IB Chemistry · Unit 9 · Reactivity

Electrons
on the
move.

Reactions where electrons change hands. Batteries, corrosion, biology — all redox, all the time.

12Lessons
4HL extensions
42Key terms
SL+HLLevel
Cu anode Zn cathode e⁻ ELECTRON TRANSFER
Unit 9 · Standard Level

8 lessons to work through.

The required syllabus content for Unit 9, in order. Each card is one lesson-sized checkpoint.

Lesson 1

Introduction to Redox

Lesson 1 of Unit 9.

Lesson 2

Constructing redox equations

Lesson 2 of Unit 9.

Lesson 3

Displacement reactions

Lesson 3 of Unit 9.

Lesson 5

Primary (voltaic) cells

Lesson 5 of Unit 9.

Lesson 7

Secondary (rechargeable) cells

In a primary cell: reactants are consumed – cannot reverse the reaction - Anode disintegrates due to oxidation

Lesson 8

Fuel cells

Fuel converted to water, any other gas, and heat

Lesson 9

Electrolytic cells (Molten)

Lesson 9 of Unit 9.

Lesson 12

Energy from fuels

Recap: Secondary cells have reversible electrochemical processes so are rechargeable

Lessons in detail

The unit, lesson by lesson.

Each lesson card below mirrors the original teacher deck — syllabus refs, content, worked examples and practice questions in order.

Lesson 1

Introduction to redox

Reactivity 3.2.1Reactivity 3.2.2

Lesson outcomes

A redox reaction is one in which electrons are transferred. OIL RIG: Oxidation Is Loss, Reduction Is Gain of electrons.

Three equivalent definitions:

OxidationReduction
ElectronsLostGained
OxygenGainedLost
HydrogenLostGained
Oxidation stateIncreasesDecreases

Agents

The oxidising agent causes oxidation by accepting electrons — and is itself reduced. The reducing agent causes reduction by donating electrons — and is itself oxidised.

Worked example

Identify oxidation and reduction

Problem. Mg + CuSO₄ → MgSO₄ + Cu. Identify what is oxidised, reduced, the oxidising agent, and the reducing agent.
Solution. Mg → Mg²⁺ + 2e⁻: Mg loses electrons. OS 0 → +2. Oxidised. Mg is the reducing agent.
Cu²⁺ + 2e⁻ → Cu: Cu²⁺ gains electrons. OS +2 → 0. Reduced. Cu²⁺ (or CuSO₄) is the oxidising agent.
SO₄²⁻ is a spectator ion (unchanged).

Try these

  1. State the three equivalent definitions of oxidation.
    Show answer
    (1) Loss of electrons. (2) Gain of oxygen / loss of hydrogen. (3) Increase in oxidation state.
  2. What is the meaning of OIL RIG?
    Show answer
    Oxidation Is Loss (of electrons); Reduction Is Gain (of electrons). The standard mnemonic.
  3. Which is the oxidising agent and which the reducing agent in: 2 KI + Cl₂ → 2 KCl + I₂?
    Show answer
    I⁻ → I₂ (loss of e⁻, oxidised). KI is the reducing agent. Cl₂ → 2 Cl⁻ (gain of e⁻, reduced). Cl₂ is the oxidising agent.
  4. Identify the oxidising agent in: 5 Fe²⁺ + MnO₄⁻ + 8 H⁺ → 5 Fe³⁺ + Mn²⁺ + 4 H₂O.
    Show answer
    MnO₄⁻ is the oxidising agent (Mn goes from +7 to +2, reduced). Fe²⁺ is the reducing agent (oxidised from +2 to +3).
  5. Assign the oxidation state of S in: (a) SO₂, (b) SO₃, (c) H₂SO₄, (d) S₂O₃²⁻.
    Show answer
    (a) +4. (b) +6. (c) +6. (d) +2 (per S atom — total +4 on two S, balanced by 3 × (−2) = −6 on O and net charge −2).
  6. Assign oxidation states in: (a) NaH, (b) H₂O₂, (c) OF₂.
    Show answer
    (a) Na = +1 (group 1, fixed). Sum = 0. H = −1 (metal hydride, exception). (b) H = +1, O = −1 (peroxide). (c) F = −1 (fixed). Sum = 0. O = +2 (the F overrides O's normal −2).
Lesson 2

Constructing redox equations · half-equations

Reactivity 3.2.3

Lesson outcomes

A half-equation shows just one half of a redox reaction (either the oxidation or the reduction), including the electrons explicitly.

