Battery Voltage from Chemical Potentials

Introduction

A battery voltage is not a fixed label printed by chemistry alone. It represents the reversible electrical work available per unit charge for a specified reaction and composition. As lithium or another ion moves between electrode hosts, the energetic cost of occupying sites changes, so equilibrium voltage can vary with state of charge. Under load, additional kinetic and transport losses change the terminal voltage from that equilibrium value.

Core explanation

For a cell reaction that transfers n moles of electrons per mole of reaction as written, the reversible relation is ΔG = −nFEcell, where F is Faraday's constant, about 96,485 C mol⁻¹. A positive galvanic discharge voltage corresponds to negative reaction ΔG. The equation concerns a reversible, equilibrium or near-equilibrium process; it is not a promise that all free energy can be recovered under a high-current load. OpenStax's electrochemical thermodynamics chapter derives the connection between cell potential and Gibbs free energy.

Chemical potential μ is the change in Gibbs free energy when a small amount of a species is added at fixed temperature, pressure and other relevant amounts. In an insertion electrode, the chemical potential of inserted lithium depends on available crystal sites, occupancy, interactions with neighboring ions, transition-metal redox and possible phase changes. Moving lithium from one host to another changes free energy. The cell's equilibrium voltage is closely related to the difference in lithium electrochemical potentials between the two electrodes per unit charge. A favorable chemical-potential difference drives lithium ions internally and electrons externally during discharge.

The Nernst equation makes composition effects explicit for a reaction in terms of activities: E = E° − (RT/nF)ln Q, where Q is the reaction quotient. For ideal solutions, activities may be approximated by concentrations; in concentrated electrolytes and solids, that shortcut can fail. Activities account for nonideal interactions. In a battery electrode, changing lithium occupancy can shift μ even when the chemical formula still names the same nominal host. Consequently a voltage-versus-state-of-charge curve may slope rather than stay flat.

A two-phase transition can produce a relatively flat equilibrium voltage plateau over a range of overall composition. The two coexisting phases have fixed compositions at a given temperature and pressure, while their relative amounts change as charge moves. The chemical potential of the transferred species remains approximately constant across that coexistence region, giving an approximately constant reversible voltage. This is a thermodynamic explanation; finite-size particles, strain, disorder and current can broaden or tilt measured plateaus.

An open-circuit voltage is measured with negligible current after sufficient relaxation toward equilibrium. Immediately after charging or discharging stops, concentration gradients and interfacial overpotentials may still relax, so a quick voltage reading need not equal the fully equilibrated value. Under discharge, terminal voltage is generally lower than reversible voltage because charge-transfer reactions, ion transport, electronic resistance and concentration gradients consume part of the available driving force. During charging, applied voltage generally must exceed the reversible voltage to drive current in the opposite direction. This voltage gap is an expression of irreversibility and heat generation.

The electrode chemistry defines an energy landscape , but the full cell defines the usable voltage. A high-potential positive electrode paired with a low-potential negative electrode can provide high cell voltage, provided both remain stable in the electrolyte and can reversibly host the required ion inventory. Raising one electrode potential may increase energy per unit charge while also increasing electrolyte oxidation or structural instability. Thus maximizing voltage alone may reduce lifetime or safety.

Voltage, capacity and energy must be kept distinct. Capacity measures total charge passed over a defined window; energy is ∫V dQ. Two batteries with identical charge capacity but different voltage profiles can deliver different energy. Likewise, a high equilibrium voltage at one composition does not establish high energy if only a tiny quantity of charge can be reversibly transferred there. Material design seeks favorable potential, accessible site count, stable redox and practical transport together.

Step-by-step reasoning

Write the discharge reaction with a clear electron count n. Determine ΔG or the relevant electrode chemical-potential difference at the specified state. Apply Erev = −ΔG/(nF) for reversible voltage and use Nernst activities if Q changes. If comparing measured voltage, ask whether the cell is open-circuit and relaxed or carrying current. Integrate a voltage curve over transferred charge to estimate energy; do not multiply maximum voltage by total capacity unless the voltage is actually near that value throughout.

Visual explanation

Draw two electrode free-energy curves against lithium composition. A lithium-transfer arrow from negative host to positive host points downhill in total G during discharge. The vertical free-energy decrease per transferred ion corresponds to electrical work per charge. Beneath, draw a voltage–state-of-charge curve with a sloping region and a flat two-phase plateau. On the same plot place a lower discharge curve and a higher charge curve to represent polarization around the equilibrium curve.

