DP Chemistry HL R3.2 Electron Transfer Reactions

DP Chemistry HL R3.2 Electron Transfer Reactions - Full notes

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DP Chemistry HL | R3.2 Electron Transfer Reactions

DP CHEMISTRY · REACTIVITY

R3.2

Electron Transfer Reactions

Full notes for HL · IB-style

Contents

1	Oxidation and reduction
2	Half-equations and balancing redox equations
3	The activity series and reactions of metals
4	Voltaic (galvanic) cells
5	Primary and secondary cells
6	Electrolytic cells
7	Redox in organic chemistry
8	Standard electrode potentials (HL only)
9	Electrolysis of aqueous solutions and electroplating (HL only)
10	Oxidation and reduction: three definitions
11	Oxidation states: practice
12	Oxidizing and reducing agents
13	Half-equations
14	Balancing half-equations in acid
15	Combining half-equations
16	The reactivity series
17	Halogens as oxidizing agents
18	Voltaic (galvanic) cells
19	Electrolytic cells
20	Comparing voltaic and electrolytic cells
21	Redox in everyday life
22	Standard electrode potentials (HL)
23	Mixed review: worked problems
24	Additional exam-style practice
25	Electrolysis of aqueous solutions (introduction)
26	Common misconceptions and links
27	Exam-style question with markscheme
28	Check your understanding
29	Summary and topic map

These are full teaching notes. Read each section in order, try the Think about it question before you read on, and do every Quick check before moving to the next section.

Before you start

Rate yourself on each statement now, and again after you've worked through the notes. Shade R (not yet), A (getting there) or G (confident).

I can…	R	A	G
define oxidation and reduction in terms of electrons, oxygen, hydrogen and oxidation states
identify oxidizing and reducing agents
write and balance half-equations and redox equations in acidic solution
use the activity series to predict reactions of metals
describe voltaic cells and the direction of electron flow
compare primary cells, secondary cells and fuel cells
describe electrolysis of molten salts
describe the oxidation of alcohols and reduction of carbonyl compounds
use standard electrode potentials to calculate E⊖cell and predict spontaneity
predict products of electrolysis of aqueous solutions and explain electroplating

Sections marked HL only (Sections 8 and 9) are Additional Higher Level content. Every other section is common to SL and HL.

Guiding questions

What happens when electrons are transferred between species? How can redox reactions be used to generate electricity or to drive non-spontaneous change?

How this topic is examined

R3.2 is examined through oxidation states, balancing redox equations, cell diagrams and electrolysis products, and oxidation of alcohols. HL questions include E⊖ calculations, aqueous electrolysis and electroplating.

Remember OIL RIG: Oxidation Is Loss of electrons, Reduction Is Gain.

Command term	What the examiner wants
State / Identify	a brief answer, no explanation
Outline / Describe	a brief or detailed account, without reasons
Explain	a detailed account with causes, reasons or mechanisms
Calculate / Determine	a numerical answer with working, correct significant figures and units
Deduce / Predict	reach a conclusion from given information
Draw / Sketch	a clear labelled diagram, structure or graph
Evaluate / Discuss	weigh strengths and limitations; reach a judgement

How to use these notes

Each section teaches one part of the syllabus in full. You'll see these boxes throughout; each has one job:

Box	What it's for
Think about it	a question to get you thinking before a section; the answer comes in the section
Key point	the one idea you must not forget
Common mistake	an error examiners see every year, and how to avoid it
In the real world	where the chemistry shows up outside the classroom
Try this	something quick you can do yourself to test the idea
Worked example	full solutions with IB-style marks (M1 method, A1 answer, R1 reasoning)
Quick check	short questions to test yourself at the end of each section

1 Oxidation and reduction

Electron loss and electron gain always happen together.

Think about it

When an iron nail is placed in copper sulfate solution, it becomes coated in copper and the solution fades. Which substance has lost electrons, and which has gained them?

Oxidation is the loss of electrons, an increase in oxidation state, gain of oxygen or loss of hydrogen. Reduction is the gain of electrons, a decrease in oxidation state, loss of oxygen or gain of hydrogen. Oxidation and reduction always occur together: a redox reaction.

An oxidizing agent (oxidant) accepts electrons and is itself reduced. A reducing agent (reductant) donates electrons and is itself oxidized.