The six-step procedure for half-equations

  1. Identify the element being oxidised/reduced.
  2. Balance that element.
  3. Balance oxygen by adding H₂O.
  4. Balance hydrogen by adding H⁺ (or H₂O / OH⁻ in basic solution).
  5. Balance charge by adding electrons (e⁻).
  6. Check: every atom and every charge balances.

Combining half-equations

Multiply each half-equation so the electrons cancel when summed. Then add and simplify (cancel any H⁺ or H₂O appearing on both sides).

Worked example

MnO₄⁻ + Fe²⁺ in acid

Problem. Construct the balanced overall equation for the oxidation of Fe²⁺ to Fe³⁺ by acidified MnO₄⁻ (which reduces to Mn²⁺).
Solution. Oxidation half-eqn: Fe²⁺ → Fe³⁺ + e⁻.
Reduction half-eqn: MnO₄⁻ → Mn²⁺. Balance O: + 4 H₂O on right. Balance H: + 8 H⁺ on left. Balance charge: + 5 e⁻ on left:
MnO₄⁻ + 8 H⁺ + 5 e⁻ → Mn²⁺ + 4 H₂O.
Combine: multiply Fe half-eqn by 5 so the electrons match (5 e⁻ on each side):
MnO₄⁻ + 8 H⁺ + 5 Fe²⁺ → Mn²⁺ + 4 H₂O + 5 Fe³⁺.

Try these

  1. Write the half-equation for the reduction of acidified Cr₂O₇²⁻ to Cr³⁺.
    Show answer
    Cr₂O₇²⁻ + 14 H⁺ + 6 e⁻ → 2 Cr³⁺ + 7 H₂O. Balance: Cr × 2; O × 7 (added 7 H₂O); H × 14 (added 14 H⁺); charge: left = −2 + 14 = +12; right = +6; need 6 e⁻ on left.
  2. Write the half-equation for the oxidation of I⁻ to I₂.
    Show answer
    2 I⁻ → I₂ + 2 e⁻.
  3. Combine the previous two to write the overall ionic equation for Cr₂O₇²⁻ oxidising I⁻.
    Show answer
    Multiply I half-equation by 3 so 6 e⁻ match: 6 I⁻ → 3 I₂ + 6 e⁻. Add to Cr₂O₇²⁻ half-equation: Cr₂O₇²⁻ + 14 H⁺ + 6 I⁻ → 2 Cr³⁺ + 7 H₂O + 3 I₂.
  4. Write the balanced equation for the disproportionation of Cu⁺ to Cu and Cu²⁺.
    Show answer
    Oxidation: Cu⁺ → Cu²⁺ + e⁻. Reduction: Cu⁺ + e⁻ → Cu. Combine (1 e⁻ each): 2 Cu⁺ → Cu + Cu²⁺. This is a disproportionation — same element both oxidised and reduced.
  5. What is a disproportionation reaction? Give an example.
    Show answer
    A redox reaction in which the same element is both oxidised and reduced. Example: 2 H₂O₂ → 2 H₂O + O₂ (O goes from −1 in peroxide to both 0 in O₂ and −2 in H₂O). Also: Cl₂ + 2 NaOH → NaCl + NaOCl + H₂O (Cl: 0 → −1 and +1).
Lesson 3

Displacement reactions · the reactivity series

Reactivity 3.2.4

Lesson outcomes

A more reactive species displaces a less reactive one from its compound. Two kinds:

Metal displacement

Reactivity series (top = most reactive): K, Na, Ca, Mg, Al, (C), Zn, Fe, Pb, (H), Cu, Ag, Au. A metal will displace any metal below it from a salt of the latter.

