Redox Reactions | ISC Class 11 Chemistry Notes

This note covers oxidation and reduction, electron transfer, oxidising and reducing agents, oxidation-number rules and calculations, types of redox reactions, applications, and balancing equations in acidic and basic media by the oxidation-number and ion-electron methods.

What do oxidation and reduction mean?

Oxidation originally meant addition of oxygen to a substance. Reduction meant removal of oxygen. These descriptions remain useful for recognising changes in many reactions, including reactions between elements and oxygen.

How do oxygen and hydrogen help identify the changes?

In equations, (s), (l), (g) and (aq) mean solid, liquid, gas and aqueous solution respectively. An aqueous solution has water as its solvent. A subscript counts atoms within a formula; a coefficient before a formula counts the corresponding chemical entities.

Magnesium, represented by Mg, combines with oxygen, O₂, to form magnesium oxide, MgO: 2Mg(s) + O₂(g) → 2MgO(s). Magnesium is oxidised because it gains oxygen. The arrow means “forms”; the plus sign separates reacting substances or products.

Removal of hydrogen is also oxidation. In 2H₂S(g) + O₂(g) → 2S(s) + 2H₂O(l), hydrogen sulphide, H₂S, loses hydrogen and forms sulphur, S. Oxygen gains hydrogen to form water, H₂O, and is reduced.

Addition of hydrogen is reduction. Ethene, C₂H₄, combines with hydrogen, H₂, to form ethane, C₂H₆: C₂H₄(g) + H₂(g) → C₂H₆(g). The two processes must be considered together when classifying the complete reaction.

How is the classical definition extended?

Electronegativity is the tendency of an atom to attract the shared electrons of a bond. Electrons are negatively charged particles. An electronegative element attracts these electrons relatively strongly; an electropositive element has a tendency to lose electrons. These ideas broaden the oxygen and hydrogen descriptions.

For example, Mg(s) + Cl₂(g) → MgCl₂(s) represents magnesium reacting with chlorine, Cl₂, to form magnesium chloride, MgCl₂. Magnesium undergoes oxidation through addition of an electronegative element. Redox joins “reduction” and “oxidation”: the two changes occur simultaneously.

How does electron transfer explain redox reactions and agents?

Definition: Oxidation is loss of electrons and reduction is gain of electrons. An electron is a negatively charged particle, written e⁻. An oxidising agent accepts electrons; a reducing agent donates electrons.

A chemical species is an atom, molecule or ion considered in a reaction. An ion carries electric charge. Superscript signs and numbers show that charge: Zn²⁺ is a zinc ion with two positive charges, whereas Cl⁻ is a chloride ion with one negative charge.

What is a half-reaction?

A half-reaction shows either the oxidation or the reduction part of a redox change, with electrons included explicitly. Zinc, Zn, reacts with copper(II) ions, Cu²⁺, to produce zinc ions and copper metal, Cu. The Roman numeral II indicates oxidation state +2, a formal value used to keep track of electron changes.

• Oxidation: Zn(s) → Zn²⁺(aq) + 2e⁻. Zinc gives up two electrons.
• Reduction: Cu²⁺(aq) + 2e⁻ → Cu(s). A copper(II) ion accepts two electrons.
• Overall reaction: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s). The electrons cancel when the two equations are added.

The reducing agent, also called the reductant, is zinc because it supplies electrons that reduce copper(II) ions. The oxidising agent, also called the oxidant, is Cu²⁺ because it accepts electrons and brings about oxidation of zinc.

Why is the agent's name different from its own change?

An agent is named for the change it causes in the other species. Consequently, the reducing agent is itself oxidised, and the oxidising agent is itself reduced. Naming the species that loses or gains electrons prevents these two roles from being reversed.

Sodium, Na, and hydrogen provide a useful check: 2Na(s) + H₂(g) → 2NaH(s). Sodium hydride, NaH, contains Na⁺ and H⁻ ions. Sodium loses electrons, while hydrogen gains them. Thus sodium is oxidised even though it combines with hydrogen.

Note: Do not apply “addition of hydrogen means reduction” mechanically to sodium hydride formation. The relative electronegativities and the electron transfer show that sodium is oxidised and hydrogen is reduced.

What do metal displacement experiments reveal?

