R3.2 Electron transfer reactions · Educerie Chemistry

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Educerie · IB Diploma · Chemistry

Reactivity 3 What are the mechanisms of chemical change? · R3.2 Electron transfer reactions

Level: SL and HL. Sections 10 to 13 are HL only. If you are SL, skip them; nothing in your papers tests them.
Themes (key concepts): reactivity, structure, and the nature of science (models and their limits). Redox is the second great mechanism of chemical change: electrons move from one species to another. Where an element sits in the periodic table, its structure, predicts how readily it gives electrons up or takes them in, and the same transfer can be run forwards to make electricity or backwards by supplying it.
The question this unit answers: what happens when electrons are transferred?
Where it is examined: Paper 1A multiple choice (oxidation states, identifying agents, cell polarity, electrolysis products); Paper 1B, where you interpret displacement data or an unfamiliar cell set-up; Paper 2 half-equations, cell diagrams and organic equations of 1 to 3 marks each. HL adds E⦵ calculations, ΔG⦵ = −nFE⦵, and the products of aqueous electrolysis, often chained into one question.

What you must be able to do

You must be able to	Level	What it looks like in the exam
Describe oxidation and reduction four ways, and deduce oxidation states	SL, HL	"Deduce the oxidation state of manganese in KMnO₄" (1 mark)
Identify what is oxidized and reduced, and the oxidizing and reducing agents; name compounds with oxidation numbers	SL, HL	"Identify the reducing agent, giving a reason" (2 marks)
Deduce half-equations and redox equations in acidic or neutral solution	SL, HL	"Deduce the half-equation for the reduction of Cr₂O₇²⁻ to Cr³⁺" (1–2 marks)
Predict the relative ease of oxidation of metals and reduction of halogens; interpret displacement data	SL, HL	Paper 1B: a results table from mixing metals with metal-ion solutions
Write equations for reactive metals with dilute HCl and H₂SO₄	SL, HL	"Formulate the ionic equation for …" (1 mark)
Label anode, cathode and their signs in voltaic and electrolytic cells; explain electron and ion flow	SL, HL	"Annotate the diagram to show the direction of electron flow" (2 marks)
Deduce charging reactions from discharge reactions; compare primary, secondary and fuel cells	SL, HL	"Deduce the reaction at the negative electrode during charging" (1 mark)
Explain conduction in an electrolytic cell and deduce the products of electrolysing a molten salt	SL, HL	"State the products at each electrode" (2 marks)
Write equations for oxidizing primary and secondary alcohols, and explain distillation against reflux	SL, HL	"Outline how the aldehyde can be obtained rather than the acid" (2 marks)
Write equations for reducing carboxylic acids, aldehydes and ketones, and for hydrogenating alkenes and alkynes	SL, HL	"Deduce the product when … is reduced" (1 mark)
Interpret standard electrode potentials in terms of ease of oxidation and reduction	HL only	"Identify the strongest oxidizing agent in the table" (1 mark)
Calculate E⦵(cell) and predict whether a reaction is spontaneous	HL only	"Determine whether Fe³⁺ will oxidize I⁻" (2 marks)
Calculate ΔG⦵ from E⦵ data	HL only	Paper 2, 2 marks, units essential
Deduce the products of electrolysing water and aqueous solutions; explain the effect of concentration and electrode	HL only	"Explain why chlorine, not oxygen, forms at the anode" (2 marks)
Deduce the electrode reactions in electroplating	HL only	"Write the half-equation at the cathode" (1 mark)

Before you start

You need ionic formulas and charges from Structure 2.1, and the group trends in atomic radius and ionization energy from Structure 3.1, because they explain which metals lose electrons easily. You need the functional groups and names of alcohols, aldehydes, ketones and carboxylic acids from Structure 3.2 for sections 8 and 9. HL sections 10 and 11 use ΔG⦵ from Reactivity 1.4.

1The idea in one paragraph

Some reactions work by moving protons; these work by moving electrons. The species that loses electrons is oxidized and the species that gains them is reduced, and the two always happen together, because electrons cannot vanish. Oxidation states are the bookkeeping that tracks where the electrons went, and half-equations write each side of the exchange separately. How willingly an element gives up or takes electrons follows from where it sits in the periodic table. If the two halves of a spontaneous redox reaction are placed in separate containers and joined by a wire, the electrons flow through the wire and you have a voltaic cell, a source of electricity. Push electricity the other way through a molten or dissolved ionic compound and you force a non-spontaneous redox reaction: electrolysis. The same ideas run organic chemistry: alcohols are oxidized to aldehydes, ketones and acids, and those are reduced back. At HL the tendency to gain electrons gets a number, the standard electrode potential E⦵, which predicts spontaneity, links to ΔG⦵, and decides what forms when you electrolyse a solution.

2Four ways to recognize oxidation and reduction

Oxidation is loss of electrons and an increase in oxidation state. Reduction is gain of electrons and a decrease in oxidation state.

The older definitions still work in the cases they were written for, and you should recognize all four:

	Oxidation	Reduction
Electrons	loss	gain
Oxidation state	increases	decreases
Oxygen	gain	loss
Hydrogen	loss	gain

The electron definition is the general one; the oxygen and hydrogen definitions are special cases that are useful in organic chemistry, where no ions are visible. "OIL RIG" (oxidation is loss, reduction is gain) is the usual way to remember the first row.

