Fuel-Cell Electrochemistry

Introduction

A battery stores its reactants inside its case; when they are used up, it must be recharged or replaced. A fuel cell is different: fuel and oxidant flow in continuously and products flow out, so it produces electricity for as long as it is supplied. Hydrogen fuel cells power some buses, lorries and backup generators, with water as the only chemical product at the point of use. This page examines the reactions, the theoretical voltage and the losses that keep real cells below it.

Core explanation

Electrode reactions. In a proton-exchange membrane (PEM) cell, the acidic electrolyte is a thin polymer membrane. Hydrogen is oxidised at the anode:

H₂ → 2H⁺ + 2e⁻

Protons cross the membrane, and electrons flow through the external circuit to the cathode, where oxygen is reduced:

O₂ + 4H⁺ + 4e⁻ → 2H₂O

In an alkaline fuel cell, hydroxide ions carry the current instead: H₂ + 2OH⁻ → 2H₂O + 2e⁻ at the anode and O₂ + 2H₂O + 4e⁻ → 4OH⁻ at the cathode. Either way, the overall reaction is 2H₂ + O₂ → 2H₂O.

Reversible voltage. For liquid water at 298 K, ΔG° = −237 kJ/mol H₂, so E° = −ΔG° ÷ nF = 237 000 ÷ (2 × 96 485) ≈ 1.23 V. Because ΔS is negative for this reaction, E falls slightly as temperature rises, and higher gas pressures raise it through the Nernst equation.

Thermodynamic efficiency. A heat engine is limited by the Carnot factor. A fuel cell is limited instead by ΔG ÷ ΔH. With ΔH° = −286 kJ/mol, the maximum efficiency is 237 ÷ 286 ≈ 0.83. The rest, TΔS, is unavoidably released as heat.

Real losses. Practical cells operate at about 0.6–0.8 V. The polarisation curve shows three regions:

- Activation losses dominate at low current. Oxygen reduction is a slow four-electron reaction with a small exchange current density, even on platinum, so a significant overpotential (often around 0.3 V) is needed just to start current flowing. Tafel behaviour describes this region. - Ohmic losses dominate at moderate current: a linear fall caused by membrane and contact resistance. Membranes conduct well only when hydrated, so water management is crucial. - Mass-transport losses appear at high current, when gas cannot diffuse to the catalyst fast enough and liquid water floods the pores. The voltage then drops sharply towards a limiting current.

Electrode design. Reaction occurs only where gas, electronically conducting catalyst and ion-conducting electrolyte meet, the three-phase boundary. Electrodes are therefore porous layers of carbon-supported platinum nanoparticles mixed with ionomer, maximising this contact area while using as little precious metal as possible.

Fuel purity. Carbon monoxide adsorbs strongly on platinum and blocks active sites, so hydrogen for low-temperature cells must be very pure. High-temperature cells, such as solid-oxide fuel cells at several hundred degrees Celsius, are more tolerant and can use other fuels.

Formulae

E° = −ΔG° ÷ nF; ε max = ΔG ÷ ΔH; voltage efficiency = V cell ÷ E; V cell = E − η act − IR − η mt.

Step-by-step reasoning

To read a polarisation curve:

1. Mark the reversible voltage, about 1.23 V at standard conditions. 2. Note the steep initial drop at low current: activation overpotential. 3. Find the straight middle section: its slope gives the area-specific resistance. 4. Identify the sharp fall at high current: mass-transport limitation. 5. Multiply V by current density to find the power density and its maximum.

Visual explanation

Imagine a graph of voltage against current density. A dashed line at 1.23 V shows the ideal. The real curve starts near 1.0 V, dips steeply, slopes gently downwards for most of its length, then plunges near the limiting current. A second curve for power rises, peaks and falls.

Real-world analogy

A fuel cell is like a watermill on a river: as long as water (fuel) keeps flowing, the wheel turns. A battery is more like a reservoir that eventually empties and must be refilled by pumping.

Real-world example

Fuel-cell electric vehicles store hydrogen as a compressed gas in strong composite tanks and can refuel in a few minutes. Their overall efficiency depends heavily on how the hydrogen was produced: hydrogen made by electrolysis using renewable electricity gives low emissions, while hydrogen made from natural gas without carbon capture does not.

Why?

Why is the cathode, not the anode, the main source of loss? Hydrogen oxidation on platinum is very fast, with a large exchange current density. Oxygen reduction requires breaking a strong O=O bond and transferring four electrons and four protons, so it is far slower and needs a much larger overpotential.

Common misconception

"A fuel cell burns hydrogen." There is no flame and no combustion. Hydrogen and oxygen never meet directly; they react at separate electrodes, and the energy is released as electrical work rather than primarily as heat.

Worked example

Question: A PEM cell operates at 0.70 V. What is its efficiency relative to the enthalpy of reaction (ΔH° = −286 kJ/mol)?

Reasoning: The voltage corresponding to ΔH is 286 000 ÷ (2 × 96 485) ≈ 1.48 V (the thermoneutral voltage). Efficiency = 0.70 ÷ 1.48 ≈ 0.47.

Answer: About 47%, assuming all the fuel reacts.

Quick check

1. Why does a fuel cell continue to produce electricity long after a battery of similar size would be flat? Answer: Its reactants are continuously supplied from outside and products removed, so the electrodes are not used up.

Exam focus

Write balanced electrode equations for both acidic and alkaline electrolytes, calculate E° from ΔG°, and compare ΔG ÷ ΔH with the Carnot limit. Label the three regions of a polarisation curve and link each to its physical cause.

Advanced insight

Platinum's cost drives research into catalysts with less or no precious metal, such as platinum alloys with a strained surface and iron–nitrogen–carbon materials. The best oxygen-reduction catalysts bind oxygen intermediates neither too strongly nor too weakly, an example of the Sabatier principle, which produces "volcano plots" of activity against binding energy.

Summary

Fuel cells convert fuel directly into electricity while reactants are supplied continuously. The hydrogen–oxygen cell has a reversible voltage of 1.23 V and a maximum efficiency of about 83% set by ΔG ÷ ΔH. Real cells run at 0.6–0.8 V because of activation losses (mainly slow oxygen reduction), ohmic losses in the membrane and mass-transport losses at high current.

Practice questions

1. Write the cathode reaction in an alkaline fuel cell. Answer: O₂ + 2H₂O + 4e⁻ → 4OH⁻. 2. Calculate the maximum efficiency of a hydrogen fuel cell given ΔG° = −237 kJ/mol and ΔH° = −286 kJ/mol. Answer: 237 ÷ 286 ≈ 0.83, or 83%. 3. Explain why carbon monoxide in the hydrogen supply reduces cell performance. Answer: CO adsorbs strongly on the platinum catalyst, blocking the sites where hydrogen would be oxidised, which raises the overpotential. 4. A cell has a voltage of 0.65 V at 1.2 A/cm². What is its power density? Answer: 0.65 × 1.2 = 0.78 W/cm².