A displacement reaction replaces an atom or ion in a compound with an atom or ion of another element. Metal displacement provides visible evidence of electron transfer and allows the electron-releasing tendencies of metals to be compared.

What happens when zinc meets copper(II) ions?

When a zinc strip is placed in aqueous copper nitrate, you may notice a coating of reddish metallic copper and the disappearance of the blue colour of the solution. Copper nitrate supplies Cu²⁺ ions. The change is represented by Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s).

Three beakers show initial, intermediate and final stages. A zinc rod stands in blue solution; the blue colour fades across the sequence. The final drawing labels copper deposited on the zinc rod.

See Fig. 7.1 in your NCERT textbook

The reverse experiment, copper metal in zinc sulphate solution, shows no visible reaction. The equilibrium greatly favours zinc ions and copper metal in the forward reaction. Equilibrium refers to the state in which the forward and reverse reaction rates are equal; it does not imply equal amounts of reactants and products.

What happens when copper meets silver ions?

Copper placed in aqueous silver nitrate reacts with silver ions, Ag⁺, producing silver metal, Ag, and Cu²⁺ ions: Cu(s) + 2Ag⁺(aq) → Cu²⁺(aq) + 2Ag(s). Copper is oxidised, and the solution develops a blue colour as copper(II) ions form.

Three beakers show a copper rod in solution. Blue colour develops and becomes more intense. The final-stage drawing labels silver deposits on the rod.

See Fig. 7.2 in your NCERT textbook

Together, these reactions establish the electron-releasing order Zn > Cu > Ag, where > means “has a greater tendency than”. Zinc releases electrons to copper ions; copper releases electrons to silver ions. This comparison explains the reducing behaviour demonstrated in these particular reactions.

The accompanying ions need not appear in the net ionic equation, which records the species involved in the chemical change. It is the metal and the reacting metal ion, rather than the full salt formula, that reveal the electron exchange most directly.

What is an oxidation number, and which rules determine it?

An oxidation number describes an element's oxidation state using a formal electron-counting convention. In a covalent bond, where atoms share an electron pair, the pair is assigned entirely to the more electronegative atom for this calculation.

This assignment is for bookkeeping. In forming water from hydrogen and oxygen, charge transfer is only partial and is perhaps better described as an electron shift than as complete electron loss and gain. An oxidation number need not represent an actual ionic charge.

Which rules apply first?

A binary compound contains two elements. A neutral compound has no overall charge, so the sum of all its atoms' oxidation numbers is zero. In a polyatomic ion, an ion containing several atoms, the sum equals the ion's overall charge.

What exceptions must be remembered for oxygen?

Oxygen has oxidation number −2 in most compounds. Peroxides contain directly linked oxygen atoms assigned −1 each; hydrogen peroxide, H₂O₂, is an example. Superoxides contain the oxygen unit assigned an average of −½ per oxygen atom, as in potassium superoxide, KO₂.

When oxygen is bonded to fluorine, oxygen has a positive oxidation number. It is +2 in OF₂ and +1 in dioxygen difluoride, O₂F₂. The fluorine rule remains −1 in these compounds, so the usual oxygen value cannot be used.

Halogens are the Group 17 elements. Chlorine, bromine, Br, and iodine, I, have oxidation number −1 when present as halide ions, their negatively charged ions. They can have positive oxidation numbers when combined with oxygen. Fluorine must be treated separately.

The oxidation states of a metal can be indicated by Stock notation, which uses Roman numerals. Iron, Fe, has +2 in iron(II) oxide, FeO, and +3 in iron(III) oxide, Fe₂O₃. The numeral specifies the state, rather than the number of atoms.

How are oxidation numbers calculated in molecules and ions?

Write the formula, assign the known values, and equate the sum to the overall charge. In the following calculations, x denotes the unknown oxidation number, or average oxidation number where several atoms of the element occur. Oxidation numbers are dimensionless, so they have no physical unit.

Derivation: Chromium in potassium dichromate

Worked example 1. Find chromium's oxidation number in potassium dichromate, K₂Cr₂O₇, where Cr represents chromium. Potassium is +1, oxygen is −2, and the compound is neutral.

1. Two potassium atoms contribute 2 × (+1) = +2.
2. Seven oxygen atoms contribute 7 × (−2) = −14.
3. The total is zero, so 2 + 2x − 14 = 0. Therefore 2x = 12.

x = +6. Answer: chromium is in oxidation state +6. For the dichromate ion, Cr₂O₇²⁻, use 2x − 14 = −2; this gives the same value.