Oxidation states. An oxidation state is the charge an atom would have if all its bonds were ionic, with the shared electrons given to the more electronegative atom. It is written with the sign first: +2, −1. Deduce it with these rules, in this order of priority:

An uncombined element is 0 (Na, O₂, S₈).
The oxidation states in a neutral compound add up to 0; in an ion they add up to the charge.
Group 1 metals are +1, group 2 metals +2, and fluorine −1.
Hydrogen is +1, except in metal hydrides such as NaH, where it is −1.
Oxygen is −2, except in peroxides such as H₂O₂ (−1) and when bonded to fluorine.

Worked example 1. Find the oxidation state of Mn in KMnO₄, Cr in Cr₂O₇²⁻ and N in NH₄⁺.

KMnO₄: (+1) + Mn + 4(−2) = 0 → Mn = +7

Cr₂O₇²⁻: 2Cr + 7(−2) = −2 → 2Cr = +12 → Cr = +6

NH₄⁺: N + 4(+1) = +1 → N = −3

Variable oxidation states. Group 1 and 2 metals have one oxidation state each. Transition elements and most non-metals have several, which is why the same element can be oxidized or reduced in different reactions. Figure 1 lays out four common ladders.

Figure 1 · One element, many oxidation states

Naming with oxidation numbers. Where an element can take more than one oxidation state, the name gives it as a Roman numeral. For a metal cation it follows the metal: FeCl₂ is iron(II) chloride and FeCl₃ is iron(III) chloride; Cu₂O is copper(I) oxide. For an oxyanion it follows the anion's name: KMnO₄ is potassium manganate(VII), K₂Cr₂O₇ is potassium dichromate(VI), NaNO₂ is sodium nitrate(III) and NaNO₃ sodium nitrate(V). There is no gap before the bracket.

Oxidizing and reducing agents. An oxidizing agent oxidizes something else, so it takes electrons and is itself reduced. A reducing agent gives electrons and is itself oxidized. The names describe what the agent does to the other species, which is exactly why they feel back to front.

Worked example 2. In a blast furnace, Fe₂O₃ + 3CO → 2Fe + 3CO₂. Iron goes from +3 to 0, so iron(III) oxide is reduced and is the oxidizing agent. Carbon goes from +2 in CO to +4 in CO₂, so carbon monoxide is oxidized and is the reducing agent. Oxygen stays at −2 throughout and is neither.

Oxidation states are a model, and they have limits. They work well for tracking electrons in ions and simple molecules, but they are assigned by a rule, not measured: the carbon in CO₂ does not really carry a charge of +4. Treat them as bookkeeping. The surface oxidation of metals, corrosion, is redox too: iron rusting to hydrated iron(III) oxide weakens bridges and car bodies, which is why metals are painted, galvanized or given sacrificial blocks of a more reactive metal.

3Half-equations

A half-equation shows one side of a redox reaction, with the electrons written in. Oxidation half-equations have electrons on the right; reduction half-equations have them on the left.

Zn(s) → Zn²⁺(aq) + 2e⁻ (oxidation)
Cl₂(aq) + 2e⁻ → 2Cl⁻(aq) (reduction)

For an oxyanion in acidic solution, balance in four steps.

Worked example 3. Deduce the half-equation for MnO₄⁻ changing to Mn²⁺ in acid, then combine it with Fe²⁺ → Fe³⁺ + e⁻.

MnO₄⁻ → Mn²⁺1. balance the element that changes: Mn already 1 : 1

MnO₄⁻ → Mn²⁺ + 4H₂O2. balance O by adding H₂O

MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O3. balance H by adding H⁺

MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O4. balance charge: left +7, right +2, so add 5e⁻

5Fe²⁺ → 5Fe³⁺ + 5e⁻match the electrons: multiply by 5

MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O

Check the full equation twice: the atoms balance, and the charge balances (+17 on each side). The electrons cancel and never appear in the overall equation. Step 4 is also a check on the oxidation states: manganese falls from +7 to +2, which is five electrons.

In neutral solution the same method works, except that the H⁺ ions appear as a product rather than a reactant. Sulfur dioxide dissolved in water is oxidized to sulfate: SO₂(aq) + 2H₂O(l) → SO₄²⁻(aq) + 4H⁺(aq) + 2e⁻.

This reaction is the basis of a redox titration. Manganate(VII) is deep purple and Mn²⁺ is almost colourless, so as manganate(VII) is run into iron(II) solution it is decolourized drop by drop, and the first permanent pale pink shows that all the iron(II) has gone. The titration is self-indicating: the reagent is its own indicator. Like an acid–base indicator, the colour change marks the end point; unlike one, nothing needs to be added.

4Predicting reactivity from the periodic table

A metal reacts by losing electrons, so the easier it is to oxidize, the more reactive it is. Down groups 1 and 2, the outer electron is further from the nucleus and more shielded by inner shells, so it is lost more easily: potassium is oxidized more easily than sodium, and sodium more easily than lithium. Across a period, the nuclear charge rises and metals get harder to oxidize.

A halogen reacts by gaining electrons, so the easier it is to reduce, the more reactive it is. Down group 17 the incoming electron is further from the nucleus and more shielded, so it is attracted less strongly: fluorine is the strongest oxidizing agent and iodine the weakest. Figure 2 puts the two trends side by side.

