IB DP Chemistry Reactivity 3.2: Electron transfer: redox, cells and electrolysis

This guide teaches Reactivity 3.2, Electron transfer reactions, using the February 2023 IB Diploma Programme Chemistry guide (first assessment 2025) as its checklist. Understandings Reactivity 3.2.1 to 3.2.11 are covered, the part common to SL and HL. The additional higher level understandings (3.2.12 onwards) are not covered here.

Then try the practice questions or read the shorter revision notes. See the printable checklist and IB DP chemistry hub for other units; our free diagnostics show where to start.

What this unit covers

All rows are for SL and HL.

Guide reference	What you must be able to do
Reactivity 3.2.1	Deduce oxidation states; name oxidizing and reducing agents
Reactivity 3.2.2	Write half-equations and redox equations (acidic or neutral)
Reactivity 3.2.3	Predict ease of oxidation of metals and reduction of halogens
Reactivity 3.2.4	Write equations for reactive metals with dilute acids
Reactivity 3.2.5	Name and sign the electrodes in both cell types
Reactivity 3.2.6	Explain electron flow and salt bridge ion movement
Reactivity 3.2.7	Reverse discharge reactions; compare types of cell
Reactivity 3.2.8	Explain conduction; electrolyse a molten salt
Reactivity 3.2.9	Oxidize primary and secondary alcohols
Reactivity 3.2.10	Reduce carboxylic acids and ketones
Reactivity 3.2.11	Add hydrogen to alkenes and alkynes

Reactivity 3.2.1 Four ways to describe redox

Oxidation is loss of electrons, a rise in oxidation state, gain of oxygen or loss of hydrogen. Reduction is the reverse of each. The oxidizing agent is itself reduced; the reducing agent is itself oxidized.

Assigning oxidation states

Write the sign first, as in +3 or −2.

1. An uncombined element is 0; a simple ion equals its charge.
2. Oxygen is −2 (−1 in peroxides); hydrogen is +1 (−1 in metal hydrides).
3. Neutral compound: values total zero. Ion: values total its charge.

Worked example 1. Find the oxidation state of the named atom.

• Cr in CrO₄²⁻: x + 4(−2) = −2, so x = +6.
• H in CaH₂: calcium is +2, so 2x = −2, x = −1 (a metal hydride).
• N in N₂O: 2x + (−2) = 0, so x = +1.

Transition elements and most main group non-metals have variable oxidation states: nitrogen runs from −3 in NH₄⁺ to +5 in HNO₃, and manganese from +2 in Mn²⁺ to +7 in MnO₄⁻. So names carry Roman numerals: FeCl₃ is iron(III) chloride and MnO₂ is manganese(IV) oxide. Common oxyanion names such as sulfate remain acceptable.

Oxidation states treat every bond as ionic, so they are bookkeeping, not real charges. Corrosion is surface oxidation of a metal.

Worked example 2. In SO₂ + 2H₂S → 3S + 2H₂O, sulfur in SO₂ goes from +4 to 0 (reduced) and sulfur in H₂S goes from −2 to 0 (oxidized). So SO₂ is the oxidizing agent and H₂S is the reducing agent.

Reactivity 3.2.2 Half-equations

To balance a half-equation in acidic solution:

1. Match up the atom whose oxidation state changes.
2. Put in water molecules to even out O atoms.
3. Put in H⁺ ions to even out H atoms.
4. Put in electrons until the charges agree.

To combine two, multiply so the electrons match, add, and cancel.

Worked example 3. Nitrate ions in acid oxidize tin(II) ions to tin(IV), forming nitrogen monoxide.

• Reduction: add 2H₂O on the right, 4H⁺ on the left, then 3e⁻ to balance charge: NO₃⁻ + 4H⁺ + 3e⁻ → NO + 2H₂O
• Oxidation: Sn²⁺ → Sn⁴⁺ + 2e⁻
• Electrons: 3 × 2 = 6, so double the first and triple the second.
• Overall: 2NO₃⁻ + 8H⁺ + 3Sn²⁺ → 2NO + 4H₂O + 3Sn⁴⁺
• Check charge: left −2 + 8 + 6 = +12; right 3 × (+4) = +12.

In neutral solution, with no oxygen moving, no H⁺ is needed: Mg → Mg²⁺ + 2e⁻ and Ag⁺ + e⁻ → Ag give Mg + 2Ag⁺ → Mg²⁺ + 2Ag.

A titration with manganate(VII) is self-indicating: purple MnO₄⁻ is reduced to almost colourless Mn²⁺, so the first lasting pink marks the end point.

Reactivity 3.2.3 Trends in a group

Metals lose electrons when they react. Going down groups 1 and 2, atoms get larger and outer electrons are more shielded, so they are removed more easily: ease of oxidation increases.

Halogens gain an electron. Down group 17 the incoming electron is further away and more shielded, so ease of reduction decreases: Cl₂ is the strongest oxidizing agent of the three.