Example: Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s). Zn is above Cu in the series; Cu²⁺ takes Zn's electrons.

Halogen displacement

Reactivity: F > Cl > Br > I. A halogen displaces any halide ion below it.

Example: Cl₂(aq) + 2 KBr(aq) → 2 KCl(aq) + Br₂(aq). Orange Br₂ formation confirms the reaction.

Try these

  1. List the metal reactivity series from K to Ag.
    Show answer
    K > Na > Ca > Mg > Al > (C) > Zn > Fe > Pb > (H) > Cu > Ag > Au. Each metal will displace any metal below it (and react more readily with water/acid). Hydrogen is listed for reference — metals above H react with dilute acid; those below don't.
  2. Will iron displace silver from silver nitrate? Write the equation if it occurs.
    Show answer
    Yes — Fe is above Ag in the series. Fe(s) + 2 AgNO₃(aq) → Fe(NO₃)₂(aq) + 2 Ag(s).
  3. Predict whether each pair reacts: (a) Mg + ZnSO₄, (b) Cu + FeSO₄, (c) Zn + Pb(NO₃)₂, (d) Ag + CuSO₄.
    Show answer
    (a) Yes (Mg above Zn): Mg + ZnSO₄ → MgSO₄ + Zn. (b) No (Cu below Fe). (c) Yes (Zn above Pb): Zn + Pb(NO₃)₂ → Zn(NO₃)₂ + Pb. (d) No (Ag below Cu).
  4. Why does Cl₂ not displace F⁻ from KF?
    Show answer
    F is more reactive than Cl (smaller atom, stronger attraction for added electron). F⁻ is more stable than Cl⁻, so Cl₂ cannot oxidise F⁻.
  5. Predict the colour change when Cl₂(aq) is added to KI(aq).
    Show answer
    Cl₂ (yellow-green) displaces I⁻ from KI: Cl₂ + 2 KI → 2 KCl + I₂. The displaced I₂ is brown in water (or violet/purple if shaken with hexane → starch tests blue-black). Cl₂ disappears; brown I₂ appears.
Lesson 5

Primary (voltaic) cells

Reactivity 3.2.5

Lesson outcomes

A voltaic cell uses a spontaneous redox reaction to produce an electric current. Physically separating the two half-reactions forces electrons to travel through an external wire — that's what we tap for useful current.

The Daniell cell (Zn / Cu)

Cell notation

Zn(s) | Zn²⁺(aq, 1 M) || Cu²⁺(aq, 1 M) | Cu(s)

Single bar | = phase boundary. Double bar || = salt bridge. Anode on left, cathode on right.

Try these

  1. In a voltaic cell, which electrode is the anode? Which is the cathode? Which is positive, which negative?
    Show answer
    Anode: where oxidation occurs; the more reactive metal. Cathode: where reduction occurs; the less reactive metal. In a voltaic cell, the anode is negative (electrons leave it) and the cathode is positive (electrons arrive).
  2. Write the cell notation for a Daniell cell.
    Show answer
    Zn(s) | Zn²⁺(aq, 1 mol dm⁻³) || Cu²⁺(aq, 1 mol dm⁻³) | Cu(s). Single bar = phase boundary. Double bar = salt bridge. Anode on left, cathode on right.
  3. Which direction do electrons flow in the external wire of a Daniell cell? Conventional current?
    Show answer
    Electrons flow from anode (Zn, negative) to cathode (Cu, positive). Conventional current is defined opposite — from positive (Cu) to negative (Zn) externally.
  4. Why does the salt bridge contain an inert electrolyte like KNO₃ rather than CuSO₄ or ZnSO₄?
    Show answer
    K⁺ and NO₃⁻ don't react with either Zn²⁺ or Cu²⁺ and don't precipitate. Using CuSO₄ would contaminate the Zn half-cell with Cu²⁺ (which Zn would displace directly), short-circuiting the cell.
  5. Predict what happens if the salt bridge is removed during operation.
    Show answer
    Charge imbalance builds up: anode half-cell becomes positive (Zn²⁺ accumulating), cathode becomes negative (Cu²⁺ depleting). The growing potential difference opposes further electron flow → current stops within seconds.
  6. What is the role of the salt bridge?
    Show answer
    Completes the electrical circuit by allowing ion migration between half-cells. Cations migrate toward the cathode; anions toward the anode. Maintains electrical neutrality in each half-cell so electron flow can continue.
Lesson 7