Derivation: Sulphur in the thiosulphate ion

Worked example 2. Find the average oxidation number of sulphur in thiosulphate, S₂O₃²⁻. Oxygen is −2, and the total ionic charge is −2.

1. Three oxygen atoms contribute 3 × (−2) = −6.
2. Two sulphur atoms contribute 2x to the oxidation-number sum.
3. Set 2x − 6 = −2. Thus 2x = 4.

x = +2. Answer: the average sulphur oxidation number is +2. This calculation alone does not determine the separate oxidation numbers of the two sulphur atoms.

How are oxygen's exceptional values obtained?

Worked example 3. In hydrogen peroxide, H₂O₂, hydrogen is +1 and the molecule is neutral. Therefore 2(+1) + 2x = 0, giving x = −1. Answer: oxygen is −1, consistent with the peroxide rule.

Worked example 4. In potassium superoxide, KO₂, potassium is +1 and the compound is neutral. Hence 1 + 2x = 0 and x = −½. Answer: the average oxidation number of oxygen is −½.

Worked example 5. In oxygen difluoride, OF₂, fluorine is −1. Therefore x + 2(−1) = 0. Answer: oxygen is +2. In dioxygen difluoride, O₂F₂, the corresponding equation is 2x − 2 = 0, giving x = +1.

What does an average oxidation number tell us?

Worked example 6. In tetrathionate, S₄O₆²⁻, oxygen is −2. Thus 4x − 12 = −2, giving x = +2.5. Answer: the average for sulphur is +2.5; the separate sulphur values are +5, 0, 0 and +5.

Worked example 7. Carbon suboxide, C₃O₂, contains carbon, C, and oxygen at −2. The neutral formula gives 3x − 4 = 0, so x = +4/3. Answer: this is an average; the two terminal carbons are +2 and the middle carbon is 0.

Worked example 8. In sulphite, SO₃²⁻, oxygen is −2. Therefore x − 6 = −2, giving x = +4. Answer: sulphur is +4. In sulphate, SO₄²⁻, x − 8 = −2 instead gives sulphur +6.

Note: A fractional average does not mean a fraction of an electron has been transferred by an individual atom. Structural information can distinguish individual oxidation states. The superoxide value also shows why fractional oxidation numbers must be interpreted with care.

How do oxidation-number changes identify different reaction types?

An increase in oxidation number indicates oxidation; a decrease indicates reduction. The oxidising agent brings about an increase in another element's oxidation number, while the reducing agent brings about a decrease. Compare the same element before and after the reaction.

Which changes occur in combination and decomposition?

A combination reaction produces one compound from two or more substances. For example, carbon combines with oxygen to form carbon dioxide, CO₂: C(s) + O₂(g) → CO₂(g). Carbon changes from 0 to +4, and oxygen from 0 to −2.

A decomposition reaction breaks a compound into two or more components. When water decomposes, 2H₂O(l) → 2H₂(g) + O₂(g), hydrogen changes from +1 to 0 and oxygen from −2 to 0. Both oxidation and reduction occur.

However, not every decomposition reaction is redox. Calcium carbonate, CaCO₃, decomposes into calcium oxide, CaO, and carbon dioxide: CaCO₃(s) → CaO(s) + CO₂(g). Calcium, Ca, remains +2, carbon remains +4, and oxygen remains −2.

How are displacement reactions used?

Metal displacement can recover metals from their compounds. Chromium(III) oxide, Cr₂O₃, reacts with aluminium, Al: Cr₂O₃(s) + 2Al(s) → Al₂O₃(s) + 2Cr(s). Aluminium oxide, Al₂O₃, forms as aluminium changes from 0 to +3; chromium changes from +3 to 0.

Non-metal displacement includes hydrogen displacement. Zinc reacts with hydrochloric acid, HCl: Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g). Zinc chloride, ZnCl₂, contains zinc at +2. Hydrogen changes from +1 to 0, while chlorine remains −1.

Chlorine also displaces bromide ions: Cl₂(g) + 2Br⁻(aq) → 2Cl⁻(aq) + Br₂(l). Bromine changes from −1 to 0 and is oxidised. Chlorine changes from 0 to −1 and is reduced. Halogen recovery from halide ions therefore involves oxidation.

