What sets fuel cell voltage?

The theoretical voltage (about 1.19 V at 80 °C) minus three losses: activation loss to drive electrode reactions, ohmic loss from membrane and contact resistance, and concentration loss as reactants run short. Plotting this against current density gives the polarization curve, the most basic description of fuel cell performance.

Which loss is largest?

In PEM cells the oxygen reaction at the cathode is slow, so activation loss is largest. At 1.0 A/cm²: activation 0.42 V, ohmic 0.10 V, concentration 0.035 V, cell voltage 0.635 V. At high current ohmic and concentration losses grow; at 1.8 A/cm² concentration loss reaches 0.115 V.

Why does heat management matter?

All voltage loss becomes heat. Hydrogen's lower heating value corresponds to 1.25 V, so at 0.635 V about half becomes heat; an 80 kW fuel cell releases about 78 kW. Without cooling, temperature rises, the membrane dries, resistance grows and performance falls — why electrochemistry, heat and water transport are solved together.

Fuel Cell Modeling

Category: Coupled Analysis – Electrochemistry | Revised 2026-10-01

Theory: theoretical voltage minus three losses

Overview

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Why does a fuel cell deliver only about 0.6 V rather than the theoretical 1.2 V?

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Because current flow incurs several losses. First, part of the voltage drives electrode reactions (activation loss); in PEM cells the cathode oxygen reaction is slow, costing about 0.4 V. Next, resistance in the membrane and between parts drops voltage (ohmic loss). And at high current oxygen and hydrogen can't reach reaction sites fast enough, so voltage falls sharply (concentration loss). Summing these against current density gives the polarization curve, the basis of cell design and operating choices. All lost voltage becomes heat, which also follows from this curve.

Polarization-curve model

$$ V = E_0 - b\ln\frac{i}{i_0} - i\,R_{ASR} + c\,\ln\!\left(1 - \frac{i}{i_L}\right),\qquad b = \frac{RT}{\alpha F} $$

E₀ is theoretical voltage (1.19 V at 80 °C), b the Tafel slope (0.0304 V with α = 1, 70 mV per decade), i₀ exchange current density (10⁻⁶ A/cm²), R_ASR area-specific resistance (0.10 Ω·cm²), i_L limiting current density (2.0 A/cm²) and c the concentration coefficient (0.05 V).

Power, efficiency and heat

$$ P = V i,\qquad \eta_{LHV} = \frac{V}{1.25\ \mathrm{V}},\qquad Q_{heat} = (1.25 - V)\,i\ \ (\text{per area, water as vapour}) $$

Coffee Break Trivia

From spacecraft to cars

Grove demonstrated the fuel-cell principle in 1839, but practical use came with 1960s space programmes: Gemini carried PEM fuel cells and Apollo alkaline ones, supplying electricity and drinking water. Membrane and catalyst advances later raised power density, and fuel-cell cars went on sale in 2014. Cutting platinum catalyst while limiting cathode activation loss remains a major research challenge.

Worked examples

Example 1: voltage and losses by current density (80 °C)

Current density	Cell voltage	Activation	Ohmic	Concentration	Power	Efficiency (LHV)
0.1 A/cm²	0.827 V	0.350 V	0.010 V	0.003 V	0.083 W/cm²	66%
0.5 A/cm²	0.726 V	0.399 V	0.050 V	0.014 V	0.363 W/cm²	58%
1.0 A/cm²	0.635 V	0.420 V	0.100 V	0.035 V	0.635 W/cm²	51%
1.5 A/cm²	0.538 V	0.433 V	0.150 V	0.069 V	0.807 W/cm²	43%
1.8 A/cm²	0.457 V	0.438 V	0.180 V	0.115 V	0.822 W/cm²	37%

Peak power is 0.831 W/cm² at 1.70 A/cm² (0.488 V).

Example 2: a drying membrane raises resistance

Current density	ASR 0.10 Ω·cm²	ASR 0.15 Ω·cm²
1.0 A/cm²	0.635 V	0.585 V
1.5 A/cm²	0.538 V	0.463 V

Delivering 80 kW at 1.0 A/cm² (0.635 V) needs about 12.6 m² of total cell area and releases about 78 kW of heat in operation (water leaving as vapour).

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If it releases about as much heat as power, cooling must be as hard as for an engine car.

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Less heat than an engine, but in some ways harder to handle. Fuel cells run near 80 °C, so the temperature difference to ambient is small and rejecting the same heat needs a large radiator. Engines dump much heat in exhaust, but fuel-cell exhaust is cool, so most heat must go through coolant. Temperature is also tightly tied to membrane water: too hot and the membrane dries, resistance rises and, as in Example 2, voltage drops and heating grows; too cold and product water condenses, blocking channels and raising concentration loss. Fuel cell models therefore couple electrochemical reactions, heat transfer and water transport (vapour and liquid) to resolve distributions within the cell.

Modelling workflow

Fit exchange current density, resistance and limiting current to measured polarization curves.
Include channels, diffusion layers, catalyst layers and membrane to solve gas and water flow.
Derive heat generation from current distribution and solve temperature including cooling.
Update membrane resistance with temperature and water content, coupling electrochemistry with heat and water.
Compare predicted current, temperature and water distributions with measurements.

Coffee Break Trivia

“A cell whose outlet side underperformed”

Current-distribution measurements on one fuel cell showed current falling sharply only near the air outlet. Coupled analysis revealed humidity rising along the channel from product water until it condensed near the outlet, flooding the diffusion layer and starving oxygen, while the inlet ran dry with high membrane resistance. Revising air flow, humidification and coolant direction to even out temperature and humidity made current uniform and raised cell voltage.

Common mistakes

Mistakes and fixes

Mistake	Effect	Fix
Constant temperature	Drying and flooding missed	Couple heat and water
Constant membrane resistance	Wrong voltage at high current	Function of water content
Average current only	Local degradation missed	Resolve distributions
Not deriving heat from losses	Insufficient cooling	Heat from voltage loss
Fitting a single point	Wrong at other currents	Fit the whole curve

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