Worked examples — Fuel cells — H₂ - O₂ fuel cell (spacecraft relevance)

This page is a drill. The parent note told you what a fuel cell is. Here we grind through every kind of number problem it can throw at you — from "just add two voltages" to "the tank is empty, what happens?".

Before we start, three tiny reminders so no symbol appears unexplained:

One more tool we will use repeatedly:

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The scenario matrix

Every fuel-cell numbers question is one of these cells. Each worked example below is labelled with the cell it hits.

#	Case class	What makes it tricky	Example
A	Add two half-cell potentials	Sign bookkeeping (cathode − anode)	Ex 1
B	Voltage → energy per mole	Get $n$ right, keep units	Ex 2
C	Mass in → water/energy out	Stoichiometry + molar mass	Ex 3
D	Thermodynamic efficiency ceiling	$\Delta G$ vs $\Delta H$, the entropy tax	Ex 4
E	Real efficiency vs Carnot rival	Compare two machines fairly	Ex 5
F	Real-world mission sizing	Multi-step word problem (Apollo)	Ex 6
G	Degenerate / zero input	Fuel runs out → voltage collapses	Ex 7
H	Limiting / sign-flip twist	Run it backwards = electrolysis	Ex 8

Cases A–F are the "everyday" cells. G and H are the edge cases most students never see — the empty tank and the reversed cell.

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Case A — Adding two half-cells

Forecast: guess before reading — will the answer be near $0.4$, near $1.2$, or near $0.0$ V?

1. Identify who is cathode, who is anode. The oxygen electrode has the higher (more positive) potential $+0.40$ V, so it wants electrons → it is the cathode (reduction). Hydrogen is the anode (oxidation). Why this step? By definition the electrode with higher reduction potential pulls electrons and becomes the cathode. This is the core rule of Standard electrode potentials.

2. Apply the master formula. $E_{\text{cell}}^{\circ }=E_{\text{cathode}}^{\circ }-E_{\text{anode}}^{\circ }$ Why this step? Both numbers are quoted as reduction potentials. Subtracting the anode value automatically "flips" it into an oxidation, so we never rewrite signs by hand.

3. Plug in. $E_{\text{cell}}^{\circ }=0.40-(-0.83)=0.40+0.83=1.23\text{V}$

Verify: the result is positive → spontaneous → a real battery would light up. ✓. Notice subtracting a negative added the magnitudes — a classic sign trap that lands you near $1.2$ V, not $0.4$ V.

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Case B — Voltage into energy per mole

Forecast: will $\Delta G$ come out positive or negative? (Spontaneous cell ⇒ ?)

1. Count the electrons $n$. For this "per one water" reaction, H₂ gives up $2$ electrons. So $n=2$. Why this step? $n$ is the number of electron-moles for the equation as written. Write $2{\text{H}}_2\text{O}$ instead and $n=4$; you must match $n$ to the equation.

2. Apply the bridge equation. $\Delta G^{\circ }=-nF{E}^{\circ }=-(2)(96,485)(1.23)$ Why this step? This is the only route from volts to joules per mole (see the bridge formula box).

3. Compute. $\Delta G^{\circ }=-237,353\text{J/mol}\approx -237\text{kJ/mol}.$ Why this step? Divide by $1000$ to go from J to kJ.

Verify: negative ✓ (spontaneous), and it matches the parent note's $\Delta G=-237$ kJ/mol exactly. Units: $[{\text{mol}}^{-1}]\cdot [\text{C/mol}]\cdot [\text{V}]=\text{C}\cdot \text{V}=\text{J}$ per mole. ✓ ($1\text{C⋅V}=1\text{J}$.)

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Case C — Mass of fuel → water and charge

Forecast: guess the water mass — is it around 3 kg, 9 kg, or 18 kg?

1. Moles of H₂. $\frac{1000\text{g}}{2\text{g/mol}}=500$ mol H₂. Why this step? Everything downstream is counted in moles, so convert mass first.