Figure 2 · Down a group: metals oxidize more easily, halogens reduce less easily

Displacement reactions show these orders directly. A more easily oxidized metal gives electrons to the ions of a less easily oxidized one:

Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

You see a pink-brown coating of copper on the zinc, the blue colour of the solution fading, and the solution warming. Silver does nothing in zinc sulfate solution, because silver is harder to oxidize than zinc. A halogen that is more easily reduced takes electrons from the halide ions of one below it:

Cl₂(aq) + 2Br⁻(aq) → 2Cl⁻(aq) + Br₂(aq)

The colourless bromide solution turns yellow-orange as bromine forms. Bromine turns iodide solution brown (iodine), but iodine does nothing to bromide. You are not expected to memorize the order of metals; an exam gives you the data and asks you to deduce the order from it (Try it Q3).

Metals with acids. Metals that are more easily oxidized than hydrogen react with dilute hydrochloric and sulfuric acids to release hydrogen:

Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)
Zn(s) + H₂SO₄(aq) → ZnSO₄(aq) + H₂(g)
ionic equation for both: M(s) + 2H⁺(aq) → M²⁺(aq) + H₂(g)

The metal is oxidized and the hydrogen ions are reduced, so this is redox, not neutralization, even though a salt forms. Copper and silver do not react with dilute acids in this way, because H⁺ cannot oxidize them.

5Electrochemical cells: the voltaic cell

In a displacement reaction the electrons pass directly from zinc atoms to copper ions touching them, and the energy appears as heat. Separate the two halves and make the electrons go through a wire, and the same reaction drives a current. Any device that links a redox reaction and an electric current is an electrochemical cell. Two words apply to every kind:

Oxidation happens at the anode. Reduction happens at the cathode.

A primary or voltaic cell converts the energy of a spontaneous redox reaction into electrical energy. Figure 3 shows one built from two half-cells, each a metal dipping into a solution of its own ions.

Figure 3 · A zinc–copper voltaic cell

Anode. Zinc is the more easily oxidized metal, so zinc atoms lose electrons: Zn(s) → Zn²⁺(aq) + 2e⁻. The zinc electrode slowly dissolves.
Cathode. The electrons travel through the external wire to the copper electrode, where copper ions gain them: Cu²⁺(aq) + 2e⁻ → Cu(s). Copper plates onto the electrode.
Electron flow. Electrons flow from anode to cathode through the external circuit, never through the solution.
Polarity. The anode is where electrons are released, so in a voltaic cell it is the negative electrode; the cathode is positive.
Salt bridge. Without it, the zinc solution would build up positive charge and the copper solution would lose it, and the current would stop at once. A salt bridge, a tube or strip of paper soaked in an unreactive electrolyte such as KNO₃(aq), completes the circuit: anions move towards the anode half-cell to balance the new Zn²⁺, and cations move towards the cathode half-cell to replace the Cu²⁺ used up. Ions carry the current inside; electrons carry it outside.

Burning a fuel and running a cell are both exothermic redox reactions with a negative ΔG. The difference is where the electrons go: in combustion they move directly and the energy is released as heat, which then has to be turned into electricity inefficiently; in a cell they are made to do electrical work on the way.

6Primary, secondary and fuel cells

A primary cell uses a redox reaction that cannot practically be reversed; when the reactants are used up it is thrown away. A secondary cell (a rechargeable cell) uses reactions that can be reversed by passing a current through it the opposite way. The lead–acid battery in a car is one. On discharge:

negative electrode (anode): Pb(s) + SO₄²⁻(aq) → PbSO₄(s) + 2e⁻
positive electrode (cathode): PbO₂(s) + 4H⁺(aq) + SO₄²⁻(aq) + 2e⁻ → PbSO₄(s) + 2H₂O(l)

To deduce the charging reactions, reverse each half-equation. The electrode that was the anode during discharge now has reduction happening at it, so it becomes the cathode:

PbSO₄(s) + 2e⁻ → Pb(s) + SO₄²⁻(aq) (the same electrode, now the cathode)
PbSO₄(s) + 2H₂O(l) → PbO₂(s) + 4H⁺(aq) + SO₄²⁻(aq) + 2e⁻ (now the anode)

Rechargeable cells work because the products of discharge (here solid PbSO₄) stay on the electrodes, ready to be turned back. Lithium-ion and nickel–cadmium cells follow the same logic.

A fuel cell is a voltaic cell whose reactants are supplied continuously from outside. In a hydrogen fuel cell in acidic electrolyte: anode H₂(g) → 2H⁺(aq) + 2e⁻; cathode O₂(g) + 4H⁺(aq) + 4e⁻ → 2H₂O(l); overall 2H₂ + O₂ → 2H₂O.