Note: Some elements can keep the same oxidation number during a redox reaction. Classify the reaction by the changes that occur, rather than expecting every element in the equation to undergo oxidation or reduction.

What makes disproportionation a special redox reaction?

Disproportionation occurs when an element in one initial oxidation state is simultaneously oxidised and reduced. The reacting element occupies an intermediate state, and the products contain it in both a higher and a lower oxidation state.

How does hydrogen peroxide illustrate the process?

Hydrogen peroxide decomposes according to 2H₂O₂(aq) → 2H₂O(l) + O₂(g). Oxygen starts at −1. In water it becomes −2, which is reduction; in oxygen gas it becomes 0, which is oxidation. Hydrogen remains at +1.

Place oxygen at −1 in H₂O₂ in the centre. Draw one arrow towards oxygen at −2 in H₂O, labelled reduction, and another towards oxygen at 0 in O₂, labelled oxidation. Write the balanced decomposition equation underneath.

The same equation is a decomposition reaction because one compound breaks down, and a disproportionation reaction because the same element undergoes both changes. These descriptions identify different features of the reaction and can apply together.

What happens to chlorine in alkaline solution?

An alkaline or basic medium supplies hydroxide ions, OH⁻. Chlorine reacts as Cl₂(g) + 2OH⁻(aq) → ClO⁻(aq) + Cl⁻(aq) + H₂O(l). In hypochlorite, ClO⁻, chlorine is +1; in chloride it is −1, compared with 0 in Cl₂.

This reaction explains the formation of household bleaching agents. The hypochlorite ion oxidises colour-bearing stains to colourless compounds. Fluorine differs from chlorine: it cannot exhibit a positive oxidation state and does not show a disproportionation tendency.

An intermediate oxidation state is necessary for both upward and downward changes. In perchlorate, ClO₄⁻, chlorine is already at its highest oxidation state, +7, so this ion does not disproportionate. Merely finding several oxidation states for an element does not establish that every one can disproportionate.

How is a redox equation balanced by oxidation numbers in acidic medium?

The oxidation-number method equates the total increase and total decrease in oxidation numbers. The formulas of the reactants and products must be known first. Balancing adjusts coefficients; changing a formula would change the chemical substance being described.

An acidic medium supplies hydrogen ions, H⁺, for balancing. Consider dichromate ions reacting with sulphite ions to give chromium(III) ions and sulphate ions. The unbalanced or skeletal equation is Cr₂O₇²⁻ + SO₃²⁻ → Cr³⁺ + SO₄²⁻.

Which steps establish the coefficients?

1. Assign oxidation numbers. Chromium changes from +6 in dichromate to +3 in Cr³⁺. Sulphur changes from +4 in sulphite to +6 in sulphate.
2. Count the total change. Each chromium decreases by 3, so the two chromium atoms decrease by 6. Each sulphur increases by 2.
3. Use three sulphite ions for each dichromate ion. Write Cr₂O₇²⁻ + 3SO₃²⁻ → 2Cr³⁺ + 3SO₄²⁻.
4. Balance charge with hydrogen ions. The left side has charge −8 and the right side 0, so add 8H⁺ to the left.
5. Balance hydrogen with water. Add 4H₂O to the right, then verify oxygen and the remaining atoms.

Worked example 9. Answer: Cr₂O₇²⁻(aq) + 3SO₃²⁻(aq) + 8H⁺(aq) → 2Cr³⁺(aq) + 3SO₄²⁻(aq) + 4H₂O(l).

How is the result checked?

There are two chromium atoms, three sulphur atoms, sixteen oxygen atoms and eight hydrogen atoms on each side. The total ionic charge is zero on each side. Both the atom count and the charge count must agree.

Dichromate is the oxidising agent because its chromium is reduced. Sulphite is the reducing agent because its sulphur is oxidised. The factor of three arises from matching total changes, including both chromium atoms, rather than comparing single atoms alone.

How does oxidation-number balancing work in basic medium?

In basic medium, hydroxide ions and water complete the balance after the oxidation-number changes are matched. The medium does not alter the requirement that total increase equals total decrease. It determines which additional species are appropriate in the final equation.

