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

This page teaches the higher level end of Reactivity 3.2, “Electron transfer reactions”, for IB DP Chemistry. It follows the IB’s Diploma Programme Chemistry guide (first assessment 2025, published February 2023) and covers understandings Reactivity 3.2.12 to 3.2.16. All five are additional higher level (AHL) material, so everything here is HL only, for exam sessions in 2025 and later years.

Use it alongside the condensed notes and a set of marked practice questions. Browse other units from the DP Chemistry hub, mark progress on the DP Chemistry checklist, and take a free diagnostic quiz to see where to start. Gibbs energy itself is taught in Reactivity 1.4: entropy and spontaneity.

Scope of this unit

Understanding	Skills	Level
Reactivity 3.2.12	Explain why the hydrogen half-cell is set at 0 V and how it is used to measure E⦵; read E⦵ data as a guide to how easily species are oxidized or reduced	HL only
Reactivity 3.2.13	Calculate E⦵cell from two electrode potentials; use its sign to decide whether a reaction is spontaneous forwards or in reverse	HL only
Reactivity 3.2.14	Use ΔG⦵ = −nFE⦵cell to find ΔG⦵ (or E⦵cell)	HL only
Reactivity 3.2.15	Deduce the products of electrolysing water and aqueous solutions, including the effect of concentration (NaCl(aq)) and of the electrode material (CuSO₄(aq))	HL only
Reactivity 3.2.16	Write the electrode equations for electroplating and describe the set-up	HL only

The guide notes that standard reduction potentials, the ΔG⦵ = −nFE⦵cell equation and the value of F are given in the data booklet. Here, use these values (E⦵ / V, for the reduction of the first species to the second):

Couple	E⦵ / V	Couple	E⦵ / V
K⁺/K	−2.93	Cu²⁺/Cu	+0.34
Mg²⁺/Mg	−2.37	Ag⁺/Ag	+0.80
H₂O/H₂, OH⁻	−0.83	Br₂/Br⁻	+1.09
Ni²⁺/Ni	−0.26	O₂, H⁺/H₂O	+1.23
Sn²⁺/Sn	−0.14	Cr₂O₇²⁻, H⁺/Cr³⁺	+1.33
H⁺/H₂	0.00	Cl₂/Cl⁻	+1.36

F = 96 500 C mol⁻¹ and R = 8.31 J K⁻¹ mol⁻¹.

Reactivity 3.2.12: the reference electrode and E⦵ (HL only)

A voltmeter only reads the difference between two electrodes, so one half-cell’s potential cannot be measured alone. By convention, the hydrogen half-cell is the zero:

H⁺(aq) + e⁻ ⇌ ½H₂(g)        E⦵ = 0.00 V (by convention)

In the standard hydrogen electrode, hydrogen at 100 kPa bubbles over an inert platinum electrode in 1.00 mol dm⁻³ H⁺(aq) at 298 K.

A half-cell’s standard electrode potential, E⦵, is the voltage it gives against this reference under standard conditions (298 K, 100 kPa, 1.00 mol dm⁻³ ions). A salt bridge and a high-resistance voltmeter join the two half-cells. A metal dips into a solution of its own ions; for two ions of one element, such as Fe³⁺ and Fe²⁺, platinum is used. If the test electrode is the negative terminal, its E⦵ is negative.

Reading E⦵ data

Every E⦵ value is quoted for the reduction, with electrons on the left.

• More positive E⦵: the species on the left gains electrons readily. It is easily reduced, so it is a strong oxidizing agent.
• More negative E⦵: the species on the right loses electrons readily. It is easily oxidized, so it is a strong reducing agent.

Worked example 1: a magnesium half-cell

A clean magnesium ribbon in 1.00 mol dm⁻³ Mg²⁺(aq) is connected to a standard hydrogen electrode. The voltmeter reads 2.37 V and the magnesium is the negative terminal.

Electrons leave the magnesium, so it is oxidized far more readily than hydrogen: E⦵(Mg²⁺/Mg) = −2.37 V.

Among Mg, Sn, Ag⁺ and Br₂, the strongest oxidizing agent is Br₂ (+1.09 V) and the strongest reducing agent is Mg (−2.37 V). Ease of oxidation of the metals: Mg > Sn > Ag.

