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Reactivity 3.2 · Electron transfer reactions

1Oxidation, reduction and oxidation states 3.2.1 SL + HL

The syllabus statement: Oxidation and reduction can be described in terms of electron transfer, change in oxidation state, oxygen gain/loss or hydrogen loss/gain. The skills: deduce oxidation states of an atom in a compound or an ion and identify the oxidized and reduced species and the oxidizing and reducing agents in a chemical reaction, including variable oxidation states and oxidation numbers in names.

A car left out in the rain rusts; a phone battery powers a screen for a day; a breathalyser turns from orange to green when a driver has been drinking. All three are redox reactions: chemical changes in which electrons move from one species to another. Reactivity 3.1 followed protons; this sub-topic follows electrons.

Four descriptions of one process

Oxidation was first defined as gaining oxygen, as when magnesium burns to magnesium oxide. Later it was recognized that removing hydrogen from a compound has the same character, and finally that the underlying event in both is the loss of electrons. The four descriptions are summarized below; the electron and oxidation-state definitions are the general ones, and the oxygen and hydrogen definitions are convenient shortcuts, especially in organic chemistry.

Oxidation and reduction always happen together, because the electrons lost by one species must be gained by another. The species that is reduced takes electrons from the other: it is the oxidizing agent (oxidant). The species that is oxidized gives electrons away: it is the reducing agent (reductant). The oxidizing agent is therefore itself reduced, and the reducing agent is itself oxidized.

Zinc loses two electrons (oxidized; the reducing agent). Copper(II) ions gain two electrons (reduced; the oxidizing agent). The sulfate ions present do not change and are spectators.

Oxidation states

For ions such as Zn²⁺ the electron transfer is obvious. In covalent species, such as SO₂ or MnO₄⁻, electrons are shared, so a book-keeping device is needed. The oxidation state (oxidation number) of an atom is the charge it would carry if all its bonds were fully ionic, with the shared electrons assigned to the more electronegative atom. It is written with the sign first (+2, −1), whereas an ionic charge is written with the number first (2+, 1−).

A calculated oxidation state that is fractional, such as +2.5 for S in S₄O₆²⁻ or +8/3 for Fe in Fe₃O₄, is an average over atoms in different environments. It is still useful for tracking electron transfer, which is one of the limitations of oxidation states the guide asks students to appreciate: they are a formal convention, not a measured charge.

Variable oxidation states and names

Transition elements and most main-group non-metals show several oxidation states. Iron is +2 in FeSO₄ and +3 in Fe₂O₃; manganese ranges from +2 in Mn²⁺ to +7 in MnO₄⁻; nitrogen from −3 in NH₃ to +5 in NO₃⁻; sulfur from −2 in H₂S to +6 in H₂SO₄. To avoid ambiguity, names include the oxidation state as a Roman numeral: iron(II) sulfate, iron(III) oxide, potassium manganate(VII), potassium dichromate(VI), copper(I) oxide, sulfate(VI).

Exam focus · what the published papers show

2Half-equations 3.2.2 SL + HL

The syllabus statement: Half-equations separate the processes of oxidation and reduction, showing the loss or gain of electrons. The skill: deduce redox half-equations and equations in acidic or neutral solutions.

A half-equation shows one half of a redox reaction with the electrons written explicitly. For the reaction of magnesium with copper(II) sulfate:

Each half-equation must balance in atoms and in charge. Adding the two, with the electrons cancelled, gives the full ionic equation. When the numbers of electrons differ, multiply each half-equation first so that the electrons lost equal the electrons gained.

Balancing in acidic solution

Oxyanions such as MnO₄⁻ and Cr₂O₇²⁻ are reduced in acid, and their oxygen ends up in water. A reliable sequence works for any such half-equation.

Deep purple MnO₄⁻ is reduced to almost colourless Mn²⁺, so a manganate(VII) titration needs no indicator: the first permanent pink colour marks the end point. Such titrations are called self-indicating, which is one of the practical links the guide draws.

Exam focus · what the published papers show

3Relative ease of oxidation and reduction 3.2.3 SL + HL

The syllabus statement: The relative ease of oxidation and reduction of an element in a group can be predicted from its position in the periodic table. The reactions between metals and aqueous metal ions demonstrate the relative ease of oxidation of different metals. The skills: predict the relative ease of oxidation of metals and of reduction of halogens, and interpret data on metal–metal ion reactions.

Down a group

Metals react by losing electrons, so their reactivity is their ease of oxidation. Down group 1 or group 2 the outer electrons are further from the nucleus and more shielded, so they are lost more easily: potassium is oxidized more readily than sodium, and calcium more readily than magnesium. Non-metals such as the halogens react by gaining electrons, so their reactivity is their ease of reduction. Down group 17 the incoming electron is further from the nucleus and less strongly attracted: fluorine is the strongest oxidizing agent and iodine the weakest. Metal reactivity therefore increases, and non-metal reactivity decreases, down the main groups (Structure 3.1).

Displacement reactions

A more reactive metal displaces a less reactive metal from a solution of its ions, because the more reactive metal is more easily oxidized.

The observations are a brown–pink coating of copper on the zinc, fading of the blue colour of the solution, and a rise in temperature. No reaction occurs the other way round: copper does not displace zinc from zinc sulfate. Carrying out every combination of a set of metals and their ion solutions, and counting the reactions each metal gives, ranks the metals by ease of oxidation. The guide states that the order need not be memorized: data are supplied.

Halogens behave in the same way: a more reactive halogen displaces a less reactive one from its halide ions. Chlorine water turns aqueous potassium bromide orange–yellow as bromine forms, and potassium iodide brown as iodine forms; iodine does not react with bromide or chloride ions.

