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Quantitative Electrochemistry: ΔG°, E°, and K

You’ll be able to

Three numbers, one reaction

A reaction has exactly one thermodynamic identity, but electrochemistry lets you read it three ways: as a free energy (ΔG°), as a voltage (E°cell), and as an equilibrium position (K). All three answer the same question — how far, and in which direction, does the reaction want to go? Because they describe one reality, any one of them fixes the other two. The whole unit hinges on the two bridges that connect them.

Free energy from cell potential
ΔG° = −nFE°
n = moles of electrons transferred in the balanced reaction, F = 96 485 C·mol⁻¹. E° > 0 ⇒ ΔG° < 0 ⇒ spontaneous. E° is intensive — never scale it by n or by balancing coefficients; the n out front is what accounts for reaction size.
Free energy from the equilibrium constant
ΔG° = −RT ln K
R = 8.314 J·mol⁻¹·K⁻¹, T in kelvin. K > 1 ⇒ ΔG° < 0 (products favored); K < 1 ⇒ ΔG° > 0 (reactants favored).

Setting the two bridges equal

Both expressions equal the same ΔG°, so set them equal: −nFE° = −RT ln K. Cancel the minus signs and solve for the potential: E° = (RT/nF) ln K. At 298 K the clump of constants (RT/F)·(ln 10) collapses to the famous 0.0592, giving the log-form shortcut E° = (0.0592/n) log K, or rearranged, log K = nE°/0.0592. This one line lets you jump straight from a table of cell voltages to an equilibrium constant without ever computing ΔG° in between.

Voltage and equilibrium (298 K shortcut)
log K = nE° / 0.0592
Derived by equating ΔG° = −nFE° with ΔG° = −RT ln K at 25 °C. A modest E° of a few tenths of a volt produces an enormous K — voltage and equilibrium are exponentially related.
Worked example

The cell Cu(s) | Cu²⁺ || Ag⁺ | Ag(s) has E°cell = +0.46 V. The reaction is Cu(s) + 2 Ag⁺ → Cu²⁺ + 2 Ag(s). Find ΔG° and the equilibrium constant K at 298 K.

  1. 1.Count electrons: copper gives up 2 e⁻ and each of the two silver ions gains 1 e⁻, so n = 2. E° stays +0.46 V — it is never scaled.
  2. 2.ΔG° = −nFE° = −(2)(96 485 C·mol⁻¹)(0.46 V) = −88 766 J ≈ −88.8 kJ·mol⁻¹. Negative, so the reaction is spontaneous as written.
  3. 3.For K, use the shortcut log K = nE°/0.0592 = (2)(0.46)/0.0592 = 0.92/0.0592 = 15.54.
  4. 4.K = 10¹⁵·⁵⁴ = 10⁰·⁵⁴ × 10¹⁵ ≈ 3.5 × 10¹⁵.
  5. 5.Sanity check via the other bridge: ln K = −ΔG°/RT = 88 766 ÷ (8.314 × 298) = 35.8, and 35.8 ÷ 2.303 = 15.6 — the same log K, confirming both bridges agree.
Answer: ΔG° ≈ −88.8 kJ·mol⁻¹ and K ≈ 3.5 × 10¹⁵ — a small positive voltage corresponds to an overwhelmingly product-favored equilibrium.

Electrolysis: charge is a reactant you can meter

In an electrolytic cell you pay for the reaction with electricity, and the bill is measured in coulombs. Faraday's laws state that the amount of substance transformed at an electrode is directly proportional to the charge passed. Charge is current × time (Q = I·t, coulombs = amperes × seconds), and one mole of electrons carries 96 485 C. So the chain is: I·t → Q → mol e⁻ (÷F) → mol product (÷ electrons per ion) → grams (× molar mass).

Faraday's laws of electrolysis
mol e⁻ = Q ÷ F = (I·t) ÷ 96 485
Then divide by the electrons-per-ion from the half-reaction: Cu²⁺ + 2e⁻ → Cu needs 2 mol e⁻ per mol Cu; Ag⁺ + e⁻ → Ag needs 1.
Worked example

A current of 5.00 A is passed through aqueous CuSO₄ for 1930 s. What mass of copper (63.55 g·mol⁻¹) plates onto the cathode? The half-reaction is Cu²⁺ + 2e⁻ → Cu.

  1. 1.Total charge: Q = I·t = (5.00 A)(1930 s) = 9650 C.
  2. 2.Moles of electrons: mol e⁻ = Q ÷ F = 9650 ÷ 96 485 = 0.100 mol e⁻.
  3. 3.The half-reaction needs 2 e⁻ per Cu, so mol Cu = 0.100 ÷ 2 = 0.0500 mol.
  4. 4.Mass = 0.0500 mol × 63.55 g·mol⁻¹ = 3.18 g.
Answer: 3.18 g of copper deposits at the cathode.
Checkpoint

A galvanic cell has E°cell = +0.85 V. Which set of conclusions about ΔG° and K is correct?

Checkpoint

A cell has E°cell = +0.30 V with n = 2 electrons transferred at 298 K. Approximately what is the equilibrium constant K?

Checkpoint

How long must a 2.00 A current flow to deposit 0.0500 mol of copper from Cu²⁺ (Cu²⁺ + 2e⁻ → Cu)?

On the exam

Memorize the triangle: E° ↔ ΔG° ↔ K, joined by ΔG° = −nFE° and ΔG° = −RT ln K. Given any one vertex you can reach the other two. The single most common error is scaling E° by n or by a coefficient — E° is intensive and never changes; the n lives in the −nFE° expression, not inside the voltage.

Watch out

Unit discipline wins these problems. F = 96 485 C·mol⁻¹ and R = 8.314 J·mol⁻¹·K⁻¹ both use joules, so ΔG° from −nFE° comes out in joules — divide by 1000 to report kJ·mol⁻¹. And in Faraday problems, carry A × s = C through every arrow so a stray factor of n never slips in.

Answer the 3 checkpoints as you read.

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