining this reaction using our three different redox definitions, we can prove it is a redox reaction: 1. In terms of Oxygen: * CuO loses oxygen to become copper (Cu) metal. Loss of oxygen is reduction. * H₂ gains oxygen to become water (H₂O). Gain of oxygen is oxidation. * Since both reduction and oxidation are taking place in the same system, this is a redox reaction. 2. In terms of Oxidation States: * Let us assign oxidation states to all atoms in the equation: * In CuO: Cu is +2, O is -2. * In H₂: H is 0 (elemental form). * In Cu: Cu is 0 (elemental form). * In H₂O: H is +1, O is -2. * Now, let us trace the changes: * Copper changes from +2 to 0. This is a decrease in oxidation state, which is reduction. * Hydrogen changes from 0 to +1. This is an increase in oxidation state, which is oxidation. * Oxygen remains at -2 on both sides; its oxidation state has not changed (it is not oxidised or reduced). Both models lead us to the exact same scientific conclusion: the copper is reduced and the hydrogen is oxidised.
IDENTIFYING THE CHARACTERISTICS OF OXIDISING AND REDUCING AGENTS In every redox reaction, there are two key players: the oxidising agent and the reducing agent. These act as chemical "helpers" that make the reaction happen. • Oxidising Agent: This is a substance that causes oxidation in another substance. In the process of doing this, the oxidising agent itself gets reduced. • Reducing Agent: This is a substance that causes reduction in another substance. In the process of doing this, the reducing agent itself gets oxidised. To understand this, think of a travel agent. A travel agent does not go on a trip; they help you go on a trip. Similarly, an oxidising agent does not get oxidised; it helps another substance get oxidised by taking its electrons or giving it oxygen. We can compare the characteristics of both agents using this clear table:

Characteristic / Property	Oxidising Agent	Reducing Agent
Oxygen Transfer	Gives or supplies oxygen to another substance.	Removes or takes oxygen from another substance.
Hydrogen Transfer	Removes hydrogen from another substance.	Gives hydrogen to another substance.
Electron Transfer	Gains (accepts) electrons from another substance.	Loses (donates) electrons to another substance.
Oxidation State Change	Its oxidation state decreases during the reaction.	Its oxidation state increases during the reaction.
What happens to it?	It undergoes reduction.	It undergoes oxidation.

Common Examples in the Laboratory: • Oxidising Agents: Oxygen (O₂), Chlorine (Cl₂), Acidified Potassium Manganate(VII) (KMnO₄), Acidified Potassium Dichromate(VI) (K₂Cr₂O₇), Concentrated Nitric Acid (HNO₃). • Reducing Agents: Hydrogen (H₂), Carbon (C), Carbon Monoxide (CO), Reactive metals (like Zinc, Magnesium), Potassium Iodide (KI) solution.
OXIDISING AGENTS (IDENTIFIED USING POTASSIUM IODIDE SOLUTION AS A REDUCING AGENT IN THE PRESENCE OF STARCH OR ACIDIFIED POTASSIUM IODIDE PAPER) In our laboratory, we often need to test if an unknown chemical solution or gas is an oxidising agent. We do this by reacting the unknown substance with a known, reliable reducing agent. The standard laboratory reagent used for this test is Potassium Iodide (KI). The Chemistry behind the Test: Potassium iodide contains colourless iodide ions (I^-). • If we add a suspected oxidising agent to potassium iodide, the oxidising agent will remove electrons from the iodide ions (oxidising them). • When iodide ions lose electrons, they are turned into molecular iodine (I₂). 2I^-(aq) → I₂(aq) + 2e^- • While iodide ions are colourless, molecular iodine in water has a distinct brown colour. • To make this test extremely sensitive, we add a few drops of starch solution. Starch reacts with even tiny traces of iodine to form a highly visible, intense blue-black colour.

Miniature apparatus for the determination of manganese in pyrolusite or of iodine in iodides (Alessandri 1895.43)

Image: valeg96 · Public Domain · Wikimedia Commons

How to Perform the Tests in the School Laboratory: Method 1: Testing a Liquid/Solution 1. Pour about 2 cm³ of colourless potassium iodide (KI) solution into a clean test tube. 2. Add a few drops of the suspected liquid oxidising agent (such as chlorine water or hydrogen peroxide). 3. Observe the colour change: the solution will turn from colourless to brown. 4. To confirm, add 2 drops of starch solution. The mixture will turn blue-black. Method 2: Testing a Gas using Acidified Potassium Iodide Paper 1. Take a strip of white filter paper and moisten it with acidified potassium iodide solution. 2. Hold the damp paper inside the gas jar containing the suspected oxidising gas (such as chlorine gas). 3. The damp white paper will quickly turn blue-black (or dark brown-speckled). This is because the oxidising gas oxidises the iodide on the paper to iodine, which immediately reacts with the starch in the filter paper.
COMMON MISTAKES MADE BY PUPILS 1. The "Agent" Confusions: * Mistake: Writing that an oxidising agent gets oxidised. Correction: Remember that an oxidising agent oxidises something else and in turn gets reduced. Similarly, a reducing agent reduces something else* and gets oxidised. 2. Confusing OIL RIG: * Mistake: Thinking "Oxidation Is Gain of electrons". * Correction: "OIL" means Oxidation Is Loss of electrons. "RIG" means Reduction Is Gain of electrons. 3. Incorrect Sign for Oxidation States: * Mistake: Writing an oxidation state as "2+" or "2-". * Correction: Ionic charges are written as "2+" or "2-". Oxidation states must always be written with the charge sign before the number (e.g., +2 or -2). 4. Assuming Oxidation Can Happen Alone: * Mistake: Writing a chemical equation where electrons are lost but nowhere are they shown to be gained. * Correction: In any balanced chemical system, the total electrons lost must equal the total electrons gained.
KEY TERMS