In Fe(s) + CuSO4(aq) → FeSO4(aq) + Cu(s): iron goes from 0 to +2 (oxidized; reducing agent); copper goes from +2 to 0 (reduced); Cu2+ is the oxidizing agent.

Common mistake

Saying the oxidizing agent is oxidized. The oxidizing agent causes oxidation of something else and is itself reduced.

Quick check

1. Define oxidation in terms of oxidation state.

2. In Zn + 2H+ → Zn2+ + H2, which is the reducing agent?

Answers: 1. increase in oxidation state 2. Zn

2 Half-equations and balancing redox equations

Balance atoms, then charge with electrons.

A half-equation shows the oxidation or reduction separately, with electrons. Oxidation: Fe → Fe2+ + 2e−. Reduction: Cu2+ + 2e− → Cu.

Balancing in acidic solution: (1) balance the atoms other than O and H; (2) balance O by adding H2O; (3) balance H by adding H+; (4) balance charge by adding electrons. To combine half-equations, multiply so that electrons lost = electrons gained, then add and cancel.

WORKED EXAMPLE 1 Balancing a redox equation

Write the overall equation for the reaction of MnO4− with Fe2+ in acidic solution.

Solution

Reduction: MnO4− + 8H+ + 5e− → Mn2+ + 4H2OM1

Oxidation: Fe2+ → Fe3+ + e− (multiply by 5)M1

MnO4− + 8H+ + 5Fe2+ → Mn2+ + 4H2O + 5Fe3+A1

Redox titrations use such reactions; potassium manganate(VII) acts as its own indicator (purple to colourless), and iodine-thiosulfate titrations use starch.

Quick check

1. What is added to balance O in acidic half-equations?

2. Write the half-equation for Cl2 forming Cl−.

Answers: 1. H2O 2. Cl2 + 2e− → 2Cl−

3 The activity series and reactions of metals

More reactive metals are stronger reducing agents.

The activity series ranks metals by their tendency to lose electrons (be oxidized): K > Na > Ca > Mg > Al > Zn > Fe > Pb > (H) > Cu > Ag > Au. More reactive metals are stronger reducing agents.

A more reactive metal displaces a less reactive metal from a solution of its ions: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s). Metals above hydrogen react with dilute acids to release hydrogen: Mg + 2H+ → Mg2+ + H2; metals below hydrogen (Cu, Ag, Au) do not.

Halogens also form an activity series as oxidizing agents (S3.1): F2 > Cl2 > Br2 > I2.

Quick check

1. Will copper displace silver from silver nitrate?

2. Why doesn't copper react with dilute HCl?

Answers: 1. yes: Cu more reactive than Ag 2. Cu is below hydrogen in the activity series

4 Voltaic (galvanic) cells

Spontaneous redox produces electricity.

A voltaic (galvanic) cell converts chemical energy into electrical energy using a spontaneous redox reaction. The two half-reactions occur in separate half-cells, each a metal electrode in a solution of its ions, connected by a wire (electrons flow) and a salt bridge (ions flow, completing the circuit and maintaining charge balance).

In the Daniell cell (Zn | Zn2+ || Cu2+ | Cu): zinc, the more reactive metal, is oxidized at the anode, which is the negative electrode: Zn → Zn2+ + 2e−. Copper(II) ions are reduced at the cathode, the positive electrode: Cu2+ + 2e− → Cu. Electrons flow through the external wire from anode to cathode (zinc to copper). In the salt bridge, anions move towards the anode and cations towards the cathode.

The cell voltage depends on the difference in reactivity of the two metals: the further apart in the activity series, the larger the voltage.

Figure 1. A Daniell cell: electrons flow from zinc (anode) to copper (cathode).

Key point

In every cell: oxidation at the anode, reduction at the cathode (An Ox, Red Cat). In a voltaic cell the anode is negative.

Quick check

1. What is the role of the salt bridge?

2. In which direction do electrons flow in the Daniell cell?

Answers: 1. completes circuit; ions move to balance charge 2. from Zn to Cu through the wire

5 Primary and secondary cells

Disposable, rechargeable and continuous.

Primary cells (such as alkaline zinc-manganese dioxide batteries) cannot be recharged: the reaction is not easily reversed. Secondary (rechargeable) cells, such as lithium-ion and lead-acid batteries, can be recharged by applying an external current that reverses the reactions. Fuel cells (R1.3) produce electricity continuously as long as fuel and oxidant are supplied.