4Acids with reactive metals 3.2.4 SL + HL

The syllabus statement: Acids react with reactive metals to release hydrogen. The skill: deduce equations for reactions of reactive metals with dilute HCl and H2SO4.

A reactive metal is oxidized by the H⁺ ions of a dilute acid, which are reduced to hydrogen gas. This is a redox reaction, not a neutralization: the metal gives electrons directly to H⁺.

Magnesium is oxidized from 0 to +2; hydrogen is reduced from +1 to 0. The observations are effervescence, the metal dissolving, and a rise in temperature; the gas “pops” with a lighted splint. Metals below hydrogen in reactivity, such as copper, silver and gold, do not release hydrogen from dilute acids: their atoms are less easily oxidized than hydrogen molecules. This is the chemical basis for the rate comparisons in Reactivity 2.2 and one of the tests for strong and weak acids in 3.1.6.

Past-paper practice · Practice set R3E · Oxidation states, half-equations and reactivity

Attempt these before opening the solutions below. References give the session, level, paper and question number of the original examination; items marked “Practice” were written for these notes.

R3E.1IB · November 2020 · SL Paper 1 · Q23 · [1]

R3E.2IB · November 2023 · SL Paper 1 · TZ1 · Q21 · [1]

R3E.3IB · May 2022 · SL Paper 1 · TZ1 · Q23 · [1]

R3E.4IB · May 2021 · SL Paper 1 · TZ1 · Q22 · [1]

R3E.5IB · November 2022 · SL Paper 1 · Q22 · [1]

R3E.6IB · May 2021 · SL Paper 1 · TZ1 · Q21 · [1]

R3E.7IB · November 2023 · SL Paper 1 · TZ1 · Q22 · [1]

R3E.8 HL paperIB · May 2023 · HL Paper 1 · TZ2 · Q28 · [1]

R3E.9IB · November 2020 · SL Paper 2 · Q1(b)(v),(vi) · [3]

Chlorine is produced in the reaction MnO₂(s) + 4HCl(aq) → MnCl₂(aq) + Cl₂(g) + 2H₂O(l). (v) State the oxidation state of manganese in MnO₂ and MnCl₂. [2] (vi) Deduce, referring to oxidation states, whether MnO₂ is an oxidizing or reducing agent. [1]

R3E.10IB · May 2021 · SL Paper 2 · TZ1 · Q3(d) · [2]

In acidic solution, hydrogen peroxide, H₂O₂, will oxidize Fe²⁺: Fe²⁺(aq) → Fe³⁺(aq) + e⁻. (i) Write the half-equation for the reduction of hydrogen peroxide to water in acidic solution. [1] (ii) Deduce a balanced equation for the oxidation of Fe²⁺ by acidified hydrogen peroxide. [1]

R3E.11IB · May 2022 · SL Paper 2 · TZ1 · Q3(b) · [2]

Magnesium is a reactive metal. Suggest an experiment that shows that magnesium is more reactive than zinc, giving the observation that would confirm this. [2]

R3E.12Practice · written for these notes · [3]

Deduce the half-equation for the reduction of dichromate(VI) ions, Cr₂O₇²⁻, to Cr³⁺ in acidic solution, and hence the overall equation for the oxidation of I⁻ to I₂ by acidified dichromate(VI). [3]

Solutions and mark-scheme guidance · Set R3E

R3E.1 C

O₂: element, 0. OF₂: F is −1, so O is +2. H₂O₂: peroxide, −1.

R3E.2 B

V(SO₄)₂: two SO₄²⁻ give −4, so V is +4. A is +2, C: 3x = +15 so +5, D: 3x = +6 so +2.

R3E.3 C

NO₃⁻ + 4H⁺ + 3e⁻ → NO + 2H₂O. N changes from +5 to +2: three electrons; charge check: left −1 + 4 − 3 = 0, right 0.

R3E.4 A

Mn goes from +4 to +2: MnO₂ is reduced. I⁻ goes from −1 to 0: oxidized, so I⁻ is the reducing agent.

R3E.5 D

S rises from +4 in SO₂ to +6 in HSO₄⁻: SO₂ is oxidized, so it is the reducing agent. MnO₄⁻ is the oxidizing agent.

R3E.6 D

W displaces none of the others, so it is the least reactive of those tested; X displaces W and Y but not Z. Z is displaced by nothing, so Z is the most reactive. (X does not react with Z²⁺, and W, the least reactive, cannot.)

R3E.7 A

The H in H₂S (+1) is reduced to H₂ (0) by the metal, so H₂S is the oxidizing agent and hydrogen is the product, exactly as for any acid with a reactive metal.

R3E.8 C

Bi goes from +5 in BiO₃⁻ to +3 in Bi³⁺: reduced, so BiO₃⁻ is the oxidizing agent. Mn²⁺ is oxidized to MnO₄⁻ (+2 to +7).

R3E.9 [3]

(v) MnO₂: +4 ✔; MnCl₂: +2 ✔. (vi) Oxidizing agent, because the oxidation state of Mn decreases from +4 to +2 ✔.

R3E.10 [2]

(i) H₂O₂(aq) + 2H⁺(aq) + 2e⁻ → 2H₂O(l) ✔. (ii) H₂O₂(aq) + 2H⁺(aq) + 2Fe²⁺(aq) → 2H₂O(l) + 2Fe³⁺(aq) ✔.

R3E.11 [2]

Any one method with its observation: place Mg in Zn²⁺(aq) ✔, a grey/black layer of zinc forms on the magnesium ✔; or place both metals in the same acid ✔, bubbles form faster with Mg ✔; or build a cell with Mg and Zn electrodes ✔, electrons flow from Mg to Zn (positive voltage with Zn as the positive terminal) ✔.