Secondary (rechargeable) cells

Reactivity 3.2.6

Lesson outcomes

Primary cells are designed to be used once (the redox reaction proceeds in one direction). Secondary cells are rechargeable — applying an external current drives the redox reaction in reverse, regenerating the reactants.

Lead-acid battery (car battery)

Anode (discharge): Pb(s) + SO₄²⁻(aq) → PbSO₄(s) + 2 e⁻
Cathode (discharge): PbO₂(s) + 4 H⁺(aq) + SO₄²⁻(aq) + 2 e⁻ → PbSO₄(s) + 2 H₂O(l)
Overall (discharge): Pb(s) + PbO₂(s) + 2 H₂SO₄(aq) → 2 PbSO₄(s) + 2 H₂O(l)

Recharge reverses every step. E°cell ≈ 2 V per cell — six cells in series make a 12 V battery.

Lithium-ion cell

Lithium ions shuttle between a graphite anode and a metal-oxide cathode (e.g. LiCoO₂). Discharge: Li migrates from anode lattice to cathode. Charge: reversed. Very high energy density and long cycle life — universal in phones, laptops and EVs.

Try these

  1. What is the difference between a primary and a secondary cell? Give an example of each.
    Show answer
    Primary: designed for single use — the redox reaction proceeds in one direction and is not easily reversed (e.g. zinc-carbon dry cell, alkaline AA). Secondary: rechargeable — applying an external current reverses the discharge reaction (e.g. lead-acid car battery, Li-ion phone battery).
  2. Write the overall discharge equation for a lead-acid battery.
    Show answer
    Pb(s) + PbO₂(s) + 2 H₂SO₄(aq) → 2 PbSO₄(s) + 2 H₂O(l). On charging, the reverse occurs.
  3. Why do lead-acid batteries lose performance in cold weather?
    Show answer
    Lower T slows the diffusion of ions and slows the electrode kinetics. The internal resistance rises, so the battery can deliver less current and the voltage drops under load.
  4. Why are Li-ion batteries preferred over lead-acid for laptops and phones?
    Show answer
    Much higher energy density (more J per kg and per dm³). Lighter, smaller, longer-lasting. Higher cell voltage (~3.7 V vs 2 V for lead-acid). No memory effect. Drawback: more expensive and Li resources are limited.
Lesson 8

Fuel cells

Reactivity 3.2.6

Lesson outcomes

A fuel cell generates electricity by continuously supplying fuel (typically H₂) and oxidant (O₂) — unlike a battery, the reactants are not stored within the cell.

The H₂-O₂ alkaline fuel cell

Anode: 2 H₂(g) + 4 OH⁻(aq) → 4 H₂O(l) + 4 e⁻
Cathode: O₂(g) + 2 H₂O(l) + 4 e⁻ → 4 OH⁻(aq)
Overall: 2 H₂(g) + O₂(g) → 2 H₂O(l). E°cell ≈ +1.23 V.

Only product is water — no CO₂, no NOx. Efficiency can exceed 60% (vs ~30% for a combustion engine doing the same chemistry).