	Advantages	Disadvantages
Primary cell	cheap, light, ready to use, holds its charge for years	single use, so more waste and more metal mining per unit of energy
Secondary cell	recharged hundreds of times, lower cost per use	higher initial cost; capacity falls with age; some contain toxic lead or cadmium
Fuel cell	runs as long as fuel is fed; hydrogen cells emit only water; more efficient than burning the fuel	hydrogen is hard to store and transport; platinum catalysts are expensive; hydrogen is mostly made from fossil fuels today

7Electrolytic cells and molten salts

An electrolytic cell converts electrical energy into chemical energy, forcing a non-spontaneous redox reaction to happen. It needs a DC power supply, two electrodes (usually inert, such as graphite or platinum) and an electrolyte: an ionic compound that is molten or dissolved, so its ions can move. A solid ionic compound does not conduct, because its ions are locked in the lattice.

Current is carried in two ways. In the wires, electrons flow. In the electrolyte, ions move: cations towards the negative electrode and anions towards the positive one. At each electrode surface, electrons pass between the electrode and the ions, and that is where the chemistry happens. Figure 4 shows molten lead(II) bromide.

Figure 4 · Electrolysis of molten lead(II) bromide

The power supply pulls electrons out of one electrode, making it positive, and pushes them into the other, making it negative. The rule for names does not change: oxidation at the anode, reduction at the cathode. So in an electrolytic cell:

anode (positive): anions lose electrons, 2Br⁻(l) → Br₂(g) + 2e⁻
cathode (negative): cations gain electrons, Pb²⁺(l) + 2e⁻ → Pb(l)

For any molten binary salt the products are the metal at the cathode and the non-metal at the anode. Molten sodium chloride gives sodium and chlorine: 2Na⁺ + 2e⁻ → 2Na and 2Cl⁻ → Cl₂ + 2e⁻.

	Voltaic cell	Electrolytic cell
Energy change	chemical → electrical	electrical → chemical
Reaction	spontaneous	non-spontaneous, driven
Anode	oxidation, negative	oxidation, positive
Cathode	reduction, positive	reduction, negative

The signs swap; the names do not. Decide by the reaction, never by the sign.

8Oxidizing alcohols

In organic chemistry it is easier to follow oxygen and hydrogen than electrons. The oxidizing agent is written as [O], and you are not asked to name it or give a mechanism. The standard reagent in the laboratory is acidified potassium dichromate(VI), which changes from orange to green as Cr₂O₇²⁻ is reduced to Cr³⁺: that colour change shows an alcohol has been oxidized.

Primary alcohols are oxidized in two steps, first to an aldehyde and then to a carboxylic acid. With butan-1-ol:

CH₃CH₂CH₂CH₂OH + [O] → CH₃CH₂CH₂CHO + H₂O (butanal)
CH₃CH₂CH₂CHO + [O] → CH₃CH₂CH₂COOH (butanoic acid)

Secondary alcohols are oxidized once, to a ketone: CH₃CH(OH)CH₂CH₃ + [O] → CH₃COCH₂CH₃ + H₂O (butan-2-ol to butanone). A ketone has no hydrogen on its carbonyl carbon, so it is not oxidized further under these conditions.

Tertiary alcohols, such as 2-methylpropan-2-ol, (CH₃)₃COH, are not oxidized under the same conditions. The carbon holding the OH has no hydrogen atom on it, and one must be removed to form C=O.

Which product you get from a primary alcohol depends on the apparatus, and Figure 5 shows both.

Figure 5 · Distillation takes the aldehyde out; reflux keeps everything in

To stop at the aldehyde, distil. Aldehyde molecules cannot hydrogen-bond to each other, so an aldehyde has a lower boiling point than both the alcohol and the acid. Heat gently with the oxidizing agent in a distillation set-up, and the aldehyde boils off as it forms and is condensed and collected before it can be oxidized further.
To reach the carboxylic acid, heat under reflux with excess oxidizing agent. The condenser is vertical, so every vapour condenses and drips back; nothing escapes, and the aldehyde stays in the flask until it is oxidized to the acid.

Controlled oxidation is not combustion. Combustion breaks every bond in the molecule and gives CO₂ and water; oxidation with [O] changes only the functional group and keeps the carbon chain intact.

9Reduction in organic chemistry

Reduction runs the same ladder downwards, and the reducing agent is written as [H]. Figure 6 shows carbon's oxidation state climbing from CH₄ (−4) through CH₃OH (−2), HCHO (0) and HCOOH (+2) to CO₂ (+4): every bond to oxygen raises it, and every bond to hydrogen lowers it.

Figure 6 · Carbon climbs the ladder as it is oxidized

Carboxylic acid → aldehyde → primary alcohol. CH₃COOH + 2[H] → CH₃CHO + H₂O, then CH₃CHO + 2[H] → CH₃CH₂OH. The aldehyde is an intermediate; with a strong reducing agent it is reduced straight on to the alcohol.
Ketone → secondary alcohol. CH₃COCH₃ + 2[H] → CH₃CH(OH)CH₃ (propanone to propan-2-ol).

The reducing agents used supply hydride ions, H⁻: a hydrogen atom with an extra electron. The hydride ion is attracted to the carbon of the C=O group, which carries a partial positive charge because oxygen is more electronegative, and bonds to it; the oxygen then picks up H⁺ to become –OH. The guide does not assess the names of these reagents or the mechanism.

