Fuel cells — H₂ - O₂ fuel cell (spacecraft relevance): Question bank

The whole page rests on one picture worth keeping in your head, drawn below: a fuel cell is an open electrochemical box. Fuel flows in one side, oxidant flows in the other, electrons walk through the outside wire, and water walks out. Nothing is stored — everything passes through. Hold that image and most traps collapse.

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

Definition The four thermodynamic symbols this page uses

Before any trap, pin down the notation so nothing is used unearned:

$Δ H$ — enthalpy change, the total heat energy released when the reaction runs. For $H_{2} + \frac{1}{2} O_{2} \to H_{2} O (l)$ it is $- 286 kJ/mol$ (negative = energy given out).
$T Δ S$ — the entropy term: temperature $T$ times the change in disorder $Δ S$ . Because three gas molecules collapse into one liquid, disorder drops, so this term subtracts energy that is unavailable for orderly work.
$Δ G$ — Gibbs free energy change, the usable part: $Δ G = Δ H - T Δ S$ . It is the maximum energy that can become electrical work; here $- 237 kJ/mol$ . See Gibbs free energy and spontaneity.
$E_{cell}^{\circ}$ — the standard cell potential (volts), the "electrical push" the cell provides when reactants sit at standard conditions.

True or false — justify

Is a fuel cell a type of galvanic cell?

True — like any galvanic cell it turns a spontaneous redox reaction directly into electricity; the twist is that reactants are fed continuously instead of being sealed inside.

A fuel cell eventually "goes flat" like a normal battery?

False — the electrodes are catalysts, not reactants, so nothing inside gets consumed; it stops only when you cut off the H₂ or O₂ supply.

The overall reaction

2 H_{2} + O_{2} \to 2 H_{2} O

is the same as burning hydrogen?

True on paper — same reactants and products, same

Δ H

— but the path differs: electrons flow through a wire instead of the energy dumping out as heat and light.

Because the overall reaction is identical to combustion, the efficiency limit is also the same?

False — combustion is a heat engine capped by the Carnot limit (~25%), while a fuel cell converts chemical energy directly to electrical work, so its ceiling is

Δ G /Δ H \approx 83%

The standard voltage

E_{cell}^{\circ}

stays fixed no matter the gas pressures?

False — the Nernst relation

E = E^{\circ} - \frac{R T}{n F} ln Q

shifts the real voltage with reactant partial pressure, pH, and temperature; only the standard-condition value is fixed.

At the anode, oxidation occurs?

True — H₂ loses electrons (

2 H_{2} + 4 OH^{-} \to 4 H_{2} O + 4 e^{-}

); loss of electrons is the definition of oxidation, and it always happens at the anode.

In the alkaline cell, OH⁻ ions travel from anode to cathode?

False — OH⁻ is consumed at the anode and regenerated at the cathode, so it migrates cathode → anode; the net effect is that OH⁻ is recycled and never depleted.

The platinum coating is used up during operation?

False — Pt is a catalyst: it lowers the activation barrier and is returned unchanged after each cycle, which is why one coating lasts for years.

The standard cell potential of

1.23 V

is what you actually read on the terminals while drawing current?

False — under load the voltage sags to ~0.9 V because of overpotential and ohmic losses;

1.23 V

is the ideal open-circuit value.

Water is a waste product with no use on a spacecraft?

False — the product water is drinkable and doubles as coolant; producing it was one of the three reasons NASA chose this cell (power, water, reliability).

A single H₂/O₂ cell can directly power a spacecraft's multi-kilowatt bus?

False — one cell gives ~0.9 V, so cells are stacked in series (for voltage) and parallel (for current) into a module delivering the needed power.

Spot the error

"At the cathode, O₂ gives away electrons to form OH⁻."

Error — at the cathode O₂ is reduced, so it gains electrons:

O_{2} + 2 H_{2} O + 4 e^{-} \to 4 OH^{-}

. Reduction (gain) always happens at the cathode.

"The anode is the positive terminal because that's where fuel enters."

Error — in a galvanic cell the anode (oxidation, electron source) is the negative terminal; electrons leave it and travel to the positive cathode.

"You get

1.23 V

straight from

Δ H

Error — voltage comes from free energy, not total:

E^{\circ} = - Δ G / n F

, using

Δ G = - 237 kJ/mol

and

n = 2

. Using

Δ H = - 286 kJ/mol

would wrongly give ~1.48 V.

"Efficiency is capped at 83% because 17% of the energy leaks out as electrical noise."

Error — the missing ~17% is the

T Δ S

term, energy tied up as entropy that cannot become free work; see Gibbs free energy and spontaneity. It is a thermodynamic limit, not a leak.

"We add the two half-reactions without balancing electrons — they just combine."

Error — you must scale so electrons cancel: the anode releases 4e⁻ and the cathode consumes exactly 4e⁻, conserving charge before the halves sum to the overall equation.