Drawbacks

Try these

  1. What is the overall reaction in a hydrogen fuel cell? What is the only product?
    Show answer
    2 H₂(g) + O₂(g) → 2 H₂O(l). The only product is water — no CO₂, NOx, particulates or other emissions.
  2. Why are fuel cells more efficient than internal-combustion engines burning the same fuel?
    Show answer
    Combustion engines are limited by thermodynamic (Carnot) efficiency — heat → work conversion has a theoretical maximum of ~30–40% in practice. Fuel cells convert chemical energy directly to electrical energy without going through heat, achieving 60%+ efficiency.
  3. Why is hydrogen not yet a widely-used fuel despite its environmental benefits?
    Show answer
    (1) Storage: H₂ is a gas at room T with very low energy density unless cryogenic or compressed to 700+ bar — costly and dangerous. (2) Production: most H₂ comes from steam reforming of CH₄ (releasing CO₂). Truly green H₂ (electrolysis of water from renewables) is expensive. (3) Infrastructure: no widespread refuelling network.
Lesson 9

Electrolytic cells (molten)

Reactivity 3.2.10

Lesson outcomes

In electrolysis, an external voltage drives a non-spontaneous redox reaction. Key uses: extracting reactive metals (Al, Na), purifying metals (Cu), electroplating, manufacturing chlorine and NaOH.

Polarity flip vs voltaic cells

In a voltaic cell, the more reactive metal's terminal is negative. In an electrolytic cell, the battery's positive terminal is connected to the anode (oxidation site) and the negative terminal to the cathode (reduction site).

VoltaicElectrolytic
Anode polarity+
Cathode polarity+
Anode = oxidation
Cathode = reduction

Molten ionic compound

Cations migrate to the cathode and are reduced. Anions migrate to the anode and are oxidised. Example: molten NaCl → Na(l) at cathode + Cl₂(g) at anode.

Try these

  1. Compare the polarity of the anode and cathode in a voltaic cell vs an electrolytic cell.
    Show answer
    Voltaic: anode is negative, cathode is positive (electrons leave the anode driven by the spontaneous reaction). Electrolytic: anode is positive, cathode is negative (the battery's + terminal pulls electrons out of the anode). In both, anode = oxidation, cathode = reduction.
  2. Predict the products at each electrode when molten NaCl is electrolysed.
    Show answer
    Cathode (−): Na⁺ + e⁻ → Na(l). Anode (+): 2 Cl⁻ → Cl₂(g) + 2 e⁻. Overall: 2 NaCl(l) → 2 Na(l) + Cl₂(g). This is the Downs process for extracting sodium.
  3. Predict the products at each electrode when molten Al₂O₃ is electrolysed.
    Show answer
    Cathode (−): Al³⁺ + 3 e⁻ → Al(l). Anode (+): 2 O²⁻ → O₂(g) + 4 e⁻. (Used industrially in the Hall-Héroult process — Al₂O₃ dissolved in molten cryolite to lower the m.p.)
  4. Why is aluminium extracted by electrolysis rather than by reduction with carbon in a blast furnace?
    Show answer
    Al³⁺ is too strongly held by O²⁻ for carbon to reduce it (carbon's E° is not negative enough). Electrolysis forces the non-spontaneous reduction. Iron, in contrast, is less reactive — carbon does reduce iron oxides in a blast furnace.
Lesson 12

Energy from fuels

Reactivity 1.3

Lesson outcomes

A fuel releases energy on combustion or other reaction. Two key metrics:

H₂ has very high specific energy (142 MJ kg⁻¹) but very low energy density at room T (low ρ). Petrol: 47 MJ kg⁻¹, 34 MJ dm⁻³ — much better for vehicles where volume is constrained.

Fossil fuels

Coal, oil, natural gas. Finite. Produce CO₂ (greenhouse gas), NOx (smog, acid rain), SO₂ (acid rain — coal especially), particulates (lung damage).