Key Term	Definition for Pupil Notebooks
Redox	A chemical reaction in which both reduction and oxidation occur at the same time.
Oxidation	The gain of oxygen, loss of hydrogen, loss of electrons, or increase in oxidation state.
Reduction	The loss of oxygen, gain of hydrogen, gain of electrons, or decrease in oxidation state.
Oxidising Agent	A substance that provides oxygen, accepts electrons, or decreases its own oxidation state.
Reducing Agent	A substance that removes oxygen, donates electrons, or increases its own oxidation state.
Oxidation State	A number representing the charge an atom would carry in a compound based on set chemical rules.

SUMMARY • Redox is a term that combines reduction and oxidation. • The three models of redox are: Oxygen/Hydrogen exchange, Electron Transfer (OIL RIG), and Oxidation State changes. • An increase in oxidation state is oxidation; a decrease in oxidation state is reduction. • An oxidising agent causes oxidation in another substance and is itself reduced. A reducing agent causes reduction and is itself oxidised. • To test for an oxidising agent, we use colourless potassium iodide solution. The iodide ions are oxidised to brown iodine, which turns blue-black when starch solution is added.
REVISION QUESTIONS 1. Define oxidation and reduction in terms of oxygen exchange. 2. State the meaning of the acronym "OIL RIG" as used in electrochemistry. 3. Calculate the oxidation state of the underlined element in each of the following: a) SO₂ b) NO₃^- c) H₂CO₃ 4. Explain why a chemical reaction cannot consist of oxidation alone. 5. State two properties of a good reducing agent. 6. Describe what you would observe when chlorine gas is bubbled through a solution of potassium iodide, and starch is added. 7. Write a balanced ionic half-equation showing the oxidation of iodide ions to molecular iodine. 8. Refer to the diagram below showing a redox experiment to answer the questions that follow:

Using the lettered labels from the diagram above: • Identify the gas entering at label A. • Name the reactant solid at label B. • Identify the heating source shown by label C. • Name the liquid product condensing at label D. • Write a balanced chemical equation for the reaction taking place inside the tube.
PRACTICE EXERCISE SECTION A: Multiple Choice Questions 1. Which of the following processes represents oxidation? A. Gain of hydrogen B. Loss of oxygen C. Loss of electrons D. Decrease in oxidation state 2. In the reaction: Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g), which substance is the reducing agent? A. Fe₂O₃ B. CO C. Fe D. CO₂ 3. What is the oxidation state of sulphur in an ion of sulphate, SO₄^2-? A. +2 B. +4 C. +6 D. -2 4. Damp acidified potassium iodide starch paper is used to test for the presence of: A. Carbon dioxide B. Reducing agents C. Oxidising agents D. Water vapour SECTION B: Structured Questions 1. When zinc metal is placed into a blue solution of copper(II) sulphate, the blue colour fades, and a reddish-brown solid deposit of copper metal is formed on the zinc. The ionic equation is: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s) a) Identify which species has been oxidised. Explain your answer in terms of electron transfer. b) Identify which species has been reduced. Explain your answer in terms of oxidation state changes. c) Write down the half-equation showing the reduction process. 2. A pupil in a Zambian secondary school laboratory wanted to identify an unknown gas in a gas jar. She brought a strip of damp acidified potassium iodide paper near the mouth of the jar, and the paper turned dark blue-black. a) What group of chemical agents does this gas belong to? b) Explain the chemical reactions that caused the paper to turn blue-black. c) Give one common laboratory example of a gas that would show this positive result.
ANSWERS TO PRACTICE EXERCISE SECTION A 1. C — Oxidation is the loss of electrons (OIL). 2. B — Carbon monoxide (CO) is the reducing agent because it removes oxygen from iron(III) oxide and gets oxidised to carbon dioxide. 3. C — Let x be the state of S. Oxygen is -2. Sum of states must equal the charge of the ion: x + 4(-2) = -2 x - 8 = -2 x = +6. 4. C — It is a standard test for oxidising agents (like chlorine gas). SECTION B Question 1: a) Zinc (Zn) has been oxidised. It lost two electrons to form Zn²⁺ ions: Zn(s) → Zn²⁺(aq) + 2e^-. b) Copper(II) ions (Cu²⁺) have been reduced. Their oxidation state decreased from +2 in Cu²⁺ to 0 in pure Cu metal. c) Cu²⁺(aq) + 2e^- → Cu(s) Question 2: a) The gas is an oxidising agent. b) The oxidising gas reacts with the iodide ions (I^-) in the potassium iodide on the paper, oxidising them to free iodine (I₂). The iodine then immediately reacts with starch present in the paper to form a characteristic blue-black starch-iodine complex. c) Chlorine gas (Cl₂) (or Oxygen gas).

teaching-notes — Chemistry (Redox Reactions)

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