Lithium-ion batteries power phones and electric cars: lithium ions move between electrodes through an electrolyte during discharge and charging. Lithium is used because it is light and highly reactive, giving a high voltage and energy density.

Type	Rechargeable?	Example	Advantage / disadvantage
Primary	no	alkaline battery	cheap; disposed of after use
Secondary	yes	lithium-ion, lead-acid	reusable; costly; limited cycles
Fuel cell	fuel supplied	hydrogen cell	efficient; fuel storage

6 Electrolytic cells

Electricity drives non-spontaneous reactions.

An electrolytic cell uses electrical energy from an external power supply to drive a non-spontaneous redox reaction: electrolysis. The electrolyte is an ionic compound that is molten or dissolved, so ions can move. The anode is connected to the positive terminal: oxidation occurs there (anions lose electrons). The cathode is negative: reduction occurs (cations gain electrons).

Electrolysis of molten sodium chloride: at the cathode: Na+ + e− → Na(l); at the anode: 2Cl− → Cl2(g) + 2e−. Electrons flow from the anode through the external circuit to the cathode; in the electrolyte, ions carry the charge.

Electrolysis is used to extract reactive metals such as aluminium and sodium, which cannot be obtained by reduction with carbon.

Figure 2. Electrolysis of molten sodium chloride.

	Voltaic cell	Electrolytic cell
Energy conversion	chemical to electrical	electrical to chemical
Reaction	spontaneous	non-spontaneous
Anode	negative; oxidation	positive; oxidation
Cathode	positive; reduction	negative; reduction

Quick check

1. Which electrode is positive in electrolysis?

2. Write the half-equation at the cathode in molten PbBr2.

Answers: 1. the anode 2. Pb2+ + 2e− → Pb

7 Redox in organic chemistry

Oxidizing alcohols and reducing carbonyls.

Oxidation of alcohols uses an oxidizing agent such as acidified potassium dichromate(VI), which changes colour from orange (Cr2O72−) to green (Cr3+). Primary alcohols are oxidized first to aldehydes and then to carboxylic acids. To obtain the aldehyde, distil it off as it forms (aldehydes have lower boiling points, no hydrogen bonding). To obtain the carboxylic acid, heat under reflux with excess oxidizing agent. Secondary alcohols are oxidized to ketones. Tertiary alcohols are not oxidized (no H on the carbon bearing OH), so the dichromate stays orange.

Reduction: carboxylic acids, aldehydes and ketones can be reduced back to alcohols using reducing agents such as lithium aluminium hydride (LiAlH4) or sodium borohydride (NaBH4); alkenes are reduced to alkanes by hydrogen with a nickel catalyst (hydrogenation).

Figure 3. Oxidation of a primary alcohol: ethanol → ethanal → ethanoic acid.

Quick check

1. What is the colour change when dichromate oxidizes an alcohol?

2. Why is a tertiary alcohol not oxidized?

Answers: 1. orange to green 2. no hydrogen on the carbon bonded to OH

8 Standard electrode potentials (HL only)

Measuring the tendency to be reduced.

HL only

This section is Additional Higher Level content. SL students do not need it.

The standard electrode potential (E⊖) of a half-cell is its voltage measured against the standard hydrogen electrode (SHE, E⊖ = 0.00 V) under standard conditions (298 K, 100 kPa, 1 mol dm−3). Half-equations are written as reductions. A more positive E⊖ means the species is more easily reduced (stronger oxidizing agent); a more negative E⊖ means the metal is more easily oxidized (stronger reducing agent).

E⊖cell = E⊖(cathode) − E⊖(anode): the half-cell with the more positive value is the cathode (reduction). A positive E⊖cell means the reaction is spontaneous. It is related to Gibbs energy by ΔG⊖ = −nFE⊖cell, where n is the number of electrons transferred and F = 96 500 C mol−1.

Half-equation	E⊖ / V
Zn2+ + 2e− ⇌ Zn	−0.76
Fe2+ + 2e− ⇌ Fe	−0.45
2H+ + 2e− ⇌ H2	0.00
Cu2+ + 2e− ⇌ Cu	+0.34
Ag+ + e− ⇌ Ag	+0.80
Cl2 + 2e− ⇌ 2Cl−	+1.36

WORKED EXAMPLE 2 Cell potential and ΔG

Calculate E⊖cell and ΔG⊖ for the Daniell cell.