R3E.12 [3]

Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O ✔ (2 Cr each change by 3: six electrons). 2I⁻ → I₂ + 2e⁻, ×3 ✔. Cr₂O₇²⁻ + 14H⁺ + 6I⁻ → 2Cr³⁺ + 7H₂O + 3I₂ ✔ (charge check: −2 + 14 − 6 = +6 on the left; 2 × (+3) = +6 on the right).

5Anode and cathode 3.2.5 SL + HL

The syllabus statement: Oxidation occurs at the anode and reduction occurs at the cathode in electrochemical cells. The skill: identify electrodes as anode and cathode, and identify their signs/polarities in voltaic cells and electrolytic cells, based on the type of reaction occurring at the electrode.

An electrochemical cell separates the two halves of a redox reaction so that the electrons pass through an external wire. There are two kinds. In a voltaic (galvanic) cell a spontaneous reaction produces electrical energy. In an electrolytic cell electrical energy from a power supply drives a non-spontaneous reaction. The names of the electrodes are defined by the reaction, and are the same in both kinds of cell:

The signs of the electrodes, however, are opposite in the two kinds of cell, and this is the most common source of confusion in the whole sub-topic.

The signs follow from what the electrons are doing. In a voltaic cell the anode is where electrons are produced, so it builds up negative charge relative to the cathode. In an electrolytic cell the power supply determines the signs: the positive terminal removes electrons from the anode, forcing oxidation there.

Exam focus · what the published papers show

6Voltaic (primary) cells 3.2.6 SL + HL

The syllabus statement: A primary (voltaic) cell is an electrochemical cell that converts energy from spontaneous redox reactions to electrical energy. The skill: explain the direction of electron flow from anode to cathode in the external circuit, and ion movement across the salt bridge, with half-cells of metal/metal ion, anode, cathode, circuit and salt bridge.

If zinc is placed directly in copper(II) sulfate solution, electrons pass from zinc atoms to copper ions at the surface of the metal, and the energy released is simply heat. A voltaic cell harnesses the same reaction by putting the two halves in separate containers.

A half-cell is a metal electrode in a solution of its ions (a metal/metal ion couple, written Zn²⁺/Zn). In the cell above:

• Anode (−): Zn(s) → Zn²⁺(aq) + 2e⁻. The zinc electrode slowly dissolves and [Zn²⁺] rises.
• Cathode (+): Cu²⁺(aq) + 2e⁻ → Cu(s). Copper is deposited; the blue colour of the solution fades as [Cu²⁺] falls.
• External circuit: electrons flow through the wire from zinc to copper; a voltmeter of high resistance measures the cell’s potential difference, about 1.10 V under standard conditions.
• Salt bridge: a tube or strip of filter paper containing an inert electrolyte such as KNO₃(aq). Without it, the zinc half-cell would soon carry an excess of positive charge (extra Zn²⁺) and the copper half-cell an excess of negative charge (SO₄²⁻ left behind as Cu²⁺ is removed), and the current would stop. Anions (NO₃⁻) migrate towards the anode half-cell and cations (K⁺) towards the cathode half-cell, completing the circuit without mixing the two solutions.

The overall reaction, Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), is the same as the displacement reaction of 3.2.3. The further apart two metals are in reactivity, the larger the voltage of the cell they make. The more reactive metal is always the anode. A voltaic cell also relates to Reactivity 1.3: combustion and electrochemical cells both release the energy of a redox reaction, but a cell delivers it as electrical work rather than heat.

Exam focus · what the published papers show

7Secondary (rechargeable) cells 3.2.7 SL + HL

The syllabus statement: Secondary (rechargeable) cells involve redox reactions that can be reversed using electrical energy. The skill: deduce the reactions of the charging process from given electrode reactions for discharge, and vice versa, with discussion of the advantages and disadvantages of fuel cells, primary cells and secondary cells.

A primary cell, such as the zinc–carbon or alkaline cell in a torch, produces electricity until its reactants are used up and is then discarded; its reaction cannot practically be reversed. In a secondary cell the products of discharge stay on the electrodes and the reaction can be driven backwards by an external power supply. During charging, the cell works as an electrolytic cell.

The lead–acid battery of a car is the classic example. Both electrodes finish discharge coated in lead(II) sulfate:

Lead is oxidized (0 → +2) and lead(IV) oxide is reduced (+4 → +2). To deduce the charging reactions, simply reverse each electrode equation: at the negative electrode PbSO₄ + 2e⁻ → Pb + SO₄²⁻, and at the positive electrode PbSO₄ + 2H₂O → PbO₂ + 4H⁺ + SO₄²⁻ + 2e⁻. Lithium-ion cells in phones and electric cars work on the same principle, with Li⁺ ions moving between electrodes.

8Electrolytic cells and the electrolysis of molten salts 3.2.8 SL + HL

The syllabus statement: An electrolytic cell is an electrochemical cell that converts electrical energy to chemical energy by bringing about non-spontaneous reactions. The skills: explain how current is conducted in an electrolytic cell and deduce the products of the electrolysis of a molten salt, with a DC power source, anode, cathode and electrolyte.

Sodium and chlorine do not come apart by themselves: sodium chloride is far more stable than the elements. Electrolysis supplies the energy to force the change. An electrolyte is a substance that conducts electricity when molten or in aqueous solution because it contains mobile ions, and is decomposed by the current (Structure 2.1). A solid ionic compound is not an electrolyte in this sense because its ions are locked in the lattice.

How the current is carried

The current is carried by two different charge carriers. In the wires and electrodes, delocalized electrons flow (Structure 2.3). In the electrolyte, ions move: cations towards the cathode and anions towards the anode. At the surface of each electrode the two are connected by a chemical reaction. At the cathode, cations accept electrons from the electrode (reduction); at the anode, anions give electrons to the electrode (oxidation). Electrons never travel through the electrolyte.

Products of a molten binary salt

A molten salt contains only one kind of cation and one kind of anion, so the products are predictable: the metal at the cathode and the non-metal at the anode.