Consider permanganate, MnO₄⁻, reacting with bromide, Br⁻, to produce manganese dioxide, MnO₂, and bromate, BrO₃⁻. Mn represents manganese. These specified products give the skeletal equation MnO₄⁻ + Br⁻ → MnO₂ + BrO₃⁻.

How are the changes and charges matched?

1. Manganese changes from +7 in permanganate to +4 in manganese dioxide, a decrease of 3 per manganese atom.
2. Bromine changes from −1 in bromide to +5 in bromate, an increase of 6 per bromine atom.
3. Use two permanganate ions for one bromide ion: 2MnO₄⁻ + Br⁻ → 2MnO₂ + BrO₃⁻.
4. The left charge is −3 and the right charge is −1. Add 2OH⁻ to the right to make both charges −3.
5. Add one water molecule to the left to supply two hydrogen atoms, then verify the oxygen count.

Worked example 10. Answer: 2MnO₄⁻(aq) + Br⁻(aq) + H₂O(l) → 2MnO₂(s) + BrO₃⁻(aq) + 2OH⁻(aq).

Both sides contain two manganese atoms, one bromine atom, nine oxygen atoms and two hydrogen atoms. Each side has total charge −3. Permanganate is reduced and acts as the oxidant; bromide is oxidised and acts as the reductant.

Note: State the medium and products before balancing. In this reaction bromide becomes bromate and permanganate becomes manganese dioxide. Those formulas determine the oxidation-number changes and cannot be replaced by guessed products.

How does the ion-electron method balance an acidic reaction?

The ion-electron method, also called the half-reaction method, balances oxidation and reduction separately before combining them. Atoms must balance within each half-reaction, and electrons are then added to balance electric charge.

Consider iron(II) ions reacting with dichromate in acidic solution to form iron(III) and chromium(III) ions. The skeletal equation is Fe²⁺ + Cr₂O₇²⁻ → Fe³⁺ + Cr³⁺. Iron is oxidised from +2 to +3; chromium is reduced from +6 to +3.

What is the sequence for balancing the two halves?

1. Separate the changes: Fe²⁺ → Fe³⁺ and Cr₂O₇²⁻ → Cr³⁺.
2. Balance atoms other than hydrogen and oxygen. The chromium half becomes Cr₂O₇²⁻ → 2Cr³⁺.
3. Balance oxygen with water and hydrogen with H⁺. Write Cr₂O₇²⁻ + 14H⁺ → 2Cr³⁺ + 7H₂O.
4. Balance charge with electrons. The oxidation half is Fe²⁺ → Fe³⁺ + e⁻. The reduction half is Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O.
5. Multiply the entire iron half-reaction by six, so it releases the six electrons consumed by the chromium half-reaction.
6. Add the two halves and cancel electrons. Check the atom totals and net electric charge in the resulting equation.

Worked example 11. Answer: 6Fe²⁺(aq) + Cr₂O₇²⁻(aq) + 14H⁺(aq) → 6Fe³⁺(aq) + 2Cr³⁺(aq) + 7H₂O(l).

Why are six electrons added to the reduction half?

Before electrons are added, the left side of the chromium half has charge −2 + 14 = +12; the right side has +6. Six negatively charged electrons on the left reduce its charge to +6. Chromium's reduction therefore appears as electron gain.

In the combined equation, the left charge is +12 − 2 + 14 = +24 and the right charge is +18 + 6 = +24. The electrons disappear from the overall equation because their loss and gain are equal.

The coefficient multiplier applies to every term in a half-reaction. Multiplying only the electrons while leaving the iron coefficients unchanged would destroy the balance. Iron(II) ions supply electrons and are the reductant; dichromate accepts them and is the oxidant.

How is the ion-electron method adapted to basic solution?

For a basic reaction, first balance atoms as for acidic conditions. Then add an equal number of hydroxide ions to both sides for every hydrogen ion introduced. On the side containing H⁺, combine H⁺ and OH⁻ into water and cancel water where possible.

Consider permanganate oxidising iodide ions, I⁻, to molecular iodine, I₂, while forming manganese dioxide. The given basic-medium skeleton is MnO₄⁻ + I⁻ → MnO₂ + I₂. Iodine changes from −1 to 0, while manganese changes from +7 to +4.

How is the basic reduction half obtained?

