3.2 Electron transfer reactions

Syllabus
First assessment 2025
Topic
3.2
Level
SL

Learning objectives

3.2.1Oxidation and reduction• Electron transfer, oxidation state change• Oxygen gain/loss, hydrogen loss/gain• Oxidizing/reducing agents• Deduce oxidation states and identify oxidized/reduced species3.2.2Half-equations• Separate oxidation and reduction• Show electron loss/gain• Deduce redox half-equations and equations in acidic or neutral solutions3.2.3Periodic trends in redox• Metals: ease of oxidation• Halogens: ease of reduction• Metal displacement reactions• Predict metal oxidation and halogen reduction using supplied data3.2.4Acids with reactive metals• Release H₂ gas• Deduce equations for reactive metals with dilute acids3.2.5Electrodes in electrochemical cells• Oxidation at anode• Reduction at cathode• Signs/polarities in voltaic vs. electrolytic cells• Identify anode/cathode from electrode reactions3.2.6Primary (voltaic) cells• Spontaneous redox → electrical energy• Electron flow: anode → cathode (external circuit)• Ion movement across salt bridge• Include metal/metal ion half-cells, circuit, and salt bridge3.2.7Secondary (rechargeable) cells• Redox reactions reversible with electrical energy• Deduce charging reactions from discharge reactions and compare cell types3.2.8Electrolytic cells• Electrical energy → chemical energy• Non-spontaneous reactions• Electrolysis of molten salts• Explain current conduction and deduce molten salt electrolysis products3.2.9Oxidation of organic functional groups• Primary alcohols → aldehydes → carboxylic acids• Secondary alcohols → ketones• Include distillation/reflux setup and that tertiary alcohols are not oxidized similarly3.2.10Reduction of organic functional groups• Carboxylic acids → aldehydes → primary alcohols• Ketones → secondary alcohols• Include hydride ion role; specific reducing agents and mechanisms are not assessed3.2.11Reduction of unsaturated compounds• Addition of H₂ lowers unsaturation• Alkenes + H₂, alkynes + H₂• Deduce hydrogenation products of alkenes and alkynes

Oxidation and Reduction

Oxidation is loss of electrons and an increase in oxidation state; reduction is gain of electrons and a decrease. The oxidizing agent is reduced, and the reducing agent is oxidized.

Use the oxidation-state rules and total charge to identify which species changed and which agent caused the change.

In Zn + Cu²⁺ → Zn²⁺ + Cu, Zn rises from 0 to +2 and is oxidized, so it is the reducing agent; Cu²⁺ falls from +2 to 0 and is reduced, so it is the oxidizing agent. Name agents from what happens to them, not from the process they cause in the other species.

Identifying Redox Agents

2 marks

Identify the oxidising and reducing agents, and the species oxidised and reduced, in the forward reaction.

CO(g)\mathbf{C O}(\mathbf{g})H2O(g)\mathbf{H}_{\mathbf{2}} \mathbf{O}(\mathbf{g})
oxidising or reducing agent?
species oxidised or reduced?

Redox Half-Equations

Separate oxidation and reduction, balance atoms, add H2O and H+ in acidic solution as needed, balance charge with electrons, then multiply to cancel electrons before adding.

A valid full redox equation conserves atoms and charge and contains no uncancelled electrons.

For MnO₄⁻ → Mn²⁺ in acid, balance O with 4H₂O, H with 8H⁺ and charge with 5e⁻: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O. After combining halves, cancel electrons and any identical H⁺ or H₂O, then recheck both atoms and net charge.

To adapt an acidic half-equation to neutral or basic conditions, first balance it with H₂O, H⁺ and e⁻. Add the same number of OH⁻ to both sides to neutralize every H⁺, replace H⁺+OH⁻ by H₂O, then cancel water appearing on both sides. Recheck atoms and total charge; do not leave free H⁺ in a stated neutral medium unless the chemistry justifies it.

Balancing Redox Equations

2 marks

The reaction continues until the violet colour disappears. The thiosulfate ion, S2O32−\mathrm{S}_{2} \mathrm{O}_{3}{ }^{2-}, is oxidized to SO2\mathrm{SO}_{2}, and Fe3+\mathrm{Fe}^{3+} is reduced to Fe2+\mathrm{Fe}^{2+}. Deduce the oxidation half-equation, and the overall redox equation for this second step of the reaction.

