IB DP Chemistry Reactivity 3.2: Electron transfer: standard electrode potentials and Gibbs energy – Revision Notes

These notes assume you have worked through the study guide for this unit, which explains each idea in full.

These notes condense Reactivity 3.2.12 to 3.2.16 of the IB Diploma Programme Chemistry guide (first assessment 2025, published February 2023): standard electrode potentials, cell potentials, Gibbs energy, aqueous electrolysis and electroplating. Every one of these understandings is additional higher level material, so the whole page is HL only. The notes apply to exam sessions from 2025 onwards.

Afterwards, attempt the marked practice set. Other units are on the DP Chemistry hub, the DP Chemistry checklist lets you record what you have covered, and the free diagnostic quizzes point out weak spots. Gibbs energy background is in the Reactivity 1.4 notes. Cell basics (anode, cathode, salt bridge, molten electrolysis) are in the Reactivity 3.2.1-3.2.11 notes.

Data for this page

Use these standard electrode potentials, E⦵ / V. Each couple is written oxidized form / reduced form.

Couple	E⦵ / V	Couple	E⦵ / V
K⁺/K	−2.93	Pb²⁺/Pb	−0.13
Na⁺/Na	−2.71	H⁺/H₂	0.00
Mg²⁺/Mg	−2.37	Cu²⁺/Cu	+0.34
Al³⁺/Al	−1.66	I₂/I⁻	+0.54
H₂O/H₂, OH⁻	−0.83	Fe³⁺/Fe²⁺	+0.77
Zn²⁺/Zn	−0.76	Ag⁺/Ag	+0.80
Fe²⁺/Fe	−0.45	Br₂/Br⁻	+1.09
Ni²⁺/Ni	−0.26	O₂, H⁺/H₂O	+1.23
Sn²⁺/Sn	−0.14	Cl₂/Cl⁻	+1.36

F = 96 500 C mol⁻¹. The guide notes that standard reduction potentials, the equation ΔG⦵ = −nFE⦵cell and the value of F are provided in the data booklet.

Reactivity 3.2.12: definitions (HL only)

• Standard hydrogen electrode: H⁺(aq) + e⁻ ⇌ ½H₂(g). Platinum electrode, H₂ at 100 kPa, [H⁺] = 1.00 mol dm⁻³, 298 K. Its E⦵ is 0.00 V by convention.
• Why a reference? A voltmeter can only measure a difference between two half-cells, so one half-cell has to be defined as zero.
• Standard electrode potential, E⦵: the voltage a half-cell gives when paired with the hydrogen reference, everything at standard conditions.
• Sign: a half-cell that is the negative terminal against hydrogen has a negative E⦵.
• Reading the table: values are reductions. The more positive the E⦵, the more easily the oxidized form is reduced (stronger oxidizing agent). The more negative the E⦵, the more easily the reduced form is oxidized (stronger reducing agent).

Key equations

Quantity	Equation	Units and notes
Cell potential	E⦵cell = E⦵(cathode) − E⦵(anode)	V; use both values as tabulated
Gibbs energy	ΔG⦵ = −nFE⦵cell	J mol⁻¹; divide by 1000 for kJ mol⁻¹
Rearranged	E⦵cell = −ΔG⦵ ÷ nF	ΔG⦵ in J mol⁻¹
Link to K (Reactivity 1.4)	ΔG⦵ = −RT lnK	R = 8.31 J K⁻¹ mol⁻¹

Reactivity 3.2.13 Method: is the reaction spontaneous? (HL only)

1. Split the equation into two half-equations.
2. The species gaining electrons is at the cathode. Find its E⦵.
3. The species losing electrons is at the anode. Find the E⦵ of its couple.
4. E⦵cell = E⦵(cathode) − E⦵(anode).
5. Positive: spontaneous as written. Negative: the reverse reaction is spontaneous.

Worked reminder. Magnesium and tin half-cells: E⦵cell = −0.14 − (−2.37) = +2.23 V. Mg is the anode (negative). Mg(s) + Sn²⁺(aq) → Mg²⁺(aq) + Sn(s).

Reactivity 3.2.14 Method: ΔG⦵ from E⦵ (HL only)

1. Write the balanced overall equation.
2. Count n: the moles of electrons transferred in that equation.
3. Put E⦵cell in volts and F = 96 500 C mol⁻¹ into ΔG⦵ = −nFE⦵cell.
4. Convert J mol⁻¹ to kJ mol⁻¹ and keep the sign.

Worked reminder. For the magnesium–tin reaction, n = 2: ΔG⦵ = −2 × 96 500 × 2.23 = −430 390 J mol⁻¹ = −430 kJ mol⁻¹.

The thermodynamic route (ΔG⦵ = ΔH⦵ − TΔS⦵, Reactivity 1.4) and the electrochemical route give the same quantity, so they always agree on spontaneity.

Reactivity 3.2.15 Method: products of aqueous electrolysis (HL only)

Electrolytic cell: anode positive (oxidation), cathode negative (reduction).

Cathode. Compare the cation with water, 2H₂O + 2e⁻ → H₂ + 2OH⁻ (−0.83 V).

• Cation E⦵ more positive than −0.83 V (for example Cu²⁺, Ag⁺): the metal forms.
• Cation E⦵ more negative (for example Na⁺, K⁺, Mg²⁺, Al³⁺): hydrogen forms.

Anode (inert electrode). Compare the anion with water, 2H₂O → O₂ + 4H⁺ + 4e⁻ (+1.23 V).

• I⁻ and Br⁻ are oxidized more easily: the halogen forms.
• SO₄²⁻ and NO₃⁻ are not oxidized: oxygen forms.
• Cl⁻: depends on concentration (below).