A more easily reduced halogen displaces a less easily reduced one from its halide ions:

• Cl₂(aq) + 2Br⁻(aq) → 2Cl⁻(aq) + Br₂(aq): the colourless solution turns yellow-orange.
• Br₂(aq) + 2I⁻(aq) → 2Br⁻(aq) + I₂(aq): the solution turns brown.
• I₂ does not react with Cl⁻ or Br⁻.

Metal/metal ion data. A metal reduces the ions of any metal that is less easily oxidized. Questions supply the data.

Worked example 4. Each metal meets the other ions (✓ = deposit).

Metal	Cr³⁺(aq)	Ni²⁺(aq)	Pb²⁺(aq)	Ag⁺(aq)
Cr	–	✓	✓	✓
Ni	✗	–	✓	✓
Pb	✗	✗	–	✓
Ag	✗	✗	✗	–

The metal reacting with most ions is oxidized most easily: Cr > Ni > Pb > Ag.

Reactivity 3.2.4 Reactive metals and dilute acids

Reactive metals release hydrogen from dilute hydrochloric and sulfuric acids: the metal is oxidized and H⁺ is reduced. Bubbles form and the metal dissolves. Copper and silver do not react.

• Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)
• Ni(s) + 2HCl(aq) → NiCl₂(aq) + H₂(g)
• Ionic form, for any metal M forming M²⁺: M(s) + 2H⁺(aq) → M²⁺(aq) + H₂(g)

Worked example 5. 1.308 g of zinc reacts with excess dilute sulfuric acid. Find the volume of hydrogen at STP (use 22.7 dm³ mol⁻¹ and Ar(Zn) = 65.38).

• Zn(s) + H₂SO₄(aq) → ZnSO₄(aq) + H₂(g), so 1 mol Zn gives 1 mol H₂.
• n(Zn) = 1.308 ÷ 65.38 = 0.02001 mol
• V(H₂) = 0.02001 × 22.7 = 0.454 dm³

Reactivity 3.2.5 and 3.2.6 Electrodes and primary cells

In every cell, oxidation is at the anode, reduction at the cathode. Only the signs change.

Cell	Anode	Cathode
Voltaic (primary or secondary, discharging)	negative	positive
Electrolytic	positive	negative

A voltaic anode pushes electrons into the wire (negative); an electrolytic anode has electrons pulled out by the supply (positive).

A primary (voltaic) cell turns the energy of a spontaneous redox reaction into electrical energy. Two half-cells, each a metal in a solution of its own ions, are joined by a wire and a salt bridge.

Worked example 6. A nickel half-cell (Ni in Ni²⁺(aq)) is joined to a silver half-cell (Ag in Ag⁺(aq)). Nickel is oxidized more easily than silver.

• Anode (negative): Ni(s) → Ni²⁺(aq) + 2e⁻
• Cathode (positive): Ag⁺(aq) + e⁻ → Ag(s)
• Overall: Ni(s) + 2Ag⁺(aq) → Ni²⁺(aq) + 2Ag(s)
• Electrons flow through the wire from the nickel anode to the silver cathode.
• Use KNO₃ in the bridge; chloride would precipitate AgCl.
• Anions (NO₃⁻) move toward the anode half-cell, where Ni²⁺ builds up; cations (K⁺) move toward the cathode half-cell, where Ag⁺ is removed. This keeps both solutions neutral and completes the circuit. Electrons never pass through the bridge.

Reactivity 3.2.7 Secondary cells, and comparing cells

A secondary (rechargeable) cell has reversible electrode reactions. Charging uses an external supply to drive each discharge half-equation backwards, so you deduce it by reversing that half-equation.

Lead–acid cell, discharge:

• Negative electrode: Pb(s) + SO₄²⁻(aq) → PbSO₄(s) + 2e⁻
• Positive electrode: PbO₂(s) + 4H⁺(aq) + SO₄²⁻(aq) + 2e⁻ → PbSO₄(s) + 2H₂O(l)

Charging reverses both: PbSO₄(s) + 2e⁻ → Pb(s) + SO₄²⁻(aq) at the negative electrode (now the cathode), and PbSO₄(s) + 2H₂O(l) → PbO₂(s) + 4H⁺(aq) + SO₄²⁻(aq) + 2e⁻ at the positive electrode (now the anode).

In a lithium-ion cell, discharge oxidizes lithium held in graphite (Li → Li⁺ + e⁻) and reduces a metal oxide (Li⁺ + CoO₂ + e⁻ → LiCoO₂); charging reverses both.