2. Moles of water from $2{\text{H}}_2+{\text{O}}_2\to 2{\text{H}}_2\text{O}$ (ratio $1:1$ for H₂→H₂O). So $500$ mol H₂ → $500$ mol H₂O. Why this step? Stoichiometry: one H₂ makes one H₂O.

3. Mass of water. $500\times 18=9000$ g $=9$ kg. Why this step? Convert moles back to mass with molar mass.

4. Oxygen needed. $500$ mol H₂ needs $250$ mol O₂ $=250\times 32=8000$ g $=8$ kg.

5. Total charge delivered. Each H₂ releases $2$ electrons, so $n=500\times 2=1000$ mol e⁻. $Q=nF=1000\times 96,485=9.65\times 10^7\text{C}.$ Why this step? Charge $=$ (moles of electrons) $\times$ (charge per mole) — this is how much electricity 1 kg of H₂ can push.

Verify: mass balance — in: $1{\text{kg H}}_2+8{\text{kg O}}_2=9$ kg; out: $9$ kg water. Nothing lost ✓ (matches the parent's "1 kg H₂ + 8 kg O₂ → 9 kg H₂O"). Water lands at $9$ kg, not $18$ — the forecast trap.

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Case D — The efficiency ceiling

Forecast: more or less than 100%? Why can't it be 100%?

1. Write the definition. ${\eta }_{\text{max}}=\frac{\Delta G}{\Delta H}$ Why this step? Only $\Delta G$ can become electrical work; $\Delta H$ is the total energy the fuel holds. The ratio is "useful / total". See Gibs free energy and spontaneity.

2. Use magnitudes (both are negative; the ratio is positive). ${\eta }_{\text{max}}=\frac{237}{286}=\mathrm{0.829.}$

3. As a percent: $0.829\times 100=82.9\%\approx 83\%$.

Verify: below 100% ✓. The missing $17\%$ is the entropy tax $T\Delta S=\Delta H-\Delta G=-286-(-237)=-49$ kJ/mol, which must leave as heat and can never be electricity. That is why the ceiling exists.

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Case E — Fuel cell vs Carnot rival

Forecast: which machine has the higher theoretical ceiling?

1. Carnot efficiency formula. ${\eta }_{\text{Carnot}}=1-\frac{T_C}{T_H}$ Why this step? Any heat engine is capped by Carnot; a fuel cell is not a heat engine, so it dodges this cap (Thermodynamics vs. kinetics).

2. Plug in temperatures in kelvin. ${\eta }_{\text{Carnot}}=1-\frac{400}{1500}=1-0.2667=\mathrm{0.7333.}$

3. As a percent: $73.3\%$.

Verify: fuel-cell ceiling $83\%>$ turbine ceiling $73\%$ ✓ — and this Carnot number only appears at a fierce $1500$ K; real turbines run cooler and fall to $\sim 25\%$. The fuel cell wins its ceiling even against an idealized rival.

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Case F — Mission sizing (real-world word problem)

Forecast: roughly how many kg of hydrogen — closer to 5, 25, or 100 kg?

1. Energy in kWh. $1.5\text{kW}\times 24\text{h}\times 8\text{days}=288$ kWh. Why this step? Energy = power × time; days×24 gives hours.

2. Convert to joules. $1\text{kWh}=3.6\times 10^6$ J, so $288\times 3.6\times 10^6=1.037\times 10^9\text{J}=1.037\times 10^6\text{kJ}.$

3. Account for efficiency. Only $60\%$ becomes electricity, so chemical energy needed: $E_{\text{chem}}=\frac{1.037\times 10^6}{0.60}=1.728\times 10^6\text{kJ}.$ Why this step? We must burn more fuel than the electrical output because $40\%$ is lost as heat/resistance.

4. Moles of H₂. Each mole releases $286$ kJ: $\frac{1.728\times 10^6}{286}=6042\text{mol}.$

5. Mass of H₂. $6042\times 2=12,085$ g $\approx 12.1$ kg.

Verify: order of magnitude sits between the parent's "8-day, 1.4 kW" figures — sensible for a real capsule ✓. Water produced would be $6042\times 18\approx 109$ kg, comfortably above the crew's drinking need — matching the note's "far more than crew needs" ✓.

