Battery Voltage and Capacity

Introduction

A phone battery label might read "3.85 V, 4500 mA h, 17.3 W h". Each of those three numbers answers a different question. The voltage tells you how much energy each unit of charge carries; the capacity tells you how much charge is stored; and the energy is, roughly, their product. Earlier you linked cell potential to Gibbs energy and met overpotential. This page brings those ideas together to explain what a battery label really means and why a battery never delivers quite what the label promises.

Core explanation

Open-circuit voltage. When no current flows, the voltage across a cell approaches the equilibrium cell potential E of its overall reaction. This value is set by thermodynamics: ΔG = −nFE. Two electrode reactions with a large difference in electrode potential give a high voltage. The Nernst equation shows that the open-circuit voltage also depends on the activities of the reacting species, so it drifts as the cell discharges and the composition changes. For many chemistries the open-circuit voltage therefore acts as a rough "fuel gauge", although some cells, such as lithium iron phosphate, have a very flat voltage curve because the electrode reactions involve two coexisting solid phases whose activities do not change.

Capacity. Capacity is total deliverable charge. It is fixed by the amount of active material that can react, using Faraday's constant F = 96 485 C/mol. Because 1 A h = 3600 C, one mole of electrons corresponds to 96 485 ÷ 3600 ≈ 26.8 A h. The theoretical specific capacity of an electrode material is therefore

q = nF ÷ (3.6 M) mA h/g, where M is the molar mass in g/mol and n is the number of electrons per formula unit.

A cell's capacity is limited by whichever electrode runs out first; manufacturers deliberately balance the electrodes so that the more dangerous failure (for example, lithium metal plating) is avoided.

Energy. Energy is charge multiplied by voltage. Because voltage changes during discharge, the stored energy is the area under the voltage–charge curve: E stored = ∫V dq. A convenient estimate multiplies the capacity by the average (nominal) voltage: 4.5 A h × 3.85 V ≈ 17.3 W h. One watt-hour is 3600 J.

Voltage under load. When current flows, the terminal voltage V falls below the open-circuit value:

V = E oc − I R int − η act − η conc

The ohmic term I R int comes from electrolyte and contact resistance; the activation and concentration overpotentials come from slow electron transfer and from depletion of reactants near the electrode surfaces. During charging the signs reverse and the applied voltage must exceed E oc. The gap between charging and discharging voltages is lost as heat, which is why energy efficiency is lower than charge (coulombic) efficiency.

Power. Power is P = VI. High currents deliver more power but increase losses, so there is a trade-off between power and usable energy. A cell discharged at a high C-rate reaches its cut-off voltage sooner and appears to have a smaller capacity.

Formulae

ΔG = −nFE; capacity Q = nFN (N moles of formula units reacting); 1 A h = 3600 C; energy ≈ Q × V average; V = E oc − IR int − overpotentials; P = VI.

Step-by-step reasoning

To estimate the energy of a cell from its chemistry:

1. Write the overall cell reaction and find E from standard electrode potentials, adjusting with the Nernst equation if needed. 2. Identify the limiting electrode and calculate its moles of reactable material. 3. Multiply by n and F to obtain charge, then convert coulombs to ampere-hours. 4. Multiply by the average voltage to estimate energy in watt-hours. 5. Reduce the estimate for losses at the intended current.

Visual explanation

Imagine a discharge curve with voltage on the vertical axis and charge delivered on the horizontal axis. At low current the curve sits high and runs far to the right. At high current the whole curve shifts downwards by IR and ends earlier at the cut-off line. The shaded area beneath each curve is the energy delivered.

Real-world analogy

A battery is like a water tower. The height of the tower is the voltage, the volume of water is the capacity and the energy is height multiplied by volume. Pumping water through a narrow pipe quickly wastes pressure in friction, just as high current wastes voltage in internal resistance.

Real-world example

Electric vehicle packs connect many cells in series to raise the voltage (often around 400 V or 800 V) and in parallel to raise the capacity. Higher pack voltage lets the same power flow at lower current, which cuts I²R heating in cables and cells and allows faster charging.

Why?

Why does a battery seem to "recover" after a rest? During heavy use, concentration gradients build up in the electrolyte and inside the electrode particles. When the current stops, diffusion evens out these gradients, the concentration overpotential disappears and the measured voltage rises again towards its open-circuit value.

Common misconception

"A higher voltage battery stores more charge." Voltage and capacity are independent: voltage depends on which reactions occur, capacity on how much active material is present. A 1.5 V alkaline D cell stores far more charge than a 3.7 V lithium coin cell.

Worked example

Question: A zinc anode contains 6.54 g of zinc (M = 65.4 g/mol), which is oxidised to Zn²⁺. If the cell's average voltage is 1.3 V, estimate the capacity and energy.

Reasoning: Moles of Zn = 6.54 ÷ 65.4 = 0.100 mol. Each Zn gives 2 electrons, so moles of electrons = 0.200 mol. Charge = 0.200 × 96 485 ≈ 19 300 C = 19 300 ÷ 3600 ≈ 5.36 A h. Energy ≈ 5.36 × 1.3 ≈ 7.0 W h.

Answer: About 5.4 A h and 7.0 W h, if zinc is the limiting electrode.

Quick check

1. Why does the terminal voltage of a battery fall when a large current is drawn from it? Answer: Because of the IR drop across the internal resistance and the activation and concentration overpotentials at the electrodes.

Exam focus

Keep capacity (A h), energy (W h) and power (W) clearly separate, and always state which quantity you are calculating. Convert with 1 A h = 3600 C and 1 W h = 3600 J. Link open-circuit voltage to ΔG = −nFE, and explain lower voltage under load in terms of IR drop and overpotential.

Advanced insight

The entropy change of the cell reaction, ΔS = nF(dE/dT), means that even a perfectly reversible cell exchanges heat TΔS with its surroundings. In some lithium-ion chemistries this reversible heat is comparable to the irreversible I²R heating at moderate rates, so thermal management of large packs must account for both the thermodynamic and the kinetic heat terms.

Summary

Open-circuit voltage reflects the Gibbs energy of the cell reaction and changes with composition through the Nernst equation. Capacity is the deliverable charge, fixed by the limiting amount of active material and Faraday's constant. Energy is the area under the voltage–charge curve, roughly capacity multiplied by average voltage. Under load, IR drop and overpotentials lower the voltage, reducing usable energy and efficiency.

Practice questions

1. Convert a capacity of 2.0 A h into coulombs. Answer: 2.0 × 3600 = 7200 C. 2. A cell has an open-circuit voltage of 4.0 V and an internal resistance of 0.050 Ω. Ignoring overpotentials, what is its terminal voltage when delivering 10 A? Answer: V = 4.0 − 10 × 0.050 = 3.5 V. 3. Calculate the theoretical specific capacity of lithium metal (M = 6.94 g/mol, n = 1) in mA h/g. Answer: q = 1 × 96 485 ÷ (3.6 × 6.94) ≈ 3860 mA h/g. 4. Explain why a cell appears to have a lower capacity when discharged quickly. Answer: Larger losses lower the voltage so the cut-off voltage is reached before all the active material has reacted, and concentration gradients leave some material unused.