3.2 Electron transfer reactions
- Syllabus
- First assessment 2025
- Topic
- 3.2
- Level
- SL
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.
2 marks
Identify the oxidising and reducing agents, and the species oxidised and reduced, in the forward reaction.
| CO(g) | H2O(g) | |
|---|---|---|
| oxidising or reducing agent? | ||
| species oxidised or reduced? |
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.
2 marks
The reaction continues until the violet colour disappears. The thiosulfate ion, S2O32−, is oxidized to SO2, and Fe3+ is reduced to Fe2+. Deduce the oxidation half-equation, and the overall redox equation for this second step of the reaction.
Oxidation half-equation:
Overall redox equation:
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.
2 marks
Discuss how the relative reactivity of copper and thallium could be established using the metals and aqueous solutions of their sulfates.
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)
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.
2 marks
Outline, using an ionic equation, what is observed when magnesium powder is added to a solution of ammonium chloride.
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.

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

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.
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 and Ni2+(aq)/Ni half-cells.
Draw the missing components and fully label the diagram to show how the cell potential can be measured.
| 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.
1 mark
Outline how a rechargeable battery differs from a primary cell.
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.

M(n+)+ne−→Matcathode;2X−→X2+2e−atanode
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.
2 marks
Deduce the products of the electrolysis of molten cobalt(II) bromide, CoBr2(l).
Product at anode:
Product at cathode:
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.




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.
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:
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.


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.
1 mark
Which product may be obtained by the reduction of CH3CH2COOH ?
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.



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