Hydrogenation. Adding hydrogen across a C=C or C≡C bond, with a nickel catalyst and heat, lowers the degree of unsaturation:

CH₃CH=CH₂ + H₂ → CH₃CH₂CH₃ (propene to propane)
HC≡CH + H₂ → H₂C=CH₂, then H₂C=CH₂ + H₂ → CH₃CH₃ (ethyne to ethene to ethane)

It counts as reduction because each carbon gains hydrogen and its oxidation state falls (from −2 to −3 for each carbon of ethene). The same reaction is also an addition reaction; which name you use depends on what you are describing. Adding HBr to an alkene is called electrophilic addition rather than reduction because the carbons, taken together, gain one H and one Br and their total oxidation state does not change (Reactivity 3.4).

10HLStandard electrode potentials

SL students can skip to section 14.

Every half-cell has some tendency to gain electrons. It cannot be measured alone, because a single half-cell does not make a circuit, so each one is measured against a reference: the standard hydrogen electrode, shown in Figure 7. By convention

H⁺(aq) + e⁻ ⇌ ½H₂(g) E⦵ = 0.00 V

Figure 7 · Measuring E⦵ against the standard hydrogen electrode (HL)

It is platinum dipping into 1.00 mol dm⁻³ H⁺(aq) with hydrogen gas bubbled over it at 100 kPa, all at 298 K. Platinum is used because it is inert and conducts, so it can carry electrons to and from the H⁺/H₂ pair without taking part. The standard electrode potential, E⦵, of any half-cell is the voltage of a cell made from it and a standard hydrogen electrode under standard conditions (298 K, 1.00 mol dm⁻³ ions, 100 kPa gases), with the sign of the half-cell's own electrode. In Figure 7 zinc is the negative electrode and the meter reads 0.76 V, so E⦵(Zn²⁺/Zn) = −0.76 V.

The data booklet lists E⦵ values as reduction potentials, each written as oxidized form + electrons ⇌ reduced form. Values below are typical; always use the ones printed in your booklet.

More positive E⦵: the oxidized form is more easily reduced, a stronger oxidizing agent. More negative E⦵: the reduced form is more easily oxidized, a stronger reducing agent.

Figure 8 shows how to read the scale. Chlorine, near the top on the left, is a strong oxidizing agent; magnesium, near the bottom on the right, is a strong reducing agent. This is the periodic-table trend of section 4, now measured.

Figure 8 · Reading the E⦵ scale (HL)

11HLCell potential, spontaneity and ΔG⦵

The standard cell potential is the difference between the two electrode potentials:

E_cell⦵ = E⦵(half-cell where reduction happens) − E⦵(half-cell where oxidation happens)

For a spontaneous reaction E_cell⦵ is positive. So to test any proposed reaction, identify which species is reduced, take its E⦵, and subtract the E⦵ of the species that is oxidized. E⦵ values are never multiplied by the coefficients in the equation: a potential is a property of the half-cell, not of the amount.

Worked example 4. Decide whether each reaction is spontaneous under standard conditions. E⦵ / V: Fe³⁺/Fe²⁺ +0.77, I₂/I⁻ +0.54, Cu²⁺/Cu +0.34, H⁺/H₂ 0.00.

2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂Fe³⁺ reduced, I⁻ oxidized

E⦵_cell = 0.77 − 0.54 = +0.23 Vpositive: spontaneous

Cu + 2H⁺ → Cu²⁺ + H₂H⁺ reduced, Cu oxidized

E⦵_cell = 0.00 − 0.34 = −0.34 Vnegative: not spontaneous; the reverse is

This is why copper does not dissolve in dilute acid. A positive E_cell⦵ says the reaction is thermodynamically possible; like a negative ΔG, it says nothing about how fast.

ΔG⦵ from E⦵. Both quantities measure the same drive, so they are linked:

ΔG⦵ = −nFE_cell⦵

Here n is the number of moles of electrons transferred in the equation as written, and F is the Faraday constant, the charge on one mole of electrons, 96 500 C mol⁻¹ (take the value from the booklet). Because C × V = J, the answer comes out in J mol⁻¹. The minus sign makes the two tests agree: a positive E_cell⦵ gives a negative ΔG⦵, and both mean spontaneous.

Worked example 5. Calculate ΔG⦵ for the zinc–copper cell (E_cell⦵ = +1.10 V) and for the manganate(VII)–iron(II) reaction of worked example 3 (E⦵: MnO₄⁻/Mn²⁺ +1.51 V, Fe³⁺/Fe²⁺ +0.77 V).

Zn + Cu²⁺ → Zn²⁺ + Cu: n = 2

ΔG⦵ = −2 × 96 500 × 1.10 = −212 300 J mol⁻¹ = −212 kJ mol⁻¹

MnO₄⁻ + 8H⁺ + 5Fe²⁺ → ...: E⦵_cell = 1.51 − 0.77 = +0.74 V, n = 5

ΔG⦵ = −5 × 96 500 × 0.74 = −357 000 J mol⁻¹ = −357 kJ mol⁻¹

Take n from the balanced equation, not from either half-equation alone: in the second reaction five electrons pass for each MnO₄⁻.