Renewable fuels

Biofuels (bioethanol, biodiesel): can be carbon-neutral if the next crop reabsorbs the CO₂. Drawbacks: land-use conflicts with food.
Hydrogen: carbon-free only if produced from renewable energy (electrolysis of water powered by solar/wind). Steam-reformed H₂ from CH₄ is currently the dominant source — not carbon-neutral.

Try these

  1. Compare the specific energy and energy density of petrol vs H₂. Why is petrol still favoured for cars?
    Show answer
    Petrol: 47 MJ kg⁻¹, 34 MJ dm⁻³. H₂: 142 MJ kg⁻¹, ~10 MJ dm⁻³ (liquid). H₂ is better per kg but much worse per dm³. Cars are volume-constrained (fuel tank size), so petrol's higher volumetric density wins. H₂ vehicles need enormous compressed-gas or cryogenic tanks.
  2. Why is coal a worse fuel than natural gas in terms of CO₂ emissions per kJ of energy?
    Show answer
    Coal is almost pure C with little H. Natural gas (CH₄) has 4 H per C — much of the energy comes from H combustion (→ H₂O), not C combustion (→ CO₂). Per kJ released, natural gas emits much less CO₂.
  3. Why are bioethanol and biodiesel considered carbon-neutral fuels?
    Show answer
    The plants used to produce them absorb CO₂ from the atmosphere as they grow (via photosynthesis). When the biofuel is burned, that same CO₂ is released. Net atmospheric change is approximately zero — if you ignore the fossil-fuel inputs to farming, transport and processing.
  4. What environmental problems are caused by SO₂ from burning coal?
    Show answer
    SO₂ + ½O₂ → SO₃ + H₂O → H₂SO₄. This creates 'acid rain' — pH 4–5 rainfall that dissolves limestone, kills fish in lakes, leaches nutrients from soil, and damages forests. Mitigated by flue-gas desulfurisation in modern coal plants.
HL extension

The Nernst
equation,
applied.

HL: standard electrode potentials, electrochemical cells under non-standard conditions, electrolysis stoichiometry.

Lesson 4 HL only

The standard hydrogen electrode

Reactivity 3.2.7 (HL)

Lesson outcomes

  • Describe the construction and operation of the standard hydrogen electrode (SHE).
  • Define standard electrode potential (E°).
  • Explain why electrode potentials must be measured relative to a reference electrode.

Any individual half-cell's potential cannot be measured in isolation — only differences between two half-cells. The Standard Hydrogen Electrode (SHE) is the universal reference, assigned E° = 0.00 V by definition.

Construction

  • Platinum electrode (inert) coated with finely divided platinum black.
  • Immersed in 1.00 mol dm⁻³ H⁺(aq).
  • H₂(g) at 100 kPa bubbled over the electrode.
  • Temperature 298 K (25 °C).

The half-reaction is: 2 H⁺(aq) + 2 e⁻ ⇌ H₂(g), E° = 0.00 V.

Measuring E° for any other half-cell

Connect the half-cell to the SHE via a salt bridge and a high-resistance voltmeter. The voltmeter reading is the standard electrode potential of the half-cell of interest. The sign is positive if the half-cell is more strongly reducing (i.e. reduces H⁺), negative if more strongly oxidising.

E° values are tabulated in the IB data booklet — always quoted with the sign and against the SHE.

Try these

  1. E°(Zn²⁺/Zn) = −0.76 V. State whether Zn is a better reducer than H₂(g) or a worse one. Explain.
    Show answer
    Negative E° means Zn is more easily oxidised than H₂. Zn is the stronger reducing agent. (Equivalently: H⁺ can oxidise Zn to Zn²⁺.)
Lesson 6 HL only

Cell potential calculations

Reactivity 3.2.7 (HL)Reactivity 3.2.8 (HL)

Lesson outcomes

  • Calculate E°cell from tabulated standard electrode potentials.
  • Predict the spontaneity of a redox reaction from the sign of E°cell.
  • Link E°cell to ΔG° via ΔG° = −nFE°.

