---
title: "Fuel cell voltage, efficiency and hydrogen use"
subject: "Energy Resources"
chapter: 6
key: "semester_6/energy_resources:fuel-cell-stack-numerical"
source: https://www.bctengineer.com/q/semester_6/energy_resources/fuel-cell-stack-numerical
site: BCT Engineering
---

# Fuel cell voltage, efficiency and hydrogen use

Energy Resources · Chapter 6: Non conventional forms of energy and batteries

- Asked in: year not given, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/energy_resources/fuel-cell-stack-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_6/energy_resources/chapter-6

## Question

A hydrogen-oxygen fuel cell stack has 50 cells in series. Each cell operates at 0.65 V and the stack current is 40 A. Take $\Delta G = -237.2$ kJ/mol and $\Delta H = -285.8$ kJ/mol (higher heating value) for the formation of liquid water, $F = 96\,485$ C/mol. Calculate (a) the reversible cell voltage and the maximum thermodynamic efficiency, (b) the stack power and voltage efficiency, (c) the hydrogen consumption in kg/h, and (d) the heat generated in the stack.

## Short answer

Reversible voltage Erev = –ΔG/(nF) = 1.229 V; max thermodynamic efficiency = ΔG/ΔH = 83%. Stack power = 50×0.65×40 = 1300 W; voltage efficiency = 0.65/1.229 = 52.9%. H2 consumtion = (50×40)/(2×96485) × 2.016 = 0.0752 kg/h. Heat = 50×40×(1.481-0.65) = 1662 W.

## Answer

#### (a) Reversible voltage and efficiency
Two electrons are transferred per H₂ molecule ($n = 2$).

$$
E_{rev} = \frac{-\Delta G}{nF} = \frac{237\,200}{2(96\,485)} = 1.229\ \text{V}
$$

$$
\eta_{th,max} = \frac{\Delta G}{\Delta H} = \frac{237.2}{285.8} = 0.830 \;(83\ \%)
$$

#### (b) Stack power and voltage efficiency
$$
V_{stack} = 50 \times 0.65 = 32.5\ \text{V}, \qquad P = V I = 32.5 \times 40 = 1300\ \text{W}
$$

$$
\eta_V = \frac{0.65}{1.229} = 0.529 \;(52.9\ \%)
$$

Efficiency on the basis of the HHV: $\eta = \dfrac{0.65}{285\,800/(2 \times 96\,485)} = \dfrac{0.65}{1.481} = 43.9\ \%$.

#### (c) Hydrogen consumption (Faraday's law)
$$
\dot n_{H_2} = \frac{N I}{nF} = \frac{50 \times 40}{2 \times 96\,485} = 0.01036\ \text{mol/s}
$$

$$
\dot m = 0.01036 \times 2.016 = 0.02090\ \text{g/s} = 75.2\ \text{g/h} = 0.0752\ \text{kg/h}
$$

#### (d) Heat generated
The heat generated is the difference between the energy released at the thermoneutral (HHV) voltage, 1.481 V per cell, and the electrical output at 0.65 V:

$$
Q = N I (1.481 - 0.65) = 50 \times 40 \times 0.831 = 1662\ \text{W}
$$

**Answer:** $E_{rev} = 1.229$ V, $\eta_{max} = 83\ \%$; $P = 1.30$ kW, $\eta_V = 52.9\ \%$ (43.9 % on HHV); H₂ use $= 0.0752$ kg/h; heat $\approx 1.66$ kW.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is the reversible cell voltage obtained and what is it?
   A. -deltaG/(nF) with n = 2, giving 1.229 V
   B. deltaH/(nF) with n = 1, giving 2.96 V
   C. deltaG/deltaH, giving 0.83 V
   D. -deltaG/F with n = 1, giving 2.46 V
   Answer: A. Two electrons are transferred per H2, so E = 237200/(2 x 96485) = 1.229 V. (Key point: Reversible voltage)

2. What is the maximum thermodynamic efficiency of the cell?
   A. deltaH/deltaG = 120 %
   B. 0.65/1.481 = 43.9 %
   C. deltaG/deltaH = 83 %
   D. 0.65/1.229 = 52.9 %
   Answer: C. Maximum efficiency is deltaG/deltaH = 237.2/285.8 = 0.83. (Key point: Maximum thermodynamic efficiency)

3. What are the stack power and voltage efficiency?
   A. P = 0.65 x 40 = 26 W; 83 %
   B. P = 32.5 x 40 = 1300 W; 52.9 % (0.65/1.229)
   C. P = 50 x 40 = 2000 W; 100 %
   D. P = 32.5 x 1.229 = 40 W; 43.9 %
   Answer: B. Stack voltage is 50 x 0.65 = 32.5 V, power is VI, and voltage efficiency is the cell voltage over reversible voltage. (Key point: Stack power and voltage efficiency)

4. Which expression gives the hydrogen consumption rate?
   A. I/(F) per cell only
   B. N I/(n F) = 50 x 40/(2 x 96485) mol/s, about 75 g/h
   C. N I x 2/F
   D. V I / deltaH
   Answer: B. All 50 cells in series carry the same current; each H2 gives 2 electrons, so 0.01036 mol/s, about 0.0752 kg/h. (Key point: Faraday's law hydrogen use)

## Related in Energy Resources

- [Working principle and types of fuel cells](https://www.bctengineer.com/question.md?key=semester_6%2Fenergy_resources%3Apem-fuel-cell-working-types) — Energy Resources, chapter 6
- [Calorific value and air from ultimate analysis](https://www.bctengineer.com/question.md?key=semester_6%2Fenergy_resources%3Adulong-calorific-value-fuel-oil) — Energy Resources, chapter 1
- [Hydrogen production processes](https://www.bctengineer.com/question.md?key=semester_6%2Fenergy_resources%3Ahydrogen-production-methods) — Energy Resources, chapter 6

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