Solution

E⊖cell = +0.34 − (−0.76) = +1.10 VA1

ΔG = −nFE = −2 × 96 500 × 1.10 = −212 300 JM1

= −212 kJ mol−1: spontaneousA1

9 Electrolysis of aqueous solutions and electroplating (HL only)

Water competes with the ions.

HL only

This section is Additional Higher Level content. SL students do not need it.

In aqueous solutions, water can also be oxidized or reduced: at the cathode, 2H2O + 2e− → H2 + 2OH−; at the anode, 2H2O → O2 + 4H+ + 4e−. Which product forms depends on E⊖ values, concentration and the electrode material.

At the cathode: if the metal ion is less easily reduced than water (reactive metals such as Na+, K+, Mg2+), hydrogen forms; ions of less reactive metals (Cu2+, Ag+) are reduced to the metal. At the anode: with concentrated chloride solutions, chlorine forms; with dilute chloride or sulfate and nitrate solutions, oxygen forms.

Electroplating coats an object with a thin layer of metal: the object is the cathode, the anode is the plating metal, and the electrolyte contains ions of the plating metal. For copper plating: at the cathode Cu2+ + 2e− → Cu; at the copper anode Cu → Cu2+ + 2e−; the concentration of Cu2+ stays constant.

Electrolyte (aqueous)	Cathode product	Anode product
concentrated NaCl	H2	Cl2
dilute NaCl	H2	O2
CuSO4 (inert electrodes)	Cu	O2
H2SO4 (dilute)	H2	O2

Quick check

1. What forms at the cathode in electrolysis of aqueous NaCl?

2. In electroplating, which electrode is the object to be plated?

Answers: 1. hydrogen 2. the cathode

10 Oxidation and reduction: three definitions

Oxygen, electrons, oxidation states.

Oxidation: gain of oxygen, loss of electrons, or increase in oxidation state. Reduction: loss of oxygen, gain of electrons, or decrease in oxidation state ('OIL RIG'). Oxidation and reduction always occur together (redox).

WORKED EXAMPLE 3 Identifying (standard)

In Zn + Cu2+ → Zn2+ + Cu, identify what is oxidized and reduced.

Solution

Zn oxidized (loses 2e−; 0 → +2)A1

Cu2+ reduced (gains 2e−; +2 → 0)A1

Quick check

1. Identify the species reduced in 2Fe3+ + 2I− → 2Fe2+ + I2.

2. Is Cl2 → 2Cl− oxidation or reduction?

Answers: 1. Fe3+ 2. reduction

11 Oxidation states: practice

Systematic assignment.

Rules: elements 0; F −1; O −2 (except peroxides −1, OF2 +2); H +1 (−1 in metal hydrides); group 1 +1, group 2 +2; sum = overall charge.

WORKED EXAMPLE 4 Assigning (standard)

Find the oxidation state of Cl in ClO3−, ClO− and Cl2O7.

Solution

ClO3−: +5; ClO−: +1A1

Cl2O7: +7A1

Quick check

1. Find the oxidation state of N in NO2, N2O and HNO3.

2. Find the oxidation state of O in H2O2.

Answers: 1. +4, +1, +5 2. −1

12 Oxidizing and reducing agents

Agents are changed oppositely.

An oxidizing agent causes oxidation and is itself reduced (gains electrons). A reducing agent causes reduction and is itself oxidized. Common oxidizing agents: O2, Cl2, MnO4−/H+, Cr2O72−/H+. Common reducing agents: reactive metals, C, CO, H2.

WORKED EXAMPLE 5 Agents (standard)

Identify the oxidizing and reducing agents in Fe2O3 + 3CO → 2Fe + 3CO2.

Solution

Oxidizing agent: Fe2O3 (Fe +3 → 0)A1

Reducing agent: CO (C +2 → +4)A1

Quick check

1. Identify the reducing agent in Mg + 2HCl → MgCl2 + H2.

Answers: 1. Mg

13 Half-equations

Showing electron transfer.

Split a redox reaction into an oxidation half-equation (electrons on the right) and a reduction half-equation (electrons on the left). Balance atoms and charges in each.

WORKED EXAMPLE 6 Half-equations (standard)

Write half-equations for Mg + 2Ag+ → Mg2+ + 2Ag.