For any cell, the standard cell potential is:

cell = E°(cathode) − E°(anode)

Both values are looked up as reduction potentials in the data booklet. The half-reaction that occurs as reduction is the cathode; the one that runs in reverse (oxidation) is the anode.

If E°cell is positive, the reaction as written is spontaneous.

Link to thermodynamics

ΔG° = −n F E°cell

n = moles of electrons transferred per mole of reaction. F = Faraday constant (96 485 C mol⁻¹). A positive E° gives a negative ΔG° — same conclusion, different formalism.

Worked example

Spontaneity of Zn + Cu²⁺

Problem. E°(Cu²⁺/Cu) = +0.34 V; E°(Zn²⁺/Zn) = −0.76 V. Calculate E°cell for Zn + Cu²⁺ → Zn²⁺ + Cu, and ΔG° in kJ mol⁻¹.
Solution. Cu²⁺ is reduced (cathode), Zn is oxidised (anode).
cell = E°cathode − E°anode = 0.34 − (−0.76) = +1.10 V. Positive → spontaneous.
ΔG° = −nFE° = −(2)(96 485)(1.10) = −2.12 × 10⁵ J mol⁻¹ = −212 kJ mol⁻¹. Very negative — strongly spontaneous.
Worked example

Will silver displace copper?

Problem. E°(Ag⁺/Ag) = +0.80 V; E°(Cu²⁺/Cu) = +0.34 V. Will the reaction Ag(s) + Cu²⁺ → ? proceed?
Solution. As written, Ag is oxidised (Ag → Ag⁺) and Cu²⁺ is reduced. E° = 0.34 − 0.80 = −0.46 V. Negative — not spontaneous.
The reverse direction is spontaneous: Cu(s) + 2 Ag⁺ → Cu²⁺ + 2 Ag with E°cell = +0.46 V. This is the basis of "silver tree" experiments.

Try these

  1. Use the data: E°(MnO₄⁻/Mn²⁺) = +1.51 V; E°(Fe³⁺/Fe²⁺) = +0.77 V. Calculate E°cell for the titration of Fe²⁺ with KMnO₄ in acid, and confirm whether the titration is feasible.
    Show answer
    MnO₄⁻ is reduced (cathode); Fe²⁺ is oxidised (anode). E°cell = 1.51 − 0.77 = +0.74 V. Positive → feasible. (This is why permanganate is a powerful and common oxidant in titrations.)
Lesson 10 HL only

Electrolytic cells (aqueous)

Reactivity 3.2.11 (HL)

Lesson outcomes

  • Predict the products of electrolysis of aqueous solutions using E° values.
  • Recognise the role of water as a possible alternative reactant at each electrode.
  • Apply concentration effects (e.g. brine) where they override E° predictions.

In aqueous solution, water is always present — and it can be electrolysed too. At the cathode, the candidates are the metal cation OR water. At the anode, the candidates are the anion OR water.

Cathode rule

The species with the more positive E° (more easily reduced) wins. Water: 2 H₂O + 2 e⁻ → H₂(g) + 2 OH⁻, E° = −0.83 V (at standard alkaline conditions; effectively ~0 V at pH 7).

So Na⁺ (E° = −2.71 V) cannot beat water → H₂ produced. Cu²⁺ (E° = +0.34 V) beats water → Cu deposited.

Anode rule

The species with the more negative E° (more easily oxidised) wins. Water: O₂(g) + 4 H⁺ + 4 e⁻ → 2 H₂O, E° = +1.23 V.

So OH⁻ in dilute solutions usually wins → O₂. But concentrated halide solutions give the halogen instead due to high concentration overriding E° predictions.