12HLElectrolysis of aqueous solutions

In a solution there is always a second candidate at each electrode: water. It can be reduced at the cathode or oxidized at the anode, and whichever reaction is easier wins.

at the cathode: 2H₂O(l) + 2e⁻ → H₂(g) + 2OH⁻(aq) E⦵ = −0.83 V
at the anode: 2H₂O(l) → O₂(g) + 4H⁺(aq) + 4e⁻ (reverse of the half-equation with E⦵ = +1.23 V)

At the cathode the species with the more positive E⦵ is reduced. Cu²⁺ (+0.34 V) and Ag⁺ (+0.80 V) are above water, so the metal is deposited. Na⁺ (−2.71 V), K⁺, Mg²⁺ and Al³⁺ are far below it, so hydrogen forms instead and these ions stay in solution.

At the anode the species that is most easily oxidized, the one whose reduction half-equation has the less positive E⦵, is oxidized. Sulfate and nitrate ions are not oxidized at all in aqueous solution, so oxygen forms. Chloride is the interesting case, because its E⦵ (+1.36 V) is only a little above water's (+1.23 V). The values are close enough that concentration decides: in concentrated chloride solution the chloride ions win and chlorine forms; in dilute solution mostly oxygen forms. Figure 9 draws the whole decision.

Figure 9 · Predicting the products of aqueous electrolysis (HL)

Water itself needs a little dissolved electrolyte to conduct, such as dilute sulfuric acid or sodium sulfate, whose ions are not discharged. Hydrogen forms at the cathode and oxygen at the anode, in a 2 : 1 volume ratio: 2H₂O(l) → 2H₂(g) + O₂(g).

The guide limits the effects of concentration and of the electrode to two solutions.

Electrolyte and electrodes	Cathode (−)	Anode (+)	What you see
concentrated NaCl(aq), inert	2H₂O + 2e⁻ → H₂ + 2OH⁻	2Cl⁻ → Cl₂ + 2e⁻	solution around the cathode becomes alkaline (NaOH)
dilute NaCl(aq), inert	2H₂O + 2e⁻ → H₂ + 2OH⁻	2H₂O → O₂ + 4H⁺ + 4e⁻ (mostly)	little or no chlorine
CuSO₄(aq), inert (graphite or Pt)	Cu²⁺ + 2e⁻ → Cu	2H₂O → O₂ + 4H⁺ + 4e⁻	copper plates out; blue fades; solution turns acidic
CuSO₄(aq), copper electrodes	Cu²⁺ + 2e⁻ → Cu	Cu → Cu²⁺ + 2e⁻	anode dissolves, cathode grows, blue colour stays constant

The last row shows the effect of the electrode. A copper anode is itself easier to oxidize (E⦵ +0.34 V) than water or sulfate, so the anode dissolves instead of making oxygen. Each Cu²⁺ ion made at the anode replaces one removed at the cathode, so [Cu²⁺] stays the same. This is how impure copper is purified: the impure metal is the anode and pure copper builds up on the cathode.

13HLElectroplating

Electroplating coats an object with a thin layer of metal by electrolysis, to stop corrosion or to improve its look. It is the copper-electrode cell of section 12 with the object in place of the cathode. Figure 10 shows a steel key being plated with copper.

Figure 10 · Copper-plating a steel key (HL)

The object to be plated is the cathode (negative), because metal ions are reduced onto it.
The anode is made of the plating metal, which dissolves to replace the ions used.
The electrolyte contains ions of the plating metal.

For copper: anode Cu(s) → Cu²⁺(aq) + 2e⁻; cathode Cu²⁺(aq) + 2e⁻ → Cu(s). For silver plating, with a silver anode and a silver salt solution: Ag(s) → Ag⁺(aq) + e⁻ and Ag⁺(aq) + e⁻ → Ag(s). In practice the object must be cleaned and degreased first, and a small current over a longer time gives an even layer that sticks; a large current gives a rough, powdery coat.

14Where marks are lost

Getting the agents back to front. The oxidizing agent is the species that is reduced. Name the species, not the element alone, and give the change in oxidation state as the reason.

Writing oxidation states as charges. Write +6, not 6+ or 6. Roman numerals go in names only: iron(III) chloride.

Unbalanced charge in a half-equation. Atoms balancing is not enough; the total charge on each side must match. Check it as the last step, every time.

Leaving electrons in an overall equation. Scale the half-equations so the electrons cancel before adding them.

Deciding anode and cathode by sign. Oxidation is always at the anode, reduction always at the cathode. The anode is negative in a voltaic cell and positive in an electrolytic cell.

Saying electrons flow through the salt bridge or the electrolyte. Electrons flow only in the external circuit; ions carry the current in solution.

Oxidizing a primary alcohol under reflux when the aldehyde is wanted. Reflux gives the carboxylic acid. Distil to collect the aldehyde.

(HL) Multiplying E⦵ by a coefficient, or dropping n from ΔG⦵ = −nFE⦵. E⦵ is never scaled; n comes from the balanced equation. ΔG⦵ comes out in J mol⁻¹, so divide by 1000 for kJ mol⁻¹.

15Draw it right

Two apparatus diagrams come up again and again: the voltaic cell and the electrolytic cell.