"You can pour gasoline into an H₂/O₂ fuel cell and it'll run — fuel is fuel."

Error — hydrocarbons foul the Pt catalyst; they must first be reformed into H₂, an extra lossy step. The cell chemistry is specific to hydrogen.

"Alkaline cells were chosen for cars because KOH is cheap."

Error — cars use PEM cells (fast cold start, compact, solid electrolyte); alkaline was chosen for Apollo, a 1960s mission where CO₂ scrubbers were already onboard.

"A fuel cell has no efficiency loss because there are no moving parts."

Error — no mechanical loss, true, but it still loses to activation overpotential, ohmic resistance, and concentration overpotential, dropping ~83% ideal to ~60% real.

Why questions

Why does the relation

E^{\circ} = - Δ G / n F

exist at all?

Because electrical work equals charge moved times voltage;

n F

is the charge per mole of reaction and

Δ G

is the maximum work, so dividing one by the other converts free energy into volts.

Why must the electrodes be porous rather than solid plates?

Because reaction happens only where gas, electrode, and electrolyte all meet; porous carbon maximises this three-phase contact area, giving usable current instead of a trickle.

Why is a platinum catalyst needed at all?

Both half-reactions are kinetically sluggish; Pt lowers activation energy so the reaction is fast enough for practical power — this is a kinetics fix, not a thermodynamics one.

Why does a fuel cell beat a hydrogen-burning turbine on efficiency?

The turbine chain is chemical → heat → mechanical → electrical, each step lossy and Carnot-limited; the fuel cell goes chemical → electrical in one step, sidestepping the heat-engine ceiling.

Why did NASA prefer fuel cells over solar panels for the Moon?

The lunar night lasts ~14 days; batteries big enough to bridge that are too heavy, whereas cryogenic H₂/O₂ tanks store the same energy at far lower mass.

Why is the alkaline cell "sensitive to CO₂"?

CO₂ reacts with the KOH electrolyte to form solid carbonate, clogging the pores and lowering ion conductivity — a problem PEM cells avoid by using a solid proton membrane.

Why does only

Δ G

, not

Δ H

, set the electrical work available?

Δ H

is total energy released; part of it (

T Δ S

) is dictated by entropy and cannot do ordered electrical work, so the extractable maximum is the free energy

Δ G

Why does the electrolyte not get "used up" over time?

Ions consumed at one electrode are regenerated at the other (OH⁻ in alkaline, H⁺ in PEM), so the electrolyte only ferries charge — it is neither reactant nor product overall.

Edge cases

If you stop feeding H₂ but keep O₂ flowing, what happens to the current?

It falls to zero — with no fuel to oxidise at the anode there are no electrons to push, so the "open system" simply idles until fuel returns.

What happens to voltage the instant you draw zero current (open circuit)?

It rises to near the ideal

1.23 V

, because with no current there are no overpotential or ohmic drops to eat into the potential.

Drop the H₂ supply pressure toward near-zero while still drawing current — what does Nernst predict?

The reaction quotient

Q

climbs, so the

- \frac{R T}{n F} ln Q

term grows and

E_{cell}

falls; starve it enough and concentration overpotential collapses the usable voltage.

Is the product water steam or liquid inside an alkaline cell running at ~200°C?

Liquid — the cell runs under elevated pressure so water stays below its raised boiling point; boiling it would strip heat and disturb the electrolyte concentration.

What if H₂ and O₂ leaked and mixed inside the cell instead of reacting at separate electrodes?

You'd get uncontrolled combustion (or a small explosion) and no electricity — the whole point of separating the half-reactions is to force electrons through the external wire.

Run the exact same cell in reverse with an applied voltage above

1.23 V

— what do you get?

Electrolysis: you split water back into H₂ and O₂, since the reaction is now driven non-spontaneously — the fuel cell and the electrolyser are the same chemistry running opposite directions.

Δ G = 0

, could the cell still deliver electrical work?

No — zero free energy means no spontaneous drive, so

E_{cell}^{\circ} = 0

and no current flows; useful work requires a negative

Δ G

, tying back to Gibbs free energy and spontaneity.

Why can a "direct methane" fuel cell exist even though methane fouls a standard Pt H₂ cell?

It uses different catalysts and higher operating temperatures designed to oxidise methane directly — a different device, not the H₂/O₂ cell, relevant to the wider Hydrogen economy debate.

Recall One-sentence spine to re-derive any answer

A fuel cell is an open galvanic cell where catalyst electrodes stay unchanged, ions recycle, electrons do work through the wire, voltage follows $E = - Δ G / n F$ (shifted by Nernst), and the ceiling is $Δ G /Δ H$ — not Carnot. ::: Anchor every trap back to "open system, catalyst not reactant, direct energy conversion."