Solution

Mg → Mg2+ + 2e−A1

Ag+ + e− → AgA1

Quick check

1. Write the half-equation for chlorine forming chloride ions.

2. Write the half-equation for Fe2+ forming Fe3+.

Answers: 1. Cl2 + 2e− → 2Cl− 2. Fe2+ → Fe3+ + e−

14 Balancing half-equations in acid

O with water, H with H+.

1. Balance the atom being oxidized or reduced. 2. Balance O with H2O. 3. Balance H with H+. 4. Balance charge with electrons.

WORKED EXAMPLE 7 Dichromate (exam level)

Balance the half-equation Cr2O72− → Cr3+ in acidic solution.

Solution

Cr2O72− → 2Cr3+ + 7H2OM1

+ 14H+M1

Cr2O72− + 14H+ + 6e− → 2Cr3+ + 7H2OA1

Quick check

1. Balance MnO4− → Mn2+ in acid.

2. Balance H2O2 → O2 in acid.

Answers: 1. MnO4− + 8H+ + 5e− → Mn2+ + 4H2O 2. H2O2 → O2 + 2H+ + 2e−

15 Combining half-equations

Equalize the electrons.

Multiply the half-equations so that the electrons lost equal the electrons gained, add them, and cancel species appearing on both sides.

WORKED EXAMPLE 8 Full equation (exam level)

Combine MnO4− + 8H+ + 5e− → Mn2+ + 4H2O with Fe2+ → Fe3+ + e−.

Solution

Multiply Fe half-equation by 5M1

MnO4− + 8H+ + 5Fe2+ → Mn2+ + 4H2O + 5Fe3+A1

Quick check

1. Combine Cr2O72− reduction with I− → I2 oxidation.

Answers: 1. Cr2O72− + 14H+ + 6I− → 2Cr3+ + 7H2O + 3I2

16 The reactivity series

Metals as reducing agents.

Metals higher in the activity series (K, Na, Ca, Mg, Al, Zn, Fe, Pb, (H), Cu, Ag, Au) lose electrons more easily and are stronger reducing agents. A more reactive metal displaces a less reactive metal from a solution of its ions.

WORKED EXAMPLE 9 Displacement (standard)

Predict whether zinc reacts with copper(II) sulfate and with magnesium sulfate.

Solution

Zn displaces Cu: Zn + Cu2+ → Zn2+ + CuA1

No reaction with MgSO4: Mg is more reactiveA1

Quick check

1. Which metals react with dilute acids to give hydrogen: Cu, Fe, Mg, Ag?

2. Arrange metals W, X, Y given: W displaces X and Y; X displaces Y.

Answers: 1. Fe and Mg 2. W > X > Y

17 Halogens as oxidizing agents

Displacement of halides.

Oxidizing strength decreases down group 17: Cl2 > Br2 > I2. A halogen oxidizes halide ions below it: Cl2 + 2Br− → 2Cl− + Br2.

WORKED EXAMPLE 10 Halogen redox (standard)

Write half-equations for Cl2 + 2I− → 2Cl− + I2.

Solution

Cl2 + 2e− → 2Cl−A1

2I− → I2 + 2e−A1

18 Voltaic (galvanic) cells

Chemical energy to electrical energy.

A voltaic cell has two half-cells connected by a wire and a salt bridge. Oxidation occurs at the anode (negative electrode), reduction at the cathode (positive). Electrons flow from anode to cathode through the wire; ions move through the salt bridge to maintain charge balance.

WORKED EXAMPLE 11 Daniell cell (exam level)

For a Zn/Zn2+ and Cu/Cu2+ cell, identify the anode, the cathode and the direction of electron flow.

Solution

Anode: Zn (oxidized); cathode: CuA1

Electrons flow from Zn to Cu in the external circuitA1

In the real world

A smartphone battery is a lithium-ion voltaic cell; during discharge, Li is oxidized at the anode and electrons flow through the phone's circuits to the cathode.

Quick check

1. What is the function of the salt bridge?

2. Which electrode gains mass in the Daniell cell?

Answers: 1. completes the circuit; maintains electrical neutrality 2. copper cathode

19 Electrolytic cells

Electrical energy drives non-spontaneous reactions.

In electrolysis, an external power supply drives a non-spontaneous redox reaction. The cathode (negative) attracts cations, which are reduced; the anode (positive) attracts anions, which are oxidized. Molten ionic compounds give the metal at the cathode and the non-metal at the anode.

WORKED EXAMPLE 12 Molten salt (standard)

Write the electrode equations for electrolysis of molten sodium chloride.