Reactivity 3.2.13 Standard cell potential (HL only)

When two half-cells are joined, the one with the more positive E⦵ is reduced (cathode); the other is oxidized (anode).

E⦵cell = E⦵(cathode) − E⦵(anode)

Use both values exactly as tabulated. E⦵cell is positive for a spontaneous reaction. If it comes out negative for the equation you wrote, the reverse reaction is the spontaneous one.

Worked example 2: a tin–silver cell

1. Silver (+0.80 V) is more positive than tin (−0.14 V), so Ag⁺ is reduced at the cathode.
2. E⦵cell = 0.80 − (−0.14) = +0.94 V.
3. Anode: Sn(s) → Sn²⁺(aq) + 2e⁻. Cathode: Ag⁺(aq) + e⁻ → Ag(s).
4. Balance electrons by doubling the silver equation: Sn(s) + 2Ag⁺(aq) → Sn²⁺(aq) + 2Ag(s).

Doubling the silver half-equation does not double its E⦵. The tin anode is negative, and electrons flow through the wire from tin to silver.

Worked example 3: which halide can dichromate oxidize?

Acidified dichromate(VI) is added separately to bromide and to chloride under standard conditions. Dichromate would be reduced, so it is the cathode couple (+1.33 V).

1. With bromide: E⦵cell = 1.33 − 1.09 = +0.24 V. Positive, so dichromate oxidizes bromide to bromine.
2. With chloride: E⦵cell = 1.33 − 1.36 = −0.03 V. Negative, so no reaction under standard conditions.

For bromide:

Cr₂O₇²⁻(aq) + 14H⁺(aq) + 6Br⁻(aq) → 2Cr³⁺(aq) + 3Br₂(aq) + 7H₂O(l)

The same test shows which metals release hydrogen from 1.00 mol dm⁻³ acid. For tin, E⦵cell = 0.00 − (−0.14) = +0.14 V, so tin reacts. For copper, E⦵cell = 0.00 − 0.34 = −0.34 V, so copper does not. A positive E⦵cell says nothing about rate.

Reactivity 3.2.14 Gibbs energy and cell potential (HL only)

Standard Gibbs energy change and standard cell potential are linked by:

ΔG⦵ = −nFE⦵cell

n is the amount of electrons (mol) transferred in the equation as written, and F is the charge on 1 mol of electrons. With E⦵cell in volts, ΔG⦵ is in J mol⁻¹. A positive E⦵cell always gives a negative ΔG⦵.

Worked example 4: ΔG⦵ for the tin–silver cell

The equation Sn + 2Ag⁺ → Sn²⁺ + 2Ag transfers 2 mol of electrons, so n = 2.

ΔG⦵ = −2 × 96 500 × 0.94 = −181 420 J mol⁻¹ = −181 kJ mol⁻¹

Worked example 5: the role of n

In the dichromate–bromide equation, 6Br⁻ give up 6 electrons, so n = 6.

ΔG⦵ = −6 × 96 500 × 0.24 = −138 960 J mol⁻¹ = −139 kJ mol⁻¹

Halve the equation and n becomes 3. E⦵cell stays at +0.24 V, but ΔG⦵ halves to −69.5 kJ mol⁻¹.

Working backwards: if a reaction has ΔG⦵ = −96.5 kJ mol⁻¹ and n = 2, then E⦵cell = 96 500 ÷ (2 × 96 500) = +0.500 V.

Linking to Reactivity 1.4

Thermodynamic data give ΔG⦵ = ΔH⦵ − TΔS⦵; electrochemical data give ΔG⦵ = −nFE⦵cell. Either predicts spontaneity, and ΔG⦵ = −RT lnK then gives the equilibrium position. For the tin–silver cell:

lnK = 181 420 ÷ (8.31 × 298) = 73.3        K ≈ 7 × 10³¹

A large positive E⦵cell means the reaction goes essentially to completion.

Reactivity 3.2.15 Electrolysis of aqueous solutions (HL only)

In an electrolytic cell a DC supply drives a non-spontaneous reaction. Reduction happens at the negative cathode, oxidation at the positive anode. In aqueous solution, water competes with the ions at both electrodes:

• Cathode: the species with the more positive E⦵ is reduced. Compare each cation with water (−0.83 V). K⁺ (−2.93 V) and Mg²⁺ (−2.37 V) are not reduced, so hydrogen forms; Cu²⁺ and Ag⁺ are reduced to the metal.
• Anode: the species that is easiest to oxidize reacts, from the couple with the least positive E⦵. Compare each anion with water (+1.23 V). Iodide and bromide give the halogen; sulfate and nitrate are not oxidized, so oxygen forms.