1. Balance the iodine atoms and charge: 2I⁻ → I₂ + 2e⁻.
2. Balance the manganese half initially as MnO₄⁻ + 4H⁺ → MnO₂ + 2H₂O.
3. Add 4OH⁻ to each side. On the left, 4H⁺ and 4OH⁻ form 4H₂O. Cancel two water molecules from each side.
4. The atom-balanced basic half is MnO₄⁻ + 2H₂O → MnO₂ + 4OH⁻. Add three electrons to the left to balance charge.
5. Multiply the iodine half by three and the manganese half by two. Each then involves six electrons.
6. Add the halves, cancel all six electrons, and check both atoms and charge.

The balanced reduction half-reaction is MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻. Its left and right charges are both −4. The electrons are on the reactant side because manganese gains electrons.

Worked example 12. Answer: 6I⁻(aq) + 2MnO₄⁻(aq) + 4H₂O(l) → 3I₂(s) + 2MnO₂(s) + 8OH⁻(aq).

What should the final verification establish?

Each side contains six iodine atoms, two manganese atoms, twelve oxygen atoms and eight hydrogen atoms. Each side has charge −8. The final basic-medium equation contains hydroxide ions, and the temporary hydrogen ions have been converted into water.

Both balancing methods enforce the same electron accounting. The oxidation-number method matches changes directly; the ion-electron method displays the electrons in separate equations. In either method, supplied products, correct formulas, atom conservation and charge conservation determine a valid result.

Glossary

• Oxidation — Loss of electrons by a species, or an increase in an element's oxidation number.
• Reduction — Gain of electrons by a species, or a decrease in an element's oxidation number.
• Redox reaction — A reaction in which oxidation and reduction occur simultaneously through related electron changes.
• Oxidising agent — A species that accepts electrons and causes oxidation of another species while itself undergoing reduction.
• Reducing agent — A species that donates electrons and causes reduction of another species while itself undergoing oxidation.
• Oxidation number — The oxidation state assigned using rules that allocate a covalent electron pair to the more electronegative atom.
• Half-reaction — An equation representing the oxidation or reduction part separately, with electrons explicitly included.
• Displacement reaction — A reaction in which an atom or ion replaces an atom or ion of another element.
• Disproportionation — Simultaneous oxidation and reduction of an element from one initial oxidation state to higher and lower states.
• Peroxide — An oxygen compound with directly linked oxygen atoms assigned an oxidation number of −1 each.
• Superoxide — An oxygen compound whose superoxide unit has an average oxidation number of −½ per oxygen atom.
• Stock notation — A notation that indicates an element's oxidation number using a Roman numeral in parentheses.

Common errors and misconceptions

• Misconception: Oxidation requires oxygen. Correct: Electron loss or an increase in oxidation number also defines oxidation, including reactions such as magnesium combining with chlorine.
• Misconception: The oxidising agent is oxidised. Correct: It causes oxidation of the other species and is itself reduced by accepting electrons.
• Misconception: Oxygen has oxidation number −2 in every compound. Correct: This holds in most compounds; peroxides, superoxides and compounds with fluorine require different assignments.
• Misconception: A neutral compound contains only atoms with oxidation number zero. Correct: Its algebraic total is zero; the individual oxidation numbers can be positive and negative.
• Misconception: The calculated average specifies every atom's individual oxidation state. Correct: Tetrathionate has average sulphur +2.5, while its individual sulphur states are +5, 0, 0 and +5.
• Misconception: Every decomposition reaction is redox. Correct: Calcium carbonate decomposition leaves the oxidation numbers of calcium, carbon and oxygen unchanged.
• Misconception: Equal atom totals are enough to balance an ionic equation. Correct: Net electric charge must also match, and the medium determines how H⁺, OH⁻ and water are used.
• Misconception: The electron multiplier can be applied only to the electrons. Correct: Multiply every coefficient in the relevant half-reaction before adding the halves and cancelling electrons.

Exam-style questions with model answers

Q1. For Zn → Zn²⁺ + 2e⁻ and Cu²⁺ + 2e⁻ → Cu, identify the reducing agent and oxidising agent, giving a reason for each. [2 marks]

1. Zinc is the reducing agent because it donates two electrons and is itself oxidised to Zn²⁺.
2. The copper(II) ion, Cu²⁺, is the oxidising agent because it accepts two electrons and is reduced to copper metal.