Solution

Cathode: Na+ + e− → NaA1

Anode: 2Cl− → Cl2 + 2e−A1

Quick check

1. Predict the products of electrolysis of molten lead(II) bromide.

Answers: 1. lead at the cathode, bromine at the anode

20 Comparing voltaic and electrolytic cells

Opposite energy changes.

	Voltaic cell	Electrolytic cell
energy change	chemical → electrical	electrical → chemical
reaction	spontaneous	non-spontaneous
anode	negative, oxidation	positive, oxidation
cathode	positive, reduction	negative, reduction

WORKED EXAMPLE 13 Common feature (standard)

What is always true of the anode in any cell?

Solution

Oxidation occurs at the anodeA1

21 Redox in everyday life

Corrosion, bleach, respiration.

Rusting is the oxidation of iron in the presence of water and oxygen: 4Fe + 3O2 + 2xH2O → 2Fe2O3·xH2O. It is prevented by painting, oiling, galvanizing (zinc coating, sacrificial protection) and alloying. Bleach (ClO−) is an oxidizing agent; respiration oxidizes glucose.

WORKED EXAMPLE 14 Sacrificial protection (standard)

Explain how zinc protects iron even when the coating is scratched.

Solution

Zn is more reactive than FeR1

Zn is oxidized in preference to FeR1

Quick check

1. Why do ships have magnesium or zinc blocks attached to their hulls?

Answers: 1. sacrificial protection: these metals corrode instead of the steel

22 Standard electrode potentials (HL)

Predicting cell voltages.

E⦵ values are measured against the standard hydrogen electrode (0.00 V). The more positive E⦵, the stronger the oxidizing agent. E⦵cell = E⦵(cathode) − E⦵(anode); a positive value means the reaction is spontaneous (ΔG⦵ = −nFE⦵).

WORKED EXAMPLE 15 Cell potential (exam level)

Find E⦵cell for Zn/Cu (Zn2+/Zn −0.76 V, Cu2+/Cu +0.34 V) and ΔG⦵.

Solution

0.34 − (−0.76) = +1.10 VA1

ΔG = −2 × 96 500 × 1.10 = −212 kJ mol−1M1 A1

23 Mixed review: worked problems

Bringing the ideas together.

WORKED EXAMPLE 16 Mixed problem (exam level)

In acidic solution, hydrogen peroxide oxidizes iodide ions to iodine. (a) Write both half-equations. (b) Write the overall equation. (c) State the change in oxidation state of oxygen.

Solution

(a) H2O2 + 2H+ + 2e− → 2H2O; 2I− → I2 + 2e−A1 A1

(b) H2O2 + 2H+ + 2I− → 2H2O + I2A1

Quick check

1. Find the volume of 0.0200 mol dm−3 KMnO4 that reacts with 25.0 cm3 of 0.100 mol dm−3 Fe2+.

Answers: 1. n(Fe2+) = 2.50 × 10−3; n(MnO4−) = 5.00 × 10−4: 25.0 cm3

24 Additional exam-style practice

Paper 1 and Paper 2.

WORKED EXAMPLE 17 Paper 1 question (standard)

What is the oxidation state of sulfur in SO32−? A. +2 B. +4 C. +6 D. −2

Solution

x − 6 = −2: +4 (B)A1

WORKED EXAMPLE 18 Paper 2 question (exam level)

Aluminium is extracted by electrolysis of molten alumina. (a) Write the cathode equation. (b) Explain why the carbon anodes must be replaced. (c) Explain why electrolysis, not carbon reduction, is used.

Solution

(a) Al3+ + 3e− → AlA1

(b) oxygen produced reacts with carbon to form CO2R1

25 Electrolysis of aqueous solutions (introduction)

Water competes.

In aqueous solutions, water can be oxidized or reduced instead of the ions. At the cathode, metals below hydrogen in the reactivity series are deposited; otherwise hydrogen forms. At the anode, halide ions (if concentrated) form halogens; otherwise oxygen forms.

WORKED EXAMPLE 19 Brine (standard)

Predict the products of electrolysis of concentrated aqueous NaCl.