The overall reaction is the sum of the two half-equations with the electrons balanced, for example 2NaCl(l) → 2Na(l) + Cl₂(g). The metal is molten at the temperature of the cell, so its state symbol is (l); the non-metal is usually a gas.

Exam focus · what the published papers show

Past-paper practice · Practice set R3F · Electrochemical cells

Attempt these before opening the solutions below. References give the session, level, paper and question number of the original examination; items marked “Practice” were written for these notes.

R3F.1IB · May 2019 · SL Paper 1 · TZ1 · Q21 · [1]

R3F.2IB · May 2019 · SL Paper 1 · TZ2 · Q22 · [1]

R3F.3IB · November 2022 · SL Paper 1 · Q21 · [1]

R3F.4IB · November 2023 · SL Paper 1 · TZ1 · Q23 · [1]

R3F.5 HL paperIB · May 2017 · HL Paper 1 · TZ1 · Q29 · [1]

R3F.6IB · May 2022 · SL Paper 1 · TZ1 · Q22 · [1]

R3F.7IB · May 2023 · SL Paper 1 · TZ1 · Q23 · [1]

R3F.8IB · November 2021 · SL Paper 1 · Q23 · [1]

R3F.9IB · November 2022 · SL Paper 2 · Q3(c)(ii)–(iv) · [4]

A voltaic cell is based on the reaction Cu²⁺(aq) + Fe(s) → Fe²⁺(aq) + Cu(s). (ii) Write the half-equation for the reaction occurring at the anode (negative electrode). [1] (iii) The salt bridge is filled with a saturated solution of KNO₃. Outline the function of the salt bridge. [1] (iv) Predict the movement of all ionic species through the salt bridge. [2]

R3F.10IB · May 2022 · SL Paper 2 · TZ2 · Q5 · [4]

Molten zinc chloride undergoes electrolysis in an electrolytic cell at 450 °C. (a) Deduce the half-equations for the reaction at each electrode. [2] (b) Deduce the overall cell reaction including state symbols. [2]

R3F.11IB · November 2019 · SL Paper 2 · Q5(c) · [3]

An electrolysis cell was assembled using graphite electrodes in molten copper(I) chloride. (i) State how current is conducted through the wires and through the electrolyte. [2] (ii) Write the half-equation for the formation of gas bubbles at the positive electrode. [1]

R3F.12Practice · written for these notes · [3]

The discharge reactions of a nickel–cadmium cell are: negative electrode Cd + 2OH⁻ → Cd(OH)₂ + 2e⁻; positive electrode NiO(OH) + H₂O + e⁻ → Ni(OH)₂ + OH⁻. (a) Deduce the overall discharge equation. (b) Write the reaction at the negative electrode during charging. (c) State one advantage of this cell over a primary cell. [3]

Solutions and mark-scheme guidance · Set R3F

R3F.1 B

Oxidation happens at the anode, which in a voltaic cell is the negative electrode (it releases electrons into the wire).

R3F.2 B

Zn²⁺ builds up in the zinc (anode) half-cell and Cu²⁺ is removed in the copper half-cell, so anions (SO₄²⁻) move to the zinc half-cell and cations (Na⁺) to the copper half-cell.

R3F.3 B

Cr is oxidized (anode), so electrons move from Cr to Fe in the wire; Fe²⁺ is reduced at the Fe cathode, so Fe²⁺ ions move towards the Fe electrode.

R3F.4 C

Cu is oxidized to Cu²⁺, so the blue colour deepens (I is false); NO₃⁻ moves towards the copper anode half-cell (II); silver is deposited at the cathode (III).

R3F.5 C

Pb: 0 → +2 (oxidized); PbO₂: Pb +4 → +2, so PbO₂ is the oxidizing agent. H⁺ stays +1: it is not reduced.

R3F.6 A

In both cells oxidation releases electrons. The signs are opposite in the two cells, so B and C are true for only one; electrons never flow through the electrolyte.

R3F.7 B

Electrolysis uses electrical energy to drive a non-spontaneous reaction.

R3F.8 C

Br⁻ ions lose electrons (are oxidized) at the positive anode; Pb²⁺ ions gain electrons at the negative cathode.

R3F.9 [4]

(ii) Fe(s) → Fe²⁺(aq) + 2e⁻ ✔ (Cu → Cu²⁺ + 2e⁻ is not accepted). (iii) Keeps each half-cell electrically neutral / balances charges / allows ion flow ✔. (iv) NO₃⁻ to the anode (Fe) half-cell ✔; K⁺ to the cathode (Cu) half-cell ✔.

R3F.10 [4]

(a) Cathode: Zn²⁺ + 2e⁻ → Zn(l) ✔; anode: 2Cl⁻ → Cl₂(g) + 2e⁻ ✔. (b) ZnCl₂(l) → Zn(l) + Cl₂(g): balanced ✔, state symbols ✔ (zinc melts at 420 °C, so it is liquid at 450 °C).

R3F.11 [3]

(i) Wires: delocalized electrons ✔; electrolyte: mobile ions ✔. (ii) 2Cl⁻ → Cl₂(g) + 2e⁻ ✔.

R3F.12 [3]

(a) Cd + 2NiO(OH) + 2H₂O → Cd(OH)₂ + 2Ni(OH)₂ ✔. (b) Cd(OH)₂ + 2e⁻ → Cd + 2OH⁻ ✔. (c) It can be recharged and reused many times, producing less waste ✔.

9Oxidation of alcohols 3.2.9 SL + HL

The syllabus statement: Functional groups in organic compounds may undergo oxidation. The skill: deduce equations to show changes in the functional groups during oxidation of primary and secondary alcohols, including the two-step reaction in the oxidation of primary alcohols, with the experimental set-up for distillation and reflux, and the fact that tertiary alcohols are not oxidized. Names and formulas of specific oxidizing agents, and mechanisms, are not assessed.

