Electrochemical Thermodynamics Review

Introduction

An electrochemical cell converts chemical free-energy change into electrical work. Its reversible voltage reflects a difference in chemical potentials, while the voltage observed under appreciable current includes kinetic and transport losses. Advanced problems begin by separating this equilibrium thermodynamic limit from how a real device performs.

Core explanation

A cell reaction combines oxidation at an anode and reduction at a cathode. Electrons released by the anode travel through an external conductor to the cathode in a galvanic cell. In both galvanic and electrolytic cells, “anode” means oxidation and “cathode” means reduction; the signs of the electrodes can differ with mode. Balance atoms and charge in each half-reaction, then cancel the same number of electrons to obtain the overall reaction.

If n moles of electrons flow per mole of balanced reaction and E is the reversible cell potential, the reaction Gibbs energy is ΔᵣG=−nFE. Faraday constant F is about 96,485 C mol⁻¹ of electrons. Because 1 V=1 J C⁻¹, nFE has units J per mole of reaction as written. A positive reversible E gives a negative ΔᵣG for the forward reaction under the stated conditions. The result depends on how the reaction is scaled: doubling all stoichiometric coefficients doubles n and ΔᵣG, but not E.

Under standard-state conditions, ΔᵣG°=−nFE°. For nonstandard composition, ΔᵣG=ΔᵣG°+RT ln Q, leading to E=E°−(RT/nF)ln Q. Q uses activities of reacting species, with pure solids and liquids usually assigned activity one in the chosen standard state. At equilibrium ΔᵣG=0 and E=0 for the reaction as written, although individual electrode potentials relative to a reference need not be zero.

The voltage is an intensive quantity: it does not double when a larger cell contains twice as much material. Capacity and total available energy do depend on amount of reactant. A larger battery can deliver more charge at a similar open-circuit voltage, while connecting identical cells in series can increase total voltage. Keep physical scaling separate from the thermodynamic potential of one reaction.

When current flows, measured terminal voltage differs from reversible open-circuit voltage because of internal resistance, charge-transfer overpotential and concentration gradients. A galvanic battery under discharge generally delivers less voltage than its reversible value; electrolysis generally requires more applied voltage than the reversible threshold. These losses depend on rate and design, so they should not be inserted into E° or mistaken for a changed equilibrium constant.

Potential alone does not give an absolute electron energy at one isolated half-cell. Only potential differences are measured directly, and electrode potentials are tabulated relative to a reference convention. A complete cell reaction and standard states must be stated before using E° to predict thermodynamic direction.

Step-by-step reasoning

1. Balance oxidation and reduction half-reactions. 2. Count n electrons in the overall reaction. 3. Choose standard or actual conditions. 4. Calculate ΔᵣG from −nFE or E from Gibbs energy. 5. If discussing a working device, add a separate account of kinetic, resistive and transport losses.

Visual explanation

Draw two electrodes connected by a wire and salt pathway. Mark electrons leaving the oxidation electrode and reaching the reduction electrode. Beside the diagram draw a high line for reversible open-circuit voltage and a lower line for loaded galvanic terminal voltage, with the gap labeled losses.

Real-world analogy

A waterfall's height difference represents available driving force, while the amount of water determines how much total work can be done. Voltage likewise reflects a per-charge driving force, while charge capacity depends on reactant amount. Friction in a pipe resembles losses when current flows.

Real-world example

A laboratory galvanic cell may show a stable open-circuit voltage with a high-resistance meter. Connecting a low-resistance load draws current and often lowers the terminal voltage. The cell chemistry did not instantly change its standard potential; internal resistance and polarization account for much of the immediate drop.

Why?

Why does doubling a balanced reaction not double its E°? Both reaction Gibbs energy and electrons transferred double. Since E°=−ΔᵣG°/(nF), the common factor cancels, leaving the same intensive potential.

Common misconception

“A positive standard cell voltage guarantees that a device delivers that exact voltage at any current.” E° describes a reversible standard-state limit. Composition changes voltage through the Nernst relation, and operating current introduces additional losses.

Worked example

A balanced reaction transfers n=2 electrons and has E°=0.80 V. ΔᵣG°=−2(96,485 C mol⁻¹)(0.80 J C⁻¹)=−154,376 J mol⁻¹, about −154 kJ per mole of reaction. If the entire reaction equation were doubled, ΔᵣG° would be about −309 kJ and n=4, while E° stays 0.80 V.

Quick check

1. Which electrode hosts oxidation in both galvanic and electrolytic cells? Answer: The anode. 2. Is E or nFE extensive when a balanced reaction is doubled? Answer: nFE is extensive for the stated reaction amount; E remains intensive.

Exam focus

Balance the reaction before counting n, show charge-to-energy units and distinguish standard from actual composition. Use ΔG=−nFE only for reversible potential under the specified state. Treat operating losses separately from thermodynamic E°.

Advanced insight

Voltage depends on electrochemical potentials of charged species, not just an intuitive “electron pressure.” Separating electrical and chemical contributions at each interface requires reference conventions, but their complete-cell difference is measurable. This explains why isolated absolute half-cell potentials are not directly observed.

Summary

Reversible cell voltage measures chemical Gibbs-energy change per unit transferred charge: ΔᵣG=−nFE. Standard potentials refer to defined activities, and nonstandard compositions follow the Nernst equation. Real operating voltage additionally reflects resistance, electron-transfer kinetics and transport.

Practice questions

1. A cell reaction has n=1 and ΔᵣG°=−48.2 kJ mol⁻¹. Estimate E°. Answer: E°=48,200/96,485≈0.500 V. 2. If a galvanic cell is discharged rapidly, why may terminal voltage fall below open-circuit voltage? Answer: Internal resistance, charge-transfer overpotential and concentration gradients create rate-dependent losses. 3. What happens to E° if every coefficient in the cell reaction is multiplied by three? Answer: It stays the same; both ΔᵣG° and n triple.