Solution

Cathode: H2A1

Anode: Cl2; NaOH remains in solutionA1

Quick check

1. Predict the products of electrolysis of aqueous copper(II) sulfate with inert electrodes.

Answers: 1. copper at the cathode, oxygen at the anode

26 Common misconceptions and links

Students often think…	Actually…
the oxidizing agent is oxidized	it is reduced
the anode is always positive	positive in electrolysis; negative in voltaic cells
electrons flow through the salt bridge	ions flow in the salt bridge; electrons in the wire
a larger E⊖ means easier oxidation	more positive E⊖ means easier reduction

How R3.2 connects to the rest of the course

Topic	Connection
S3.1 Periodic table	oxidation states; halogen reactivity
R1.3 Energy from fuels	fuel cells
R2.1 How much?	redox titrations
S3.2 Functional groups	alcohols and carbonyls
R1.4 Entropy (HL)	ΔG

27 Exam-style question with markscheme

Try this under exam conditions (about 15 minutes) before reading the markscheme.

Figure 4. A zinc-copper voltaic cell.

(a) Identify the oxidizing agent in Zn + Cu2+ → Zn2+ + Cu. [1]

(b) Describe the direction of electron flow and the role of the salt bridge in a Zn/Cu cell. [3]

(d) Describe how ethanol can be oxidized to ethanoic acid and state the observation. [3]

(e) Calculate E⊖cell for a Zn/Ag cell and state whether the reaction is spontaneous. [2]

Markscheme

Part	Answer	Marks
(a)	Cu2+	A1
(b)	electrons from Zn to Cu in the wire	R1
	salt bridge completes circuit	R1
	ions move to balance charge	R1
(c)	cathode: K+ + e− → K	A1
	anode: 2Br− → Br2 + 2e−	A1
(d)	acidified K2Cr2O7	R1
	heat under reflux	R1
	orange to green	A1
(e)	+0.80 − (−0.76) = +1.56 V	A1
	positive: spontaneous	A1

Examiner's comments. In (b), many answers said electrons pass through the salt bridge.

28 Check your understanding

Section A: multiple choice

1. Oxidation is

A. loss of electrons B. gain of electrons C. gain of hydrogen D. decrease in oxidation state

2. In a voltaic cell the anode is

A. positive, reduction B. negative, reduction C. negative, where oxidation occurs D. positive, oxidation

3. In electrolysis of molten NaCl, chlorine forms at the

A. cathode B. anode C. salt bridge D. electrolyte surface only

4. Acidified dichromate oxidizes a secondary alcohol to

A. an aldehyde B. a carboxylic acid C. nothing D. a ketone

5. Which metal displaces copper from CuSO4?

A. silver B. zinc C. gold D. copper

6. A secondary cell is

A. rechargeable B. single use C. a fuel cell D. an electrolytic cell

7. The standard hydrogen electrode has E⊖ =

A. +1.00 V B. −0.76 V C. +0.34 V D. 0.00 V

8. In electroplating the object is the

A. anode B. electrolyte C. cathode D. salt bridge

Section B: structured

9. Define oxidizing agent. [2]

10. Balance: Cr2O72− → Cr3+ in acid. [2]

11. Compare voltaic and electrolytic cells. [4]

12. Explain why aqueous NaCl gives hydrogen, not sodium, at the cathode. [2]

Answers

1. A 2. C 3. B 4. D 5. B 6. A 7. D 8. C

9. Accepts electrons; is reduced; causes oxidation.

10. Cr2O72− + 14H+ + 6e− → 2Cr3+ + 7H2O.

11. Voltaic: spontaneous, chemical to electrical, anode negative; electrolytic: non-spontaneous, electrical to chemical, anode positive.

12. Water more easily reduced than Na+ (E⊖ Na+ very negative); H2 forms.

29 Summary and topic map

Idea	Remember
Redox	OIL RIG; oxidation state changes
Agents	oxidizing agent reduced; reducing agent oxidized
Half-equations	balance with H2O, H+, e−
Activity series	displacement; metals with acids
Voltaic cells	anode negative (oxidation); salt bridge
Electrolysis	molten salts; anode positive
Organic redox	alcohols oxidized by dichromate
HL	E⊖, Ecell, ΔG = −nFE; aqueous electrolysis; electroplating

Before the exam, make sure you…

•track oxidation states
•balance half-equations systematically
•know the sign of each electrode
•distinguish reflux and distillation

Electron Transfer on one page

Cover this page with a sheet of paper and try to rebuild it from memory. Whatever you can't recall is what to revise next.

Figure 5. Topic map for Electron Transfer.

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