Worked reminder. Aqueous copper(II) bromide, inert electrodes. Cathode: Cu²⁺ (+0.34 V) is more positive than water (−0.83 V), so Cu²⁺(aq) + 2e⁻ → Cu(s). Anode: Br⁻ (+1.09 V) is easier to oxidize than water (+1.23 V), so 2Br⁻(aq) → Br₂(aq) + 2e⁻. You would see a pink-brown coating on the cathode and an orange-brown colour near the anode.

Electrolysis of water (a little acid or sodium sulfate added): H₂ at the cathode, O₂ at the anode, 2 : 1 by volume. E⦵cell = −1.23 V, so more than 1.23 V must be applied.

The two guide-specified cases

Electrolyte and electrodes	Cathode	Anode	Change in solution
Concentrated NaCl(aq), inert	H₂	Cl₂	NaOH forms
Dilute NaCl(aq), inert	H₂	mainly O₂	becomes more concentrated
CuSO₄(aq), inert (graphite or Pt)	Cu	O₂	blue fades; becomes acidic
CuSO₄(aq), copper electrodes	Cu	Cu dissolves: Cu → Cu²⁺ + 2e⁻	colour unchanged

Reactivity 3.2.16 Electroplating (HL only)

• Object to be plated = cathode: Mⁿ⁺(aq) + ne⁻ → M(s).
• Plating metal = anode: M(s) → Mⁿ⁺(aq) + ne⁻.
• Electrolyte = a solution of the plating metal’s ions; its concentration stays roughly constant.
• Practical points: clean and degrease the object first; a low current gives an even, firmly attached coat; weigh before and after to find the mass deposited.

Must-know distinctions

• Voltaic vs electrolytic cell: in a voltaic cell the anode is negative; in an electrolytic cell the anode is positive. Oxidation is at the anode in both.
• E⦵ vs ΔG⦵ when an equation is scaled: E⦵cell is unchanged; ΔG⦵ scales with n.
• Spontaneous vs fast: a positive E⦵cell predicts direction, not rate.
• Concentrated vs dilute NaCl(aq): chlorine vs mainly oxygen at the anode.
• Inert vs copper anode in CuSO₄(aq): oxygen vs the anode dissolving.
• Standard vs non-standard: E⦵ values hold only for standard conditions; other temperatures or concentrations change the potential.

Quick self-test

Use the data at the top of the page.

1. Why must one half-cell be assigned a potential of zero?
2. Which is the stronger reducing agent, Fe or Sn?
3. Calculate E⦵cell for a cell made from Mg²⁺/Mg and Cu²⁺/Cu half-cells.
4. Is Ag(s) + Fe³⁺(aq) → Ag⁺(aq) + Fe²⁺(aq) spontaneous under standard conditions?
5. A reaction has E⦵cell = +0.50 V and n = 3. Calculate ΔG⦵.
6. A reaction has ΔG⦵ = −193 kJ mol⁻¹ and n = 1. Calculate E⦵cell.
7. Calculate E⦵cell and ΔG⦵ for 2Al(s) + 3Sn²⁺(aq) → 2Al³⁺(aq) + 3Sn(s).
8. Name the products at each electrode when aqueous sodium nitrate is electrolysed with inert electrodes.
9. Name the anode product for aqueous sodium iodide with inert electrodes.
10. Write the anode half-equation for electrolysis of CuSO₄(aq) with copper electrodes.
11. How does the anode product change when concentrated NaCl(aq) is replaced by a dilute solution?
12. A steel bolt is electroplated with zinc. Name the cathode and anode, and write the cathode half-equation.

Answers

1. Only potential differences can be measured, so a reference (the hydrogen half-cell) is defined as 0.00 V.
2. Fe (−0.45 V is more negative than −0.14 V).
3. 0.34 − (−2.37) = +2.71 V.
4. E⦵cell = 0.77 − 0.80 = −0.03 V, so no; the reverse is spontaneous.
5. −3 × 96 500 × 0.50 = −144 750 J mol⁻¹ = −145 kJ mol⁻¹.
6. 193 000 ÷ (1 × 96 500) = +2.00 V.
7. E⦵cell = −0.14 − (−1.66) = +1.52 V; n = 6, ΔG⦵ = −6 × 96 500 × 1.52 = −880 kJ mol⁻¹.
8. Cathode hydrogen (Na⁺ at −2.71 V is below water); anode oxygen (nitrate is not oxidized).
9. Iodine (+0.54 V, easier to oxidize than water).
10. Cu(s) → Cu²⁺(aq) + 2e⁻.
11. From chlorine to mainly oxygen.
12. Cathode: the bolt; anode: zinc; Zn²⁺(aq) + 2e⁻ → Zn(s).

Where marks are usually lost

• Changing the sign of the anode E⦵ and then subtracting it as well.
• Doubling or tripling an E⦵ value to match the electrons in a balanced equation.
• Taking n as the charge on one ion rather than the total electrons in the overall equation.
• Leaving ΔG⦵ in J mol⁻¹, or dropping the negative sign.
• Saying sodium, potassium or aluminium is deposited from aqueous solution.
• Giving chlorine for dilute NaCl(aq), or oxygen for a copper anode in CuSO₄(aq).
• Mixing up polarities: the anode is negative in a voltaic cell but positive in an electrolytic cell.
• Putting the object at the anode in electroplating.
• Assuming a positive E⦵cell means a fast reaction.

Official syllabus

Source document: the Diploma Programme Chemistry guide (International Baccalaureate Organization), first assessment 2025, published February 2023, sub-topic Reactivity 3.2, understandings 3.2.12 to 3.2.16 (additional higher level).