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Case G — Degenerate input (the empty tank)

Forecast: does the voltage stay at $1.23$ V, sag a bit, or crash to zero?

1. Recall the Nernst correction (voltage under non-standard conditions): $E_{\text{cell}}=E_{\text{cell}}^{\circ }-\frac{RT}{nF}\ln Q,Q\propto \frac{1}{P_{{\text{H}}_2}\sqrt{P_{{\text{O}}_2}}}.$ Why this step? Standard voltage assumes plenty of reactant; when a reactant vanishes we need the concentration-aware form. $Q$ is the reaction quotient; low reactant pressure makes $Q$ large.

2. Send $P_{{\text{H}}_2}\to 0$. Then $Q\to \infty$, so $\ln Q\to +\infty$, and the correction term $-\frac{RT}{nF}\ln Q\to -\infty$. Why this step? The math mirrors reality: no fuel, no push.

3. Physical reading. $E_{\text{cell}}$ collapses toward (and effectively to) $0$ V. The cell is dead, not "resting at 1.23 V".

Verify: consistent with everyday experience — an empty battery reads ~0 V under load ✓. This is the exact difference from a normal battery in the parent note: a fuel cell only holds voltage while fuel flows. A sanity spot-check: at standard pressures $Q=1$, $\ln 1=0$, so $E_{\text{cell}}=E^{\circ }=1.23$ V — the formula reduces correctly. ✓

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Case H — The sign-flip / limiting twist (run it backwards)

Forecast: is the applied voltage bigger or smaller than $1.23$ V, and does $\Delta G$ change sign?

1. Reverse the reaction, reverse the signs. Forward (fuel cell) had $\Delta G=-237$ kJ/mol. Running backwards flips the sign: $\Delta G_{\text{rev}}=+237\text{kJ/mol}.$ Why this step? Reversing a reaction negates every thermodynamic quantity. Positive $\Delta G$ ⇒ not spontaneous ⇒ we must supply energy. This is Electrolysis of water.

2. Minimum applied voltage. From $\Delta G=-nFE$, reversing gives the decomposition voltage: $E_{\text{min}}=\frac{\Delta G_{\text{rev}}}{nF}=\frac{237,000}{2\times 96,485}=1.23\text{V}.$ Why this step? At minimum, the applied voltage exactly cancels the cell's own spontaneous voltage — same magnitude, opposite job.

3. Interpretation. You must push at least $1.23$ V (in practice more, to beat overpotential) to force water apart. Below that, nothing splits.

Verify: the fuel cell releases $237$ kJ/mol; electrolysis demands $237$ kJ/mol back. Energy in = energy out at the ideal limit — perfectly symmetric ✓. The number $1.23$ V appears in both directions, which is the beautiful reversibility of this one reaction and links straight to the Hydrogen economy (make H₂ by electrolysis, burn it in a fuel cell).

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Recall Rapid self-test

Cathode-minus-anode of $+0.40$ and $-0.83$ V gives what cell voltage? ::: $1.23$ V (subtracting the negative adds it). For ${\text{H}}_2+\frac{1}{2}{\text{O}}_2\to {\text{H}}_2\text{O}$, how many electrons is $n$? ::: $2$. 1 kg of H₂ makes how much water? ::: $9$ kg. Why is the efficiency ceiling $83\%$ and not $100\%$? ::: the $T\Delta S$ entropy tax ($49$ kJ/mol) must leave as heat. When the H₂ tank empties, the voltage goes to…? ::: ~0 V (Nernst term diverges). Reverse the cell and you get…? ::: electrolysis of water, needing ≥1.23 V.

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See also: Catalysis (why platinum is needed to make these reactions fast at low temperature) and Galvanic cells and cell potential (the general machinery behind every example above).