Voltaic cell: two separate half-cells, each a metal in a solution of its own ions, with concentrations and states labelled (for example Zn(s) in Zn²⁺(aq)).
Join the electrodes with a wire through a voltmeter (or a bulb), and join the solutions with a labelled salt bridge dipping into both.
Label the anode and cathode by the reaction, and give their signs: anode negative, cathode positive.
Arrow the electron flow in the wire from anode to cathode, and the ion flow in the salt bridge: anions to the anode side, cations to the cathode side.
Electrolytic cell: one container of molten or aqueous electrolyte, two electrodes, and a DC supply drawn with its long (+) and short (−) plates.
Label the anode (connected to +) and the cathode (connected to −), and write the half-equation at each.
Distillation and reflux: the condenser slopes down into a collecting vessel for distillation and stands vertically, open at the top, for reflux. Water enters the condenser at the lower end.
(HL) A standard hydrogen electrode needs platinum, H₂(g) at 100 kPa, 1.00 mol dm⁻³ H⁺(aq), and 298 K, all labelled.

16Try it

Marks in brackets. Answers and marker's notes are at the end.

Q1. Oxidation states. 5 marks

(a) Deduce the oxidation state of: chromium in CrO₄²⁻; nitrogen in NO₂⁻; sulfur in S₂O₃²⁻; manganese in MnO₂. 2 marks

(b) State the name of Fe₂(SO₄)₃, using an oxidation number. 1 mark

(c) In the reaction 3Cu + 8H⁺ + 2NO₃⁻ → 3Cu²⁺ + 2NO + 4H₂O, identify the oxidizing agent, giving the change in oxidation state as your reason. 2 marks

Q2. Acidified hydrogen peroxide oxidizes iodide ions to iodine. 3 marks

(a) Deduce the half-equation for the reduction of H₂O₂ to water in acidic solution. 1 mark

(b) Deduce the overall ionic equation. 1 mark

Q3. (Data-based, invented data.) Three metals, P, Q and R, each form 2+ ions. A student adds a piece of each metal to solutions of the nitrates of the other two. ✓ means a reaction was seen; ✗ means none. 5 marks

	P²⁺(aq)	Q²⁺(aq)	R²⁺(aq)
P(s)	—	✓	✓
Q(s)	✗	—	✗
R(s)	✗	✓	—

(a) Deduce the order of the metals from most to least easily oxidized, explaining your reasoning. 2 marks

(b) Formulate the ionic equation for the reaction between R and Q²⁺(aq). 1 mark

(c) A voltaic cell is made from P in P²⁺(aq) and Q in Q²⁺(aq). Identify the anode and state the direction of electron flow in the external circuit. 2 marks

Q4. A nickel–cadmium rechargeable cell discharges by these reactions: Cd(s) + 2OH⁻(aq) → Cd(OH)₂(s) + 2e⁻ and NiO(OH)(s) + H₂O(l) + e⁻ → Ni(OH)₂(s) + OH⁻(aq). 4 marks

(a) Identify the anode during discharge. 1 mark

(b) Deduce the reaction at each electrode during charging. 2 marks

Q5. Propan-1-ol is oxidized by acidified potassium dichromate(VI). 5 marks

(a) Deduce the equation, using [O], for the formation of the organic product collected by distillation. 1 mark

(b) State the product if the mixture is instead heated under reflux with excess oxidizing agent, and explain why the apparatus changes the product. 2 marks

(d) Deduce the product formed when butanone is reduced. 1 mark

Q6 (HL). Use E⦵ / V: Ag⁺/Ag +0.80, Ni²⁺/Ni −0.26, Br₂/Br⁻ +1.07, Cl₂/Cl⁻ +1.36; F = 96 500 C mol⁻¹. 6 marks

(a) Calculate E_cell⦵ for a cell made from Ni/Ni²⁺ and Ag/Ag⁺ half-cells, and identify the positive electrode. 2 marks

(b) Calculate ΔG⦵, in kJ mol⁻¹, for Ni(s) + 2Ag⁺(aq) → Ni²⁺(aq) + 2Ag(s). 2 marks

Q7 (HL). Electrolysis of solutions. 5 marks

(a) Concentrated sodium chloride solution is electrolysed with inert electrodes. State the half-equation at each electrode. 2 marks

(b) Explain why the anode product changes when the solution is very dilute. 1 mark

(c) A student wants to silver-plate a copper medal. State what the anode and the electrolyte should be, and write the half-equation at the cathode. 2 marks

17In one breath

Oxidation is loss of electrons and a rise in oxidation state; reduction is gain and a fall; oxygen gain and hydrogen loss are older special cases. Oxidation states add to zero in a compound or to the charge in an ion, and Roman numerals put them in names. The oxidizing agent is reduced; the reducing agent is oxidized. Half-equations balance the element, then O with H₂O, then H with H⁺, then charge with electrons, and are scaled so the electrons cancel. Metals oxidize more easily down a group, halogens reduce less easily, and displacement data give the order; reactive metals give hydrogen with dilute acids. Oxidation is at the anode, reduction at the cathode: in a voltaic cell the anode is negative, electrons run through the wire from anode to cathode and ions through the salt bridge; in an electrolytic cell a DC supply drives a non-spontaneous reaction, the anode is positive, and a molten salt gives metal at the cathode and non-metal at the anode. Secondary cells recharge by reversing their half-equations; fuel cells are fed continuously. Primary alcohols give aldehydes (distil) then acids (reflux), secondary give ketones, tertiary resist; [H] runs the ladder back down using hydride ions, and hydrogen adds across C=C and C≡C. HL: E⦵ is measured against the hydrogen electrode, more positive means a stronger oxidizing agent, E_cell⦵ = E⦵(reduced) − E⦵(oxidized) must be positive for a spontaneous reaction, ΔG⦵ = −nFE_cell⦵, water competes in aqueous electrolysis so the easier reaction wins at each electrode, concentrated chloride gives chlorine, a copper anode dissolves, and electroplating makes the object the cathode.