Electrolysis of water

Pure water conducts poorly, so a little sulfuric acid or sodium sulfate is added. With inert electrodes:

Cathode (−): 2H₂O(l) + 2e⁻ → H₂(g) + 2OH⁻(aq)
Anode (+):   2H₂O(l) → O₂(g) + 4H⁺(aq) + 4e⁻
Overall:     2H₂O(l) → 2H₂(g) + O₂(g)

The volume of hydrogen is twice that of oxygen. Using the acidic couples (H⁺/H₂ and O₂, H⁺/H₂O), E⦵cell = 0.00 − 1.23 = −1.23 V, so the supply must provide more than 1.23 V.

Worked example 6: aqueous potassium bromide

1. Cathode: K⁺ (−2.93 V) against water (−0.83 V). Water has the more positive value, so hydrogen forms: 2H₂O(l) + 2e⁻ → H₂(g) + 2OH⁻(aq).
2. Anode: Br⁻ (+1.09 V) against water (+1.23 V). Bromide is oxidized more easily, so bromine forms: 2Br⁻(aq) → Br₂(aq) + 2e⁻.
3. The solution turns orange near the anode and alkaline near the cathode.

Effect of concentration: sodium chloride solution

For chloride, the two anode values (+1.36 V and +1.23 V) are close, so conditions decide the product.

• Concentrated NaCl(aq): chloride is oxidized and chlorine forms at the anode, 2Cl⁻(aq) → Cl₂(g) + 2e⁻. The high Cl⁻ concentration favours this, and oxygen is in practice slow to form at the electrode. Hydrogen forms at the cathode and sodium hydroxide remains.
• Dilute NaCl(aq): water is oxidized and oxygen is the main anode product, as E⦵ predicts. Hydrogen still forms at the cathode.

Effect of the electrode: copper(II) sulfate solution

• Inert electrodes (graphite or platinum): Cu²⁺ (+0.34 V) beats water at the cathode, so copper is deposited. Sulfate is not oxidized, so the anode gives oxygen. As Cu²⁺ is used up the solution loses its blue colour and becomes acidic.
• Copper electrodes: copper (+0.34 V) is far easier to oxidize than water (+1.23 V), so the anode dissolves: Cu(s) → Cu²⁺(aq) + 2e⁻. Copper deposits on the cathode at the same rate, so the blue colour stays the same.

Reactivity 3.2.16 Electroplating (HL only)

In electroplating, electrolysis deposits a thin film of one metal on an object:

• the object to be plated is the cathode, because metal ions are reduced onto it;
• the anode is the plating metal, which dissolves and replaces the ions used up;
• the electrolyte contains ions of the plating metal.

Worked example 7: nickel-plating a steel bicycle bell

Anode (nickel bar):   Ni(s) → Ni²⁺(aq) + 2e⁻
Cathode (steel bell): Ni²⁺(aq) + 2e⁻ → Ni(s)
Electrolyte:          nickel(II) sulfate solution

Nickel dissolves from the bar as fast as it deposits, so the electrolyte concentration stays steady.

Electroplating in the laboratory

Clean and degrease the object so the coating sticks, then weigh it. Connect it to the negative terminal of a low-voltage DC supply, with the plating-metal anode on the positive terminal. A small current gives a smoother, better-attached layer than a large one. Rinse, dry and reweigh: the mass gained is the metal deposited.

Common errors

• Reversing the sign of an E⦵ value and then also subtracting, which counts the change twice.
• Multiplying E⦵ by the coefficient used to balance electrons.
• Using the wrong n in ΔG⦵ = −nFE⦵cell: n is electrons transferred in the full equation, not the charge on one ion.
• Saying sodium or potassium is deposited from an aqueous solution.
• Predicting chlorine from dilute NaCl(aq), or forgetting that a copper anode dissolves.
• Making the object the anode in electroplating.

Next steps

Next, consolidate with the condensed notes, and check yourself on the marked questions.

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).