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3.2.14 (HL)—Gibbs energy and cell potential

Syllabus
First assessment 2025
Objective
3.2.14
Level
HL

Gibbs Energy and Cell Potential

HL only

ΔG°=−nFE°cellΔG° = −nFE°cell

n is the moles of electrons transferred and F is Faraday's constant. Positive E°cell gives negative ΔG° and a spontaneous reaction.

Find n from the balanced overall redox equation, not from a single unscaled half-equation. With E° in volts and F in C mol⁻¹, ΔG° is obtained in J mol⁻¹; convert to kJ mol⁻¹ only at the end.

Worked ΔG∘\Delta G^\circ example: for 2 HX++Zn→ZnX2++HX2\ce{2H+ + Zn -> Zn^{2+} + H2}, n=2n=2 and Ecell∘=+0.76 VE^\circ_{cell}=+0.76\,\mathrm{V}. Using F=9.65×104 C mol−1F=9.65\times10^4\,\mathrm{C\,mol^{-1}}, ΔG∘=−(2)(9.65×104)(0.76)=−1.47×105 J mol−1=−147 kJ mol−1\Delta G^\circ=-(2)(9.65\times10^4)(0.76)=-1.47\times10^5\,\mathrm{J\,mol^{-1}}=-147\,\mathrm{kJ\,mol^{-1}}. Its negative sign agrees with a spontaneous standard-cell reaction.

Calculating ΔG° for a Cell

HL only

Assessment in practice

Representative question

Question 1

[Maximum number: 2]

Calculate the standard Gibbs free energy of the cell, in kJmol−1\mathrm{kJ} \mathrm{mol}^{-1}. Use sections 1, 2 and 24 of the data booklet.

Electron Transfer Reactions Summary

Retrieve the route: assign oxidation states, balance half-equations, predict displacement, label cells, trace electrons and ions, follow organic redox pathways, calculate potentials and choose electrolysis products.

Check electron loss/gain, anode/cathode versus polarity, spontaneous sign, salt-bridge direction, ions present, organic functional-group direction and object-cathode placement.