The breathalyser used by traffic police for decades relied on a colour change: orange dichromate(VI) ions turned green as they oxidized the ethanol in a driver’s breath and were themselves reduced to Cr³⁺. In organic chemistry, oxidation usually means that a carbon atom gains a bond to oxygen or loses a bond to hydrogen, and its oxidation state rises.

In equations, the oxidizing agent is represented by [O], because its identity is not assessed; in the laboratory it is usually acidified potassium dichromate(VI), which changes from orange to green, or acidified potassium manganate(VII), which changes from purple to colourless.

Primary alcohols: two steps

A primary alcohol has its –OH on a carbon attached to at most one other carbon. It is oxidized first to an aldehyde and then further to a carboxylic acid:

Which product is obtained depends on the apparatus, because the aldehyde is much more volatile than either the alcohol or the acid: it has no O–H group, so its molecules cannot hydrogen bond to one another (Structure 3.2).

• Distillation as it forms, with the alcohol in excess, removes the aldehyde from the reaction mixture before it can be oxidized further. Ethanal (b.p. 21 °C) distils off; ethanol (78 °C) and ethanoic acid (118 °C) remain.
• Heating under reflux with excess oxidizing agent returns every vapour to the flask through a vertical condenser, so the aldehyde stays in contact with the oxidant and is oxidized completely to the carboxylic acid, which can be separated by distillation afterwards.

Secondary and tertiary alcohols

A secondary alcohol has its –OH carbon bonded to two other carbons. It is oxidized to a ketone, which is not oxidized further under these conditions because that would require breaking a C–C bond:

A tertiary alcohol has no hydrogen on the carbon bearing the –OH, so it cannot be oxidized under similar conditions: acidified dichromate(VI) stays orange. This colour test distinguishes tertiary alcohols from primary and secondary ones.

Oxidation is not the same as combustion. Combustion (Reactivity 1.3) breaks every C–C and C–H bond and produces CO₂ and H₂O; controlled oxidation changes only the functional group and leaves the carbon skeleton intact.

Exam focus · what the published papers show

10Reduction of carboxylic acids and ketones 3.2.10 SL + HL

The syllabus statement: Functional groups in organic compounds may undergo reduction. The skill: deduce equations to show reduction of carboxylic acids to primary alcohols via the aldehyde, and reduction of ketones to secondary alcohols, including the role of hydride ions. Names and formulas of specific reducing agents, and mechanisms, are not assessed.

Reduction runs the oxidation ladder in reverse. The reducing agent is represented by [H], two of which are needed for each step. In practice the reagents are sources of the hydride ion, H⁻: a hydrogen atom carrying an extra electron, which donates an electron pair to the electron-deficient carbon of the C=O group.

A carboxylic acid is reduced to a primary alcohol via the aldehyde. The aldehyde is more easily reduced than the acid, so it is normally not isolated. A ketone is reduced to a secondary alcohol. In each case the carbon of the functional group gains hydrogen and its oxidation state falls. The carbonyl carbon is attacked because it carries a partial positive charge: the hydride ion acts as a nucleophile, which links this reaction to Reactivity 3.4.

Exam focus · what the published papers show

11Reduction of unsaturated compounds 3.2.11 SL + HL

The syllabus statement: Reduction of unsaturated compounds by the addition of hydrogen lowers the degree of unsaturation. The skill: deduce the products of the reactions of hydrogen with alkenes and alkynes.

Hydrogen adds across C=C and C≡C bonds in the presence of a finely divided metal catalyst such as nickel (about 150 °C) or platinum or palladium (room temperature). Because hydrogen is gained, the reaction is a reduction; because two molecules become one, it is also an addition. Each molecule of H₂ removes one degree of unsaturation (Structure 3.2).

An alkyne can add one mole of hydrogen to give an alkene, or two to give the alkane. The most important application is the partial hydrogenation of vegetable oils to make margarine and other spreads: some C=C bonds are reduced, which raises the melting point of the fat. The reaction also illustrates the question asked in the guide: the addition of hydrogen to an alkene is classed as reduction, while the addition of bromine or HBr (Reactivity 3.4) is classed as electrophilic addition, because the classification depends on which aspect of the change is being described.

Past-paper practice · Practice set R3G · Organic oxidation and reduction

Attempt these before opening the solutions below. References give the session, level, paper and question number of the original examination; items marked “Practice” were written for these notes.

R3G.1IB · November 2023 · SL Paper 1 · TZ1 · Q27 · [1]

R3G.2IB · November 2022 · SL Paper 1 · Q26 · [1]

R3G.3IB · May 2019 · SL Paper 1 · TZ1 · Q27 · [1]

R3G.4IB · May 2023 · SL Paper 1 · TZ2 · Q25 · [1]

R3G.5 HL paperIB · May 2021 · HL Paper 1 · TZ2 · Q37 · [1]

R3G.6 HL paperIB · November 2019 · HL Paper 1 · Q37 · [1]

R3G.7 HL paperIB · May 2017 · HL Paper 1 · TZ2 · Q36 · [1]

R3G.8IB · May 2022 · SL Paper 1 · TZ2 · Q22 · [1]

R3G.9 HL paperIB · May 2021 · HL Paper 2 · TZ2 · Q5(b) · [2]

Oxidation of ethanol with potassium dichromate, K₂Cr₂O₇, can form two different organic products. Determine the names of the organic products and the methods used to isolate them. [2]

R3G.10Adapted from IB · May 2022 · SL Paper 2 · TZ1 · Q3(e) · [1]

In the original question Compound B is the tertiary alcohol 2-methylpropan-2-ol, (CH₃)₃COH. Deduce what would be observed when Compound B is warmed with acidified aqueous potassium dichromate(VI). [1]

R3G.11Practice · written for these notes · [4]

Butan-1-ol, butan-2-ol and 2-methylpropan-2-ol are isomers of C₄H₁₀O. (a) For each, state the organic product (if any) of heating under reflux with excess acidified potassium dichromate(VI). (b) Write an equation, using [O], for the reaction of butan-2-ol. [4]

Solutions and mark-scheme guidance · Set R3G

R3G.1 C

Distillation removes the aldehyde, propanal, as soon as it forms.