Oxidation half-equation:
Overall redox equation:

Redox Displacement

A more active metal more readily donates electrons to a less active metal ion. A halogen with greater reduction tendency oxidizes the halide of a weaker halogen.

Test a predicted displacement by placing one metal in the other metal's sulfate or comparing supplied electrode data.

Zinc displaces Cu²⁺ because Zn more readily oxidizes: Zn + Cu²⁺ → Zn²⁺ + Cu. Chlorine displaces Br⁻ because Cl₂ more readily reduces. Keep the metal and halogen trends in their correct electron directions instead of using one vague 'more reactive' rule.

Predicting Displacement Reactions

2 marks

Discuss how the relative reactivity of copper and thallium could be established using the metals and aqueous solutions of their sulfates.

Metals with Dilute Acids

A metal above hydrogen in the activity series can donate electrons to acid and release hydrogen gas; a metal below hydrogen, such as copper, does not react with dilute hydrochloric acid.

metal+acid→salt+H2(g)metal + acid → salt + H2(g)

Balance the electron transfer behind the molecular equation: metal atoms are oxidized and 2H⁺ + 2e⁻ → H₂ is the reduction. Use the metal charge and acid anion to construct the salt rather than assuming every metal forms a 2+ ion.

Predicting Hydrogen Release

2 marks

Outline, using an ionic equation, what is observed when magnesium powder is added to a solution of ammonium chloride.

Anodes, Cathodes and Polarity

Oxidation always occurs at the anode and reduction always occurs at the cathode. In a voltaic cell the anode is negative and cathode positive; in an electrolytic cell the anode is positive and cathode negative.

the electrolytic anode is positive and is the site of oxidation; the electrolytic cathode is negative and is the site of reduction; cations move to the cathode and anions move to the anode; the external DC source is explicit and no spontaneous operation is implied.

Name electrodes from the half-reactions before assigning signs. Electrons leave the anode and reach the cathode through the external circuit; a power supply reverses the polarities in an electrolytic cell but never changes where oxidation and reduction occur.

Labelling Electrochemical Cells

2 marks

Annotate the electrolytic cell with the terms anode and cathode, and show the direction of ion movement.

Voltaic Cells

A voltaic cell uses a spontaneous redox reaction to convert chemical energy to electrical energy. Electrons flow through the wire from anode to cathode; the salt bridge carries ions to maintain charge neutrality.

Zn is the negative anode and is oxidized by Zn(s) → Zn2+(aq) + 2e−; Cu is the positive cathode and Cu2+ is reduced by Cu2+(aq) + 2e− → Cu(s); external electrons flow from Zn through the voltmeter to Cu and never through the salt bridge; NO3− migrates toward the anode and Na+ migrates toward the cathode.

Both half-cells connect to the external circuit and the salt bridge must contact both solutions.

In a Zn|Zn²⁺ || Cu²⁺|Cu cell, Zn is oxidized at the negative anode and electrons travel through the wire to the positive Cu cathode, where Cu²⁺ is reduced. Salt-bridge anions migrate toward the anode compartment and cations toward the cathode compartment to prevent charge buildup; electrons do not flow through the bridge.

Completing a Voltaic-Cell Diagram

3 marks

Simple cells rely on differences in standard electrode potential values between different elements and their ions. The following is an incomplete diagram for measuring a cell potential between Mn2+(aq)/Mn\mathrm{Mn}^{2+}(\mathrm{aq}) / \mathrm{Mn} and Ni2+(aq)/Ni\mathrm{Ni}^{2+}(\mathrm{aq}) / \mathrm{Ni} half-cells.

Draw the missing components and fully label the diagram to show how the cell potential can be measured.

Primary, Secondary and Fuel Cells

Cell Energy direction Reuse
primary chemical → electrical not readily reversible
secondary chemical ⇌ electrical recharge by external power
fuel chemical → electrical while reactants are supplied refill fuel

Write the discharge half-equations first. Charging a secondary cell requires an external potential to drive their reverse, whereas a primary cell is not designed for safe efficient reversal and a fuel cell continues only while reactants are supplied. Rechargeability is a reaction-design property, not simply the presence of a power socket.

Explaining Rechargeability

1 mark

Outline how a rechargeable battery differs from a primary cell.

Molten-Salt Electrolysis

In molten salt there is no water: metal ions are reduced to metal at the cathode and anions are oxidized at the anode. For molten chloride, chloride forms chlorine gas.

the electrolyte is heated molten NaCl and contains no water; Cl- moves to the positive anode and Na+ moves to the negative cathode; the anode half-equation is 2Cl- to Cl2 plus 2e-; the cathode half-equation is Na+ plus e- to Na.