Worked example

Electrolysis of aqueous NaCl (brine)

Problem. Predict products at each electrode when concentrated aqueous NaCl is electrolysed with inert graphite electrodes.
Solution. Cathode: Na⁺ (E° = −2.71 V) vs H₂O. Water wins → H₂(g) evolved. Solution becomes alkaline (OH⁻ accumulates).
Anode: Cl⁻ vs H₂O. E° comparison suggests O₂ should win, but at high [Cl⁻] the kinetic preference for Cl₂ dominates → Cl₂(g) evolved.
Industrially this is the chlor-alkali process: yields Cl₂, H₂ and NaOH from a single feed.
Lesson 11 HL only

Electrolysis with non-inert electrodes

Reactivity 3.2.11 (HL)

Lesson outcomes

  • Explain copper purification (refining) by electrolysis.
  • Describe electroplating and predict the deposited material.

When the electrode itself can be oxidised, the rules change. The active anode reaction is the oxidation of the electrode metal rather than water/anion.

Copper refining

Impure Cu anode → Cu²⁺ + 2 e⁻ (anode dissolves). Cathode: Cu²⁺ + 2 e⁻ → pure Cu (deposits). Insoluble impurities (Ag, Au) fall to the bottom as "anode slime"; more reactive metals (Zn, Fe) dissolve but don't redeposit (their E° is too negative). Result: 99.99%+ pure copper.

Electroplating

The object to be plated is made the cathode. The plating metal is the anode. Electrolyte contains a salt of the plating metal. Example: silver-plating a fork — fork is cathode, Ag bar is anode, electrolyte is AgNO₃(aq). Ag dissolves from the anode, then redeposits on the fork.

Worked example

Faraday's law calculation

Problem. Electrolysis of CuSO₄ for 30 minutes at 1.5 A. What mass of Cu is deposited at the cathode?
Solution. Charge Q = I × t = 1.5 × 1800 = 2700 C.
Moles of electrons = Q / F = 2700 / 96 485 = 0.02798 mol.
Cu²⁺ + 2 e⁻ → Cu, so n(Cu) = 0.02798 / 2 = 0.01399 mol.
m(Cu) = n × M = 0.01399 × 63.55 = 0.889 g.

Try these

  1. In copper refining, why don't reactive metal impurities (Zn, Fe) end up on the cathode?
    Show answer
    Their E° values are more negative than Cu²⁺/Cu. At the cathode voltage chosen to deposit Cu, only Cu²⁺ is reduced — Zn²⁺ and Fe²⁺ remain in solution (their reduction requires a more negative cathode potential).
  2. Why are unreactive impurities (Ag, Au) found as 'anode slime'?
    Show answer
    Their E° values are more positive than Cu — they're harder to oxidise than copper. So they don't dissolve at the chosen anode voltage; they fall off the anode as the surrounding Cu dissolves around them and collect at the bottom of the cell. Often recovered as a valuable by-product.
  3. Electroplating a fork with silver: which electrode is the fork, which is silver, what is the electrolyte?
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    Fork = cathode (negative; where Ag deposits). Silver = anode (positive; dissolves as Ag⁺). Electrolyte = AgNO₃(aq) — provides Ag⁺ ions and conducts current.
  4. An electroplating bath operates at 2.0 A for 1.0 hour. Calculate the mass of Ag deposited.
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    Q = I × t = 2.0 × 3600 = 7200 C. n(e⁻) = 7200/96485 = 0.0746 mol. Ag⁺ + e⁻ → Ag, so n(Ag) = 0.0746 mol. m = 0.0746 × 107.87 = 8.05 g.
Vocabulary

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elementatomprotonelectronmolestoichiometryconcentrationsolutionbalanced equationgraphcatalystequilibriumequilibrium constantkcgibbs free energyspontaneous reactionperiodic tableperiodgrouphalogenselectronegativitycovalent bondelectronegativity differencepolymerredoxoxidationreductionoxidation stateoxidising agentreducing agenthalf-equationoverall equationspectator ionelectrochemical cellvoltaic cellelectrolytic cellanodecathodesalt bridgeelectrolyteelectrolysisstandard electrode potential