Q2. Calculate chromium's oxidation number in neutral K₂Cr₂O₇. Use potassium +1 and oxygen −2. Show the equation and calculation. [3 marks]

1. Let x be chromium's oxidation number. The two potassium atoms contribute +2 and the seven oxygen atoms contribute −14 to the total.
2. Because the compound is neutral, the algebraic sum is zero: 2(+1) + 2x + 7(−2) = 0.
3. Rearranging gives 2x = 12 and x = +6. Therefore chromium has oxidation number +6 in potassium dichromate.

Q3. Calculate the average sulphur oxidation number in S₂O₃²⁻, taking oxygen as −2. Explain whether this calculation determines the separate sulphur oxidation states. [3 marks]

1. Let x represent the average sulphur oxidation number. Three oxygen atoms contribute −6, while the two sulphur atoms together contribute 2x.
2. The total must equal the ionic charge, −2. Thus 2x − 6 = −2, giving 2x = 4 and x = +2.
3. The result is an average. It does not establish separate oxidation states for the two sulphur atoms; structural information is needed for that distinction.

Q4. Explain why 2H₂O₂ → 2H₂O + O₂ is a disproportionation reaction. Use oxygen −1 in H₂O₂, −2 in H₂O and 0 in O₂, with hydrogen remaining +1. [4 marks]

1. Oxygen begins in the single oxidation state −1 in hydrogen peroxide, which lies between the two product states given.
2. The oxygen forming water changes from −1 to −2. Its oxidation number decreases, so this portion undergoes reduction.
3. The oxygen forming O₂ changes from −1 to 0. Its oxidation number increases, so this portion undergoes oxidation.
4. The same element is therefore both oxidised and reduced from one initial state, which defines disproportionation. Hydrogen remains unchanged at +1.

Q5. Balance Cr₂O₇²⁻ + SO₃²⁻ → Cr³⁺ + SO₄²⁻ in acidic solution by the oxidation-number method. Use oxygen −2. Show oxidation-number changes, coefficients, charge balancing and the final equation. [5 marks]

1. Chromium is +6 in dichromate and +3 in Cr³⁺. Sulphur is +4 in sulphite and +6 in sulphate. Chromium is reduced while sulphur is oxidised.
2. Two chromium atoms undergo a total decrease of 6. Each sulphur increases by 2, so one dichromate ion requires three sulphite ions.
3. The partially balanced equation is Cr₂O₇²⁻ + 3SO₃²⁻ → 2Cr³⁺ + 3SO₄²⁻. The left charge is −8 and the right charge is zero.
4. Add 8H⁺ to the left to balance charge, then 4H₂O to the right to balance hydrogen and oxygen.
5. The result is Cr₂O₇²⁻ + 3SO₃²⁻ + 8H⁺ → 2Cr³⁺ + 3SO₄²⁻ + 4H₂O. Both sides contain two Cr, three S, sixteen O and eight H atoms, with zero total charge.

Q6. Balance MnO₄⁻ + Br⁻ → MnO₂ + BrO₃⁻ in basic solution by the oxidation-number method. Use oxygen −2. Show the changes and check the final atom and charge balance. [5 marks]

1. Manganese decreases from +7 in MnO₄⁻ to +4 in MnO₂. Bromine increases from −1 in Br⁻ to +5 in BrO₃⁻.
2. The decrease is 3 per manganese atom and the increase is 6 per bromine atom. Therefore two permanganate ions are required for one bromide ion.
3. Write 2MnO₄⁻ + Br⁻ → 2MnO₂ + BrO₃⁻. Add 2OH⁻ on the right because the left charge is −3 and the right charge is initially −1.
4. Add H₂O on the left. The balanced equation is 2MnO₄⁻ + Br⁻ + H₂O → 2MnO₂ + BrO₃⁻ + 2OH⁻.
5. Each side contains two Mn, one Br, nine O and two H atoms, with charge −3. Permanganate is the oxidising agent and bromide the reducing agent.