Answers

Q1. (a) Cr in CrO₄²⁻: Cr + 4(−2) = −2, so +6. N in NO₂⁻: N + 2(−2) = −1, so +3. S in S₂O₃²⁻: 2S + 3(−2) = −2, so each S is +2 (an average). Mn in MnO₂: +4. (b) Iron(III) sulfate. (c) The nitrate ion, NO₃⁻, is the oxidizing agent: nitrogen goes from +5 in NO₃⁻ to +2 in NO, so nitrate is reduced. 2 for all four oxidation states, 1 for three; 1 for the name with (III); 1 for NO₃⁻, 1 for +5 → +2. "Nitrogen" alone as the agent scores the second mark only if the oxidation states are given; "H⁺" scores 0.

Q2. (a) H₂O₂(aq) + 2H⁺(aq) + 2e⁻ → 2H₂O(l). (b) With 2I⁻ → I₂ + 2e⁻: H₂O₂ + 2H⁺ + 2I⁻ → 2H₂O + I₂. (c) Oxygen goes from −1 in H₂O₂ to −2 in H₂O. 1 for each. Electrons left in the overall equation score 0 for (b); "−2 to −1" scores 0 for (c).

Q3. (a) P > R > Q. P displaces both Q and R from solution, so it gives up electrons most readily. R displaces Q but not P. Q displaces neither. (b) R(s) + Q²⁺(aq) → R²⁺(aq) + Q(s). (c) P is the anode, because it is more easily oxidized; electrons flow through the wire from P to Q. 1 for the order, 1 for reasoning from the table; 1 for the equation; 1 for P as anode with reason, 1 for the direction. ECF from a wrong order in (a).

Q4. (a) The cadmium electrode (it is oxidized). (b) Reverse each half-equation. At the cadmium electrode: Cd(OH)₂(s) + 2e⁻ → Cd(s) + 2OH⁻(aq). At the nickel electrode: Ni(OH)₂(s) + OH⁻(aq) → NiO(OH)(s) + H₂O(l) + e⁻. (c) Any one: it must be recharged rather than refuelled and has a fixed capacity; cadmium is toxic; its capacity falls after many cycles. 1 for cadmium; 1 for each charging equation; 1 for a valid disadvantage. An equation with electrons on the wrong side scores 0.

Q5. (a) CH₃CH₂CH₂OH + [O] → CH₃CH₂CHO + H₂O (propanal). (b) Propanoic acid, CH₃CH₂COOH. Under reflux the condenser returns all vapour to the flask, so the propanal cannot escape and is oxidized further; in distillation it boils off as it forms because it has a lower boiling point (no hydrogen bonding between its molecules). (c) The carbon bearing the –OH group has no hydrogen atom, and one must be removed to form C=O. (d) Butan-2-ol, CH₃CH(OH)CH₂CH₃. 1 for the balanced equation; 1 for propanoic acid, 1 for the reflux explanation; 1 for no H on that carbon; 1 for butan-2-ol. "Tertiary alcohols are too stable" scores 0 in (c).

Q6 (HL). (a) Ag⁺ is reduced and Ni is oxidized: E_cell⦵ = 0.80 − (−0.26) = +1.06 V. The silver electrode is positive (the cathode). (b) n = 2. ΔG⦵ = −2 × 96 500 × 1.06 = −204 580 J mol⁻¹ = −205 kJ mol⁻¹. (c) For Br₂ + 2Cl⁻ → 2Br⁻ + Cl₂, bromine would be reduced and chloride oxidized: E_cell⦵ = 1.07 − 1.36 = −0.29 V. Negative, so not spontaneous: bromine does not oxidize chloride. M1 A1 for +1.06 V with silver positive; M1 for n = 2 in −nFE, A1 for −205 kJ mol⁻¹ (−204 580 J mol⁻¹ accepted, but a J value labelled kJ scores A0); M1 for the subtraction the right way round, A1 for the conclusion. Doubling the Ag⁺ potential scores M0 in (a).

Q7 (HL). (a) Cathode: 2H₂O(l) + 2e⁻ → H₂(g) + 2OH⁻(aq). Anode: 2Cl⁻(aq) → Cl₂(g) + 2e⁻. (b) The E⦵ values for oxidizing water (+1.23 V) and chloride (+1.36 V) are close; at low [Cl⁻] the oxidation of water to oxygen is favoured, so mostly oxygen forms. (c) Anode: pure silver; electrolyte: a solution of a silver salt, such as silver nitrate; cathode (the medal): Ag⁺(aq) + e⁻ → Ag(s). 1 for each half-equation in (a) (Na⁺ + e⁻ → Na scores 0); 1 for the concentration argument; 1 for silver anode and silver-ion electrolyte together, 1 for the cathode equation.

Educerie · written from the published IB Diploma Programme Chemistry guide, first assessment 2025, section Reactivity 3.2 Electron transfer reactions. Original text, examples and questions. Diagrams drawn by Educerie. Last reviewed 25 September 2026.

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