R3G.2 B

Excess oxidizing agent and reflux keep the aldehyde in contact with the oxidant until it becomes the acid.

R3G.3 B

Only a primary alcohol gives a carboxylic acid: butan-1-ol. Propan-2-ol and pentan-3-ol are secondary (ketones); 2-methylpropan-2-ol is tertiary (no reaction).

R3G.4 C

The isomers are propan-1-ol and propan-2-ol. Under reflux the primary alcohol goes all the way to propanoic acid, and the secondary gives propanone; propanal is not isolated under reflux. A report comments that since the syllabus states that reflux leads to the formation of the carboxylic acid then the answer should have been clear to candidates.

R3G.5 D

CH₃COC₂H₅ is a ketone (butanone), reduced to butan-2-ol. C is an aldehyde, which gives a primary alcohol.

R3G.6 C

A carboxylic acid is reduced to the aldehyde (and then to a primary alcohol). Ketones cannot give aldehydes, and alcohols are already fully reduced.

R3G.7 D

C₂H₄ gains hydrogen (to ethane), CH₃COOH is reduced via the aldehyde to ethanol, and CH₃CHO to ethanol.

R3G.8 D

C–Cl is replaced by C–H: substitution. Carbon gains hydrogen and its oxidation state falls from −2 to −4: reduction.

R3G.9 [2]

Ethanal AND distillation ✔; ethanoic acid AND reflux (followed by distillation) ✔. One mark is awarded for both products or both methods.

R3G.10 [1]

No change; the solution stays orange ✔. A tertiary alcohol is not oxidized.

R3G.11 [4]

(a) Butan-1-ol → butanoic acid ✔; butan-2-ol → butanone ✔; 2-methylpropan-2-ol → no reaction ✔. (b) CH₃CH(OH)CH₂CH₃ + [O] → CH₃COCH₂CH₃ + H₂O ✔.

Additional higher level · outcomes 3.2.12 to 3.2.16

12Standard electrode potentials 3.2.12 HL only

The syllabus statement: The hydrogen half-cell H+(aq) + e− ⇌ ½H2(g) is assigned a standard electrode potential of zero by convention. It is used in the measurement of standard electrode potential, E⦵. The skill: interpret standard electrode potential data in terms of ease of oxidation/reduction.

Displacement reactions put metals in order, but they do not say by how much one metal outranks another. A voltmeter can: the potential difference of a cell measures how strongly electrons are driven from one half-cell to the other. The potential of a single half-cell cannot be measured, because a voltmeter always needs two connections. So one half-cell is chosen as the reference and every other is measured against it.

A platinum electrode is used because it is inert and conducts, and its finely divided surface catalyses the equilibrium between H⁺ ions and H₂ molecules. By convention its potential is exactly 0.00 V.

Interpreting E⦵ values

A more positive E⦵ means the species on the left of the half-equation is more easily reduced: it is a stronger oxidizing agent. A more negative E⦵ means the species on the right is more easily oxidized: it is a stronger reducing agent. The table of standard electrode potentials is therefore a ranking of oxidizing and reducing strength, and the metals at the negative end are the reactive metals of 3.2.3.

Exam focus · what the published papers show

13Standard cell potential and spontaneity 3.2.13 HL only

The syllabus statement: Standard cell potential, E⦵cell, can be calculated from standard electrode potentials. E⦵cell has a positive value for a spontaneous reaction. The skill: predict whether a reaction is spontaneous in the forward or reverse direction from E⦵ data.

In a cell made from two half-cells, the half-cell with the more positive E⦵ undergoes reduction (it is the cathode, positive terminal), and the other is reversed to become an oxidation (the anode, negative terminal).

For the zinc–copper cell: E⦵_cell = +0.34 − (−0.76) = +1.10 V. The E⦵ values are not multiplied when a half-equation is multiplied to balance electrons, because a potential is an intensive property: it measures energy per unit charge, not total energy.

A positive E⦵_cell means the reaction as written is spontaneous under standard conditions. To test any proposed redox reaction, identify which species is reduced and which is oxidized, and calculate E⦵_cell: if it is positive, the forward reaction is spontaneous; if it is negative, the reverse reaction is.

E⦵ values predict feasibility, not rate: a reaction with a positive E⦵_cell may still be too slow to observe if its activation energy is high (Reactivity 2.2). They also apply strictly only under standard conditions.

14The relationship ΔG⦵ = −nFE⦵_cell 3.2.14 HL only

The syllabus statement: The equation ΔG⦵= −nFE⦵cell shows the relationship between standard change in Gibbs energy and standard cell potential for a reaction. The skill: determine the value for ΔG⦵ from E⦵ data.

Reactivity 1.4 showed that a reaction is spontaneous when ΔG⦵ is negative. The cell potential measures the same driving force electrically: the electrical work a cell can do is the charge transferred multiplied by the potential difference.

Here n is the number of moles of electrons transferred in the balanced equation, F is the Faraday constant, the charge on one mole of electrons (9.65 × 10⁴ C mol⁻¹, given in the data booklet), and E⦵_cell is in volts. Because 1 C × 1 V = 1 J, the product is in joules per mole of reaction; divide by 1000 for kJ mol⁻¹. The minus sign links the two criteria: positive E⦵_cell ⇔ negative ΔG⦵ ⇔ spontaneous. Combined with ΔG⦵ = −RT ln K (Reactivity 2.3), it also means a positive E⦵_cell corresponds to K > 1.