Q7. Use the ion-electron method to balance Fe²⁺ + Cr₂O₇²⁻ → Fe³⁺ + Cr³⁺ in acidic solution. Show the two balanced half-reactions, their combination and a charge check. [5 marks]

1. The iron oxidation half is Fe²⁺ → Fe³⁺ + e⁻. One electron on the right balances the increase in positive ionic charge.
2. Balance chromium and then oxygen and hydrogen: Cr₂O₇²⁻ + 14H⁺ → 2Cr³⁺ + 7H₂O. Add six electrons on the left to balance charge.
3. The reduction half is Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O. Multiply the entire iron half by six to equalise electron loss and gain.
4. Add the halves and cancel electrons: 6Fe²⁺ + Cr₂O₇²⁻ + 14H⁺ → 6Fe³⁺ + 2Cr³⁺ + 7H₂O.
5. The left charge is 12 − 2 + 14 = +24; the right charge is 18 + 6 = +24. Iron, chromium, oxygen and hydrogen counts also agree.

Q8. Balance MnO₄⁻ + I⁻ → MnO₂ + I₂ in basic solution by the ion-electron method. Show conversion from the temporary acidic half-reaction, electron cancellation and a final charge check. [6 marks]

1. Balance iodine and charge to obtain the oxidation half: 2I⁻ → I₂ + 2e⁻. Iodide loses electrons as molecular iodine forms.
2. Balance the manganese half temporarily as MnO₄⁻ + 4H⁺ → MnO₂ + 2H₂O, using water for oxygen and hydrogen ions for hydrogen.
3. Add 4OH⁻ to both sides, combine hydrogen and hydroxide ions into water, and cancel two water molecules. This gives MnO₄⁻ + 2H₂O → MnO₂ + 4OH⁻.
4. Balance charge by adding three electrons on the left: MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻.
5. Multiply the iodine half by three and the manganese half by two. Add and cancel six electrons to obtain 6I⁻ + 2MnO₄⁻ + 4H₂O → 3I₂ + 2MnO₂ + 8OH⁻.
6. Each side has charge −8. Each also contains six iodine, two manganese, twelve oxygen and eight hydrogen atoms, confirming the balance in basic medium.

Key takeaways

• Oxidation and reduction occur together: electron loss by one species is matched by electron gain by another.
• An oxidising agent accepts electrons and is reduced; a reducing agent donates electrons and is oxidised.
• Oxidation-number sums equal zero for neutral compounds and equal the total ionic charge for polyatomic ions.
• Oxygen is −2 in most compounds, but peroxides, superoxides and compounds containing fluorine require the exception rules.
• An average oxidation number can differ from individual atomic values, so structural information matters when assigning separate states.
• Disproportionation produces higher and lower oxidation states of the same element from a single initial oxidation state.
• The oxidation-number method matches total increases and decreases, including every changing atom in each formula.
• The ion-electron method balances separate halves, equalises electrons, combines the equations and verifies both atom and charge totals.

Test yourself

Why is magnesium oxidised in Mg + Cl₂ → MgCl₂ even though oxygen is absent?

Magnesium combines with the electronegative element chlorine. It loses electrons and changes from oxidation number 0 to +2, so the change is oxidation.

In Zn → Zn²⁺ + 2e⁻, is zinc the oxidising or reducing agent?

Zinc is the reducing agent because it supplies electrons. Its own electron loss means that zinc undergoes oxidation.

Why does oxygen have +2 in OF₂? Use fluorine −1.

The molecule is neutral. Two fluorine atoms contribute −2, so the oxygen atom must contribute +2 to make the sum zero.

What is the average sulphur oxidation number in S₂O₃²⁻ if oxygen is −2?

Writing x for the average gives 2x − 6 = −2. Therefore x = +2, without specifying the individual sulphur states.

In CaCO₃ → CaO + CO₂, calcium stays +2, carbon +4 and oxygen −2. Is this redox?

No. The reaction is decomposition, but none of the elements changes oxidation number, so it is not a redox reaction.

Why can ClO₄⁻, with chlorine already at its highest state of +7, not disproportionate?

Disproportionation requires both oxidation and reduction. Chlorine cannot undergo the necessary increase beyond its highest oxidation state of +7.

One half-reaction releases two electrons and the other consumes three. What multipliers equalise them?

Multiply the first half-reaction by three and the second by two. Each then involves six electrons, which cancel on addition.

How are four temporary H⁺ ions removed while balancing in basic medium?

Add four OH⁻ ions to both sides. Combine H⁺ and OH⁻ into water, then cancel any water appearing on both sides.