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IB Diploma Chemistry HL · Reactivity 3.2 (AHL) — electrode potentials and electrolysis
Mini-Lesson

Electrochemistry (AHL)

This Additional Higher Level mini-lesson covers Reactivity 3.2: standard electrode potentials, E°cell and feasibility, ΔG° = −nFE°, and quantitative electrolysis.

electrode potentialsE°cell & ΔG°electrolysis Reactivity 3.2 (AHL) — electron transfer, quantified

Work through each screen, answer the questions as you go (some are wordy, some are calculations) and collect ⭐ stars. Press Start when you're ready.

AHL — Reactivity 3.2

Standard electrode potentials

Each half-cell has a standard electrode (reduction) potential E°, measured against the standard hydrogen electrode (defined as 0.00 V) at 298 K, 100 kPa and 1 mol dm⁻³. A more positive E° means a stronger tendency to be reduced.

AHL — Reactivity 3.2

Cell potential and feasibility

Combine two half-cells:

E°cell = E°(cathode) − E°(anode)a reaction is feasible (spontaneous) when E°cell > 0

The half-cell with the more positive E° is reduced (cathode); the other is oxidised (anode).

AHL calculate

Cell potential

1Using E°(Cu²⁺/Cu) = +0.34 V and E°(Zn²⁺/Zn) = −0.76 V, calculate E°cell for the Zn–Cu cell.
V
Cu is the cathode: E°cell = 0.34 − (−0.76).
AHL — Reactivity 3.2

Linking E° to Gibbs energy

Cell potential connects directly to spontaneity:

ΔG° = −nFE°celln = moles of electrons, F = 96 500 C mol⁻¹

A positive E°cell gives a negative ΔG° — a feasible cell.

AHL calculate

Gibbs energy of the cell

2For the Zn–Cu cell (n = 2, E°cell = 1.10 V), calculate ΔG° in kJ (F = 96 500 C mol⁻¹).
kJ
ΔG° = −nFE° = −(2 × 96 500 × 1.10) J, then ÷ 1000.
AHL — Reactivity 3.2

Quantitative electrolysis

The amount deposited depends on charge passed:

Q = I t and n(e⁻) = Q ÷ FQ in coulombs, I in amps, t in seconds, F = 96 500 C mol⁻¹

Use the half-equation to convert moles of electrons to moles (and mass) of product.

AHL calculate

Mass deposited

3A current of 2.0 A flows for 30 minutes through CuSO₄. For Cu²⁺ + 2e⁻ → Cu (Cu = 63.55), calculate the mass of copper deposited.
g
Q = 2.0 × 1800 = 3600 C; n(e⁻) = 3600 ÷ 96 500; n(Cu) = half of that; mass = n × 63.55.
AHL check

Feasible cell

?A cell reaction is spontaneous (feasible) when E°cell is:
AHL check

Faraday relationship

?Which expression gives the moles of electrons passed in electrolysis?
Sort it

Voltaic, electrolytic or law?

Tap a phrase, then where it belongs.

🔋 Voltaic (spontaneous)

🔌 Electrolytic (driven)

🧮 Quantitative law

Match it

Match the quantity to its formula

Tap an item on the left, then its match on the right.

Quantity
Formula
Recap

The big ideas to know

Electrode potentials: E° vs the standard hydrogen electrode (0.00 V)

Cell & feasibility: E°cell = E°cathode − E°anode; feasible when > 0

Gibbs link: ΔG° = −nFE°cell

Electrolysis: Q = It, n(e⁻) = Q ÷ F, then use the half-equation

You've covered Electrochemistry (AHL) for IB Diploma Chemistry HL. Press Finish to see your score.

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