15Electrolysis of aqueous solutions 3.2.15 HL only

The syllabus statement: During electrolysis of aqueous solutions, competing reactions can occur at the anode and cathode, including the oxidation and reduction of water. The skill: deduce from standard electrode potentials the products of the electrolysis of aqueous solutions, including water; the effects of concentration and of the electrode are limited to NaCl(aq) and CuSO₄(aq).

A molten salt offers each electrode only one choice. An aqueous solution offers two, because water itself can be reduced at the cathode or oxidized at the anode:

Predicting the products

• At the cathode, the species most easily reduced (most positive E⦵) is reduced. Ions of reactive metals (K⁺, Na⁺, Ca²⁺, Mg²⁺, Al³⁺) have much more negative E⦵ than water, so hydrogen is produced. Ions of less reactive metals (Cu²⁺, Ag⁺) have more positive E⦵, so the metal is deposited.
• At the anode, the species most easily oxidized (least positive E⦵ for its reduction half-equation) is oxidized. Sulfate and nitrate ions are very hard to oxidize, so oxygen is produced from water. Bromide and iodide (E⦵ +1.09 and +0.54 V) are oxidized more easily than water, giving bromine or iodine.

The effect of concentration: sodium chloride solution

Chloride is the borderline case: E⦵(Cl₂/Cl⁻) = +1.36 V is only a little more positive than E⦵(O₂/H₂O) = +1.23 V. With dilute NaCl(aq), oxygen is the main product at the anode, as the E⦵ values predict. With concentrated NaCl(aq) (brine), the high [Cl⁻] shifts the balance and chlorine is the main product. At the cathode hydrogen is produced in both cases, and the solution around it becomes alkaline as OH⁻ accumulates, which is the basis of the chlor-alkali industry producing chlorine, hydrogen and sodium hydroxide.

The effect of the electrode: copper(II) sulfate solution

With inert (graphite or platinum) electrodes, copper is deposited at the cathode and oxygen is released at the anode; the blue colour fades as Cu²⁺ is removed and the solution becomes acidic (H⁺ forms). With copper electrodes, the copper anode is itself oxidized more easily than water: Cu(s) → Cu²⁺(aq) + 2e⁻. The anode loses mass, the cathode gains the same mass, and the colour and concentration of the solution remain unchanged. This is the basis of copper purification and of electroplating.

Exam focus · what the published papers show

16Electroplating 3.2.16 HL only

The syllabus statement: Electroplating involves the electrolytic coating of an object with a metallic thin layer. The skill: deduce equations for the electrode reactions during electroplating.

Steel car parts are coated with chromium, cutlery with silver, and connectors in electronics with gold. Electroplating deposits a thin, even layer of a metal on an object by electrolysis, for decoration, for corrosion resistance or for better electrical contact.

Three rules set up the cell:

• The object to be plated is the cathode (negative), because metal ions are reduced there: Cu²⁺(aq) + 2e⁻ → Cu(s).
• The anode (positive) is a piece of the plating metal, which is oxidized and replaces the ions used: Cu(s) → Cu²⁺(aq) + 2e⁻.
• The electrolyte is a solution containing ions of the plating metal, such as CuSO₄(aq) for copper or AgNO₃(aq) for silver.

A smooth, adherent coating requires a clean object, a low and steady current, and an electrolyte that does not deposit hydrogen instead of the metal. The mass deposited increases with current and time, which is why practical investigations of electroplating control both carefully.

Exam focus · what the published papers show

Past-paper practice · Practice set R3H · Electrode potentials, electrolysis and electroplating

Attempt these before opening the solutions below. References give the session, level, paper and question number of the original examination; items marked “Practice” were written for these notes.

R3H.1IB · May 2019 · HL Paper 1 · TZ2 · Q30 · [1]

R3H.2IB · November 2018 · HL Paper 1 · Q31 · [1]

R3H.3IB · May 2021 · HL Paper 1 · TZ1 · Q30 · [1]

R3H.4IB · May 2021 · HL Paper 1 · TZ2 · Q30 · [1]

R3H.5IB · November 2018 · HL Paper 1 · Q30 · [1]

R3H.6IB · November 2023 · HL Paper 1 · TZ1 · Q30 · [1]

R3H.7IB · May 2021 · HL Paper 1 · TZ1 · Q31 · [1]

R3H.8IB · November 2020 · HL Paper 1 · Q31 · [1]

R3H.9IB · May 2021 · HL Paper 1 · TZ2 · Q31 · [1]

R3H.10 HL paperIB · May 2023 · HL Paper 2 · TZ1 · Q4(b),(c) · [6]

A voltaic cell was constructed using a copper(II) sulfate/copper half-cell and a zinc sulfate/zinc half-cell. (b)(i) Outline why electrons flow from zinc to copper when these half-cells are connected with a wire. [1] (ii) Formulate equations for the reactions taking place at each electrode. [2] (c)(i) Calculate the standard cell potential, using E⦵(Cu²⁺/Cu) = +0.34 V and E⦵(Zn²⁺/Zn) = −0.76 V. [1] (ii) Calculate the standard Gibbs free energy change, ΔG⦵, in kJ mol⁻¹, for this reaction. [2]

R3H.11 HL paperIB · May 2021 · HL Paper 2 · TZ1 · Q3(d) · [2]

A voltaic cell is set up between the Fe²⁺(aq)|Fe(s) and Fe³⁺(aq)|Fe²⁺(aq) half-cells. Deduce the equation and the cell potential of the spontaneous reaction, using E⦵(Fe²⁺/Fe) = −0.45 V and E⦵(Fe³⁺/Fe²⁺) = +0.77 V. [2]

R3H.12 HL paperIB · May 2019 · HL Paper 2 · TZ1 · Q7 · [2]

An aqueous solution of silver nitrate, AgNO₃(aq), can be electrolysed using platinum electrodes. Formulate the half-equations for the reaction at each electrode during electrolysis. [2]

R3H.13 HL paperIB · November 2020 · HL Paper 2 · Q6(c) · [1]

Copper plating can be used to improve the conductivity of an object. State, giving your reason, at which electrode the object being electroplated should be placed. [1]

Solutions and mark-scheme guidance · Set R3H

R3H.1 A

The more positive E⦵ (−0.13 V) belongs to Pb²⁺/Pb, so Pb²⁺ is the most easily reduced species: the strongest oxidizing agent. Al is the strongest reducing agent.