Cell type	Advantages	Disadvantages
Primary	cheap; light; ready to use; long shelf life	used once; more waste; some contain toxic metals
Secondary	recharged many times; lower cost over its life; less waste	higher initial cost; capacity falls with use; lead and cadmium types are toxic and heavy
Fuel cell	runs as long as fuel is supplied; efficient; a hydrogen cell gives only water	expensive catalysts; hydrogen is hard to store; hydrogen is often made from fossil fuels

Reactivity 3.2.8 Electrolytic cells and molten salts

In an electrolytic cell, electrical energy forces a non-spontaneous redox reaction to happen: a DC supply is connected to two electrodes in an electrolyte (a molten or dissolved ionic compound). Current is conducted by:

• electrons in the external circuit, from the anode to the supply and on to the cathode;
• ions in the electrolyte, cations to the cathode and anions to the anode;
• electron transfer at each electrode surface.

Solid ionic compounds do not conduct: their ions are fixed in the lattice.

Worked example 7. Molten lithium iodide is electrolysed with inert electrodes.

• Cathode (negative), reduction: Li⁺ + e⁻ → Li
• Anode (positive), oxidation: 2I⁻ → I₂ + 2e⁻
• Overall: 2LiI(l) → 2Li(l) + I₂(g)

So a molten binary salt gives its metal at the cathode and its non-metal element at the anode.

Reactivity 3.2.9 Oxidation of alcohols

Write the oxidizing agent as [O]; the reagent is not assessed. Acidified dichromate(VI) is often used, turning from orange to green.

• Primary alcohols oxidize in two steps: alcohol → aldehyde → carboxylic acid.
• Secondary alcohols oxidize to ketones.
• Tertiary alcohols are not oxidized under the same conditions, as the carbon bearing the OH has no hydrogen atom to lose.

Worked example 8.

• Hexan-1-ol to hexanal: C₅H₁₁CH₂OH + [O] → C₅H₁₁CHO + H₂O
• Hexanal to hexanoic acid: C₅H₁₁CHO + [O] → C₅H₁₁COOH
• Hexan-3-ol to hexan-3-one: C₂H₅CH(OH)C₃H₇ + [O] → C₂H₅COC₃H₇ + H₂O
• 3-Methylpentan-3-ol: no reaction.

Choosing the set-up.

• For the aldehyde, distil as it forms. With no hydrogen bonding between its molecules, it boils below the alcohol and leaves before further oxidation.
• For the carboxylic acid, heat under reflux with excess oxidizing agent. Vapour condenses in the vertical condenser and drips back, so the aldehyde cannot escape before it is oxidized further.

Reactivity 3.2.10 Reduction of carbonyl compounds

Write the reducing agent as [H]. Reduction runs the oxidation sequence backwards:

• Carboxylic acid → aldehyde → primary alcohol
• Ketone → secondary alcohol

The reducing agents supply hydride ions, H⁻: a hydrogen carrying two electrons, which adds to the slightly positive carbonyl carbon; the oxygen then gains H⁺. Reagent names and mechanisms are not assessed.

Worked example 9.

• 2-Methylpropanoic acid to 2-methylpropanal: (CH₃)₂CHCOOH + 2[H] → (CH₃)₂CHCHO + H₂O
• 2-Methylpropanal to 2-methylpropan-1-ol: (CH₃)₂CHCHO + 2[H] → (CH₃)₂CHCH₂OH
• Pentan-2-one to pentan-2-ol: CH₃COC₃H₇ + 2[H] → CH₃CH(OH)C₃H₇

Carbon is −4 in CH₄, −2 in CH₃OH, 0 in HCHO, +2 in HCOOH and +4 in CO₂: each step along this series is an oxidation.

Reactivity 3.2.11 Hydrogenation lowers unsaturation

Adding hydrogen across C=C or C≡C is a reduction. A catalyst (usually nickel, heated) is needed. Each H₂ added removes one degree of unsaturation.

• Alkene + H₂ → alkane: CH₃CH=CH₂ + H₂ → CH₃CH₂CH₃
• Alkyne + H₂ → alkene; alkyne + 2H₂ → alkane: CH₃C≡CCH₃ + 2H₂ → CH₃CH₂CH₂CH₃

Worked example 10. How much hydrogen, at STP, fully saturates 2.70 g of but-2-yne, C₄H₆? (Use 22.7 dm³ mol⁻¹.)

• M(C₄H₆) = 4(12.01) + 6(1.01) = 54.10 g mol⁻¹
• n = 2.70 ÷ 54.10 = 0.04991 mol
• Two H₂ per molecule: n(H₂) = 0.09982 mol
• V = 0.09982 × 22.7 = 2.27 dm³

Common errors

• Leaving electrons, or unequal electron numbers, in an overall equation.
• Giving the anode the same sign in voltaic and electrolytic cells.
• Forgetting the water when going straight from alcohol to acid: RCH₂OH + 2[O] → RCOOH + H₂O.
• Claiming tertiary alcohols oxidize to ketones.

Official syllabus

Reactivity 3.2 (Electron transfer reactions), understandings 3.2.1 to 3.2.11 for SL and HL, in the Chemistry guide of the International Baccalaureate Diploma Programme, February 2023, first assessment 2025.