M(n+)+ne−→Matcathode;2X−→X2+2e−atanodeM^(n+) + ne− → M at cathode; 2X− → X2 + 2e− at anode

Molten MgCl₂ contains only Mg²⁺ and Cl⁻: Mg²⁺ + 2e⁻ → Mg at the cathode and 2Cl⁻ → Cl₂ + 2e⁻ at the anode. The melt conducts by ion migration; do not introduce H₂, O₂ or water-based competition into a molten-salt question.

Deducing Molten-Electrolysis Products

2 marks

Deduce the products of the electrolysis of molten cobalt(II) bromide, CoBr2(l)\mathrm{CoBr}_{2}(\mathrm{l}).

Product at anode:
Product at cathode:

Oxidation of Alcohols

A primary alcohol oxidizes to an aldehyde and then a carboxylic acid; a secondary alcohol oxidizes to a ketone. Reflux supports further oxidation to the acid, while distillation can remove an aldehyde.

R–CH2OH becomes R–CHO; the aldehyde retains one H on the carbonyl carbon; H2O is shown as in the approved source.
R–CHO becomes R–COOH; the same carbonyl carbon and R group are retained; no unsupported water product is added.
R–CH(OH)–R′ becomes R–C(=O)–R′; both R groups and the functional carbon are retained; H2O is shown as in the approved source.
the structure is (CH3)3C–OH; the carbon bearing OH has no H; the non-reaction claim is explicitly limited to the stated reflux/distillation oxidizing conditions.

In a primary-alcohol experiment, distil the aldehyde as it forms to limit further oxidation; heat under reflux when the carboxylic acid is required. Tertiary alcohols lack the required hydrogen on the carbon bearing –OH and are not oxidized in the same way.

Choosing Alcohol-Oxidation Products

2 marks

Deduce the organic products when butan-1-ol and butan-2-ol are separately heated under reflux with acidified potassium dichromate(VI).

Butan-1-ol:
Butan-2-ol:

Reduction of Carbonyl Compounds

A carboxylic acid can be reduced through an aldehyde to a primary alcohol; a ketone is reduced to a secondary alcohol. Hydride ions supply the reduction equivalent in these transformations.

R–CHO becomes R–CH2OH; the product carbon has R, OH and two H substituents; the product is correctly labelled primary alcohol.
R–C(=O)–R′ becomes R–CH(OH)–R′; the product is correctly labelled secondary alcohol; the carbon skeleton is unchanged.

Track the carbon functional group rather than only the reagent: an aldehyde gives a primary alcohol and a ketone gives a secondary alcohol. Hydride supplies an electron-rich H unit to the carbonyl carbon; named reducing agents and detailed mechanisms are outside this objective.

Deducing Reduction Products

1 mark

Which product may be obtained by the reduction of CH3CH2COOH\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH} ?

Hydrogenation of Alkenes and Alkynes

Hydrogenation adds H2 across π bonds. Continue addition until the required saturated product is formed; nickel, palladium or platinum catalysts with heat or pressure are typical conditions.

R–C≡C–R′ receives exactly one H2; one H is added to each alkyne carbon and C≡C becomes C=C; R and R′ remain on their original carbons; no unsupported stereochemical outcome is claimed.
one H2 is added across C=C; C=C becomes C–C while the sigma carbon skeleton remains intact; R, R′, R″ and R‴ remain on the correct original carbon; each alkene carbon gains exactly one H.
R–C≡C–R′ receives exactly 2 H2; the product is R–CH2–CH2–R′; the triple bond becomes a single bond without breaking the carbon skeleton.

Count π bonds to determine hydrogen demand: one mole of H₂ saturates one C=C, while full conversion of one C≡C to C–C needs two moles of H₂. Keep the carbon skeleton unchanged when drawing the product.

Deducing Hydrogenation Products

2 marks

State the reagent and conditions needed and draw the structural formula of the product.

Electron Transfer Reactions Summary

Retrieve the route: assign oxidation states, balance half-equations, predict displacement, label cells, trace electrons and ions, follow organic redox pathways, calculate potentials and choose electrolysis products.

Check electron loss/gain, anode/cathode versus polarity, spontaneous sign, salt-bridge direction, ions present, organic functional-group direction and object-cathode placement.