R3H.2 D

E⦵_cell = E⦵(Hg²⁺/Hg) − E⦵(Cr³⁺/Cr) = 0.85 − (−0.74) = +1.59 V. The coefficients do not multiply E⦵.

R3H.3 D

Mn is oxidized (more negative E⦵) and Ag⁺ reduced: E⦵_cell = 0.80 − (−1.18) = +1.98 V, positive for the spontaneous direction.

R3H.4 A

Measured against Fe³⁺/Fe²⁺ set to zero, every value shifts by −0.77 V: −1.18 − 0.77 = −1.95 V.

R3H.5 C

Negative E⦵ gives positive ΔG⦵ (= −nFE⦵), and positive ΔG⦵ = −RT ln K requires ln K < 0, so K < 1.

R3H.6 B

At the cathode in aqueous NaCl, water is reduced in preference to Na⁺: hydrogen. Sodium metal forms only from the molten salt.

R3H.7 C

Concentrated chloride is oxidized to chlorine at the anode; Cu²⁺ (E⦵ +0.34 V) is reduced in preference to water at the cathode.

R3H.8 C

The spoon is the cathode; the anode is silver, which dissolves to replace the Ag⁺ ions deposited. The electrolyte must contain Ag⁺, not Zn²⁺.

R3H.9 C

The copper anode is oxidized (Cu → Cu²⁺ + 2e⁻) and loses mass; copper is deposited on the cathode, which gains mass.

R3H.10 [6]

(b)(i) Zinc is more reactive / a better reducing agent / more easily oxidized (or: Zn has the more negative E⦵) ✔. (ii) Anode: Zn(s) → Zn²⁺(aq) + 2e⁻ ✔; cathode: Cu²⁺(aq) + 2e⁻ → Cu(s) ✔ (one mark only if equilibria are used or electrodes swapped). (c)(i) +0.34 − (−0.76) = +1.10 V ✔. (ii) ΔG⦵ = −2 × 9.65 × 10⁴ × 1.10 ✔ = −212 kJ mol⁻¹ ✔.

R3H.11 [2]

2Fe³⁺(aq) + Fe(s) → 3Fe²⁺(aq) ✔; E⦵_cell = +0.77 − (−0.45) = +1.22 V ✔. The reverse reaction or a negative value is not accepted.

R3H.12 [2]

Cathode: Ag⁺(aq) + e⁻ → Ag(s) ✔. Anode: 2H₂O(l) → O₂(g) + 4H⁺(aq) + 4e⁻ ✔ (4OH⁻ → 2H₂O + O₂ + 4e⁻ also accepted). Nitrate is not oxidized.

R3H.13 [1]

At the cathode (negative electrode), because Cu²⁺ ions are reduced there to copper ✔.

Review · Reactivity 3.2

17Misconceptions, the examiner’s view, and the question types

Seven question types cover the sub-topic.

18Quick check

19Summary and knowledge organiser

Essential knowledge

• Oxidation = loss of electrons, increase in oxidation state, gain of O, loss of H; reduction is the reverse; the oxidizing agent is reduced.
• Oxidation states by the rules; sum = charge; Roman numerals in names; averages can be fractional.
• Half-equations balance atoms and charge; in acid add H₂O, H⁺, e⁻; equalize electrons to combine.
• Metal reactivity (ease of oxidation) increases down a group; halogen oxidizing strength decreases down group 17; displacement reactions rank metals.
• Reactive metal + dilute acid → salt + H₂ (redox).
• Anode = oxidation, cathode = reduction; voltaic: anode −, cathode +; electrolytic: anode +, cathode −; electrons flow anode → cathode in the wire.
• Salt bridge keeps half-cells neutral; secondary cells reverse on charging; molten salts give metal at the cathode and non-metal at the anode.
• Primary alcohol → aldehyde (distil) → carboxylic acid (reflux); secondary → ketone; tertiary no reaction. Acid → aldehyde → primary alcohol and ketone → secondary alcohol by hydride; alkenes and alkynes + H₂ (Ni) → alkanes.
• HL E⦵ measured against the SHE (0.00 V); more positive = stronger oxidizing agent; E⦵_cell = E(cathode) − E(anode), positive = spontaneous; ΔG⦵ = −nFE⦵_cell.
• HL Aqueous electrolysis: H₂ unless the metal ion is less reactive than H (Cu, Ag); O₂ unless halide (concentrated Cl⁻, Br⁻, I⁻); copper anodes dissolve; electroplating: object = cathode.

Examination checklist

• Name the agent as the whole reacting species.
• Electrons on the right for oxidation, on the left for reduction; never in the overall equation.
• Single arrows, not equilibrium signs, in electrolysis equations; state symbols (l) for molten products.
• Give both charge carriers: electrons in the wire, ions in the electrolyte.
• State distillation or reflux, and acidified dichromate(VI), for alcohol oxidation.
• HL E⦵ is not multiplied; n is; convert ΔG⦵ from J to kJ.

Knowledge organiser