Real-world analogy

Water flows between reservoirs at different heights, and the height difference sets the ideal work available per unit water moved. As water levels change, the difference can change. Electrode chemical potentials play a similar role in setting work per unit charge. The analogy does not capture redox chemistry, nonideal activities or ion–electron separation, so it explains only the energy-gradient idea.

Real-world example

A new cathode material shows a flat voltage region during slow discharge but a lower, sloped terminal voltage at high rate. The flat slow region may reflect coexistence of two insertion phases. Under high current, diffusion and charge-transfer losses cause extra polarization, so the observed voltage drops before all theoretical sites are used. A test that reports only the high-rate terminal voltage cannot directly reveal the equilibrium free-energy curve.

Why?

Why can a cell's voltage recover upward after a discharge pulse ends? During the pulse, ion concentration gradients, ohmic losses and interfacial overpotentials lower terminal voltage. When current stops, those nonequilibrium gradients relax and the voltage moves toward the open-circuit value at the new state of charge. The recovered voltage is not newly created stored energy; it reflects removal of rate-dependent losses.

Common misconception

“A 4 V battery delivers exactly 4 V throughout every discharge.” Composition changes equilibrium voltage, and current-dependent polarization changes terminal voltage. Another mistake says a positive voltage means the chemistry is stable forever. It indicates a thermodynamic driving force for the specified cell reaction, not the absence of side reactions or degradation. A third confuses ΔG per mole of cell reaction with energy per mole of electrons; the factor n must be included.

Worked example

A cell reaction as written transfers two moles of electrons and has ΔG = −579 kJ mol⁻¹ of reaction at the stated composition. Using F = 96,485 C mol⁻¹, Erev = −ΔG/(nF) = 579,000 J mol⁻¹/(2 × 96,485 C mol⁻¹) ≈ 3.00 V . The units work because 1 J C⁻¹ = 1 V. At appreciable current the measured discharge voltage may be below 3.00 V, depending on resistance and reaction/transport kinetics.

Quick check

1. Why can two electrode compositions of the same battery have different equilibrium voltages? Answer: Insertion-site occupancy and interactions change the chemical potentials of stored ions and electrons, changing reversible free energy per transferred charge.

Exam focus

Use ΔG = −nFE with the reaction as written and identify n. For nonstandard conditions, express Q in activities and apply the Nernst equation. Explain a plateau by phase coexistence and a slope by composition-dependent chemical potential, with appropriate qualifications. Distinguish equilibrium/open-circuit voltage from operating terminal voltage and compute energy from the voltage profile rather than from one arbitrary point.

Advanced insight

In a solid insertion host, voltage is related to the derivative of free energy with respect to ion content rather than merely to total binding energy at one composition. Strong interactions can make the free-energy curve nonconvex, leading to phase separation and a common-tangent construction. Surface energy and strain in small particles can shift the apparent plateau and create hysteresis between charge and discharge. These effects show why microscopic site energies, mesoscopic phase behavior and macroscopic voltage are connected but not identical.

Summary

Reversible battery voltage is free-energy change per unit charge, E = −ΔG/(nF). Electrode chemical potentials and activities vary with composition, creating sloping voltage profiles or two-phase plateaus. Under current, kinetic and transport losses move terminal voltage away from equilibrium. Useful energy is the integral of voltage over delivered charge, not one voltage number alone.

Practice questions

1. A one-electron reaction has ΔG = −96.5 kJ mol⁻¹. What is its approximate reversible voltage? Answer: E ≈ 96,500/96,485 ≈ 1.00 V.

2. Why is a two-phase plateau approximately flat under equilibrium conditions? Answer: The phase compositions and transferred-species chemical potential remain approximately fixed while the relative phase amounts change.

3. Why is discharge terminal voltage usually below open-circuit voltage at the same composition? Answer: Ohmic resistance, interfacial kinetics and mass-transport gradients require additional driving force under load.

4. Two cells each deliver 2 Ah, one averaging 3 V and the other 4 V. Which delivers more energy under these approximations? Answer: The 4 V cell delivers about 8 Wh versus 6 Wh for the 3 V cell, assuming those are their actual average discharge voltages.