Equilibrium constant and Gibbs energy from cell EMF
A positive EMF means a spontaneous reaction: ΔrG° = −nFE°cell. At equilibrium E_cell = 0, so log K = nE°cell ÷ 0.059 at 298 K.
Equilibrium constant from the Nernst equation
- 1Nernst equation at 298 K: E_cell = E°cell − (0.059/n) log Q.
- 2As the cell works, Q rises and E_cell falls. At equilibrium, the cell is 'dead': E_cell = 0 and Q = K.
- 30 = E°cell − (0.059/n) log K.
- 4★So, E°cell = (0.059/n) log K, or log K = nE°cell ÷ 0.059.
- 5In general, at any temperature: E°cell = (2.303RT/nF) log K.
★ key step
Gibbs energy and electrical work
The largest electrical work a cell can do equals the fall in Gibbs energy of its reaction. Electrical work = charge × potential difference = nF × E_cell, so ΔrG = −nFE_cell. Under standard conditions, ΔrG° = −nFE°cell. E_cell is an intensive property: it does not depend on how much substance reacts. ΔrG is extensive: doubling the equation doubles n, and so doubles ΔrG. E_cell stays the same. ΔrG° is also linked to K by ΔrG° = −RT ln K.
Linking E°cell, ΔrG° and K
| E°cell | ΔrG° | K | Forward reaction |
|---|---|---|---|
| Positive | Negative | Greater than 1 | Spontaneous |
| Zero | Zero | Equal to 1 | At equilibrium |
| Negative | Positive | Less than 1 | Not spontaneous (the reverse is) |
Worked example: the Daniell cell
For the Daniell cell, E°cell = 1.10 V at 298 K. Find (a) ΔrG° and (b) K for Zn(s) + Cu²⁺(aq) ⇌ Zn²⁺(aq) + Cu(s). Take F = 96500 C mol⁻¹ and log 2 = 0.3010.
- n = 2.
- (a) ΔrG° = −nFE°cell = −2 × 96500 C mol⁻¹ × 1.10 V = −212300 J mol⁻¹.
- ΔrG° = −212.3 kJ mol⁻¹. (With F = 96487 C mol⁻¹ it is −212.27 kJ mol⁻¹.)
- (b) log K = nE°cell ÷ 0.059 = (2 × 1.10) ÷ 0.059 = 2.20 ÷ 0.059 = 37.29.
- K = 10⁰·²⁹ × 10³⁷. Since log 2 = 0.3010 is close to 0.29, 10⁰·²⁹ ≈ 2.
- K ≈ 2 × 10³⁷, so the reaction goes practically to completion.
(a) ΔrG° = −212.3 kJ mol⁻¹; (b) K ≈ 2 × 10³⁷
Worked example: a board-style numerical
The standard EMF of a cell in which two electrons are transferred is 0.295 V at 298 K. Find (a) the equilibrium constant and (b) ΔrG° of the cell reaction. (F = 96500 C mol⁻¹)
- (a) log K = nE°cell ÷ 0.059 = (2 × 0.295) ÷ 0.059 = 0.590 ÷ 0.059 = 10.
- K = 1 × 10¹⁰.
- (b) ΔrG° = −nFE°cell = −2 × 96500 × 0.295 J mol⁻¹.
- ΔrG° = −56935 J mol⁻¹ ≈ −56.9 kJ mol⁻¹.
(a) K = 1 × 10¹⁰; (b) ΔrG° ≈ −56.9 kJ mol⁻¹
An inverter battery at home or an e-rickshaw battery stores energy as the Gibbs energy of a chemical reaction. ΔrG = −nFE tells engineers the most electrical work each kilogram of chemicals can deliver.
3 marks: “For the reaction Fe(s) + Cu²⁺(aq) → Fe²⁺(aq) + Cu(s), calculate ΔrG° and log K at 298 K. E°(Fe²⁺/Fe) = −0.44 V; E°(Cu²⁺/Cu) = +0.34 V; F = 96500 C mol⁻¹.” E°cell = 0.34 − (−0.44) = 0.78 V; n = 2. ΔrG° = −2 × 96500 × 0.78 = −150540 J mol⁻¹ ≈ −150.5 kJ mol⁻¹. log K = (2 × 0.78) ÷ 0.059 = 1.56 ÷ 0.059 = 26.44.
2 marks: “Why is E_cell an intensive property while ΔrG of the cell reaction is extensive?” E_cell depends only on the nature and concentrations of the species, not on the amount. ΔrG = −nFE_cell depends on n, which doubles when the equation is doubled.
1 marks: “What is the EMF of a cell whose reaction has reached equilibrium?” Zero.
Positive E, negative G, big K: go! ΔrG° = −nFE°cell and log K = nE°cell ÷ 0.059.