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

• (a) The anode is where oxidation (loss of electrons) happens. Hydrogen, ${\text{H}}_2$, is the fuel that gives up electrons, so H₂ is fed to the anode.
• (b) The cathode is where reduction (gain of electrons) happens. Oxygen, ${\text{O}}_2$, accepts electrons, so O₂ is fed to the cathode.
• (c) Electrons leave the cell from the anode into the wire, so the anode is the negative (−) terminal; they return at the cathode, making it positive (+).
• (d) The overall reaction $2{\text{H}}_2+{\text{O}}_2\to 2{\text{H}}_2\text{O}$ makes liquid water as the only product.

Which tool? For any galvanic cell the standard cell potential is $E_{\text{cell}}^{\circ }=E_{\text{cathode}}^{\circ }-E_{\text{anode}}^{\circ }$ We subtract because both numbers are quoted as reduction potentials (from Standard electrode potentials); to turn the anode into an oxidation we flip its sign, which the minus sign does for us. $E_{\text{cell}}^{\circ }=0.40-(-0.83)=0.40+0.83=\boxed{1.23\text{V}}$ A positive $E_{\text{cell}}^{\circ }$ tells us the reaction is spontaneous as written — the cell really does push current.

Which tool? A fuel cell is not a heat engine, so Carnot's efficiency does not apply. Instead, only the free energy $\Delta G$ (the part of the released energy that can do electrical work) is usable; the total energy released is $\Delta H$. The efficiency $\eta$ (defined in the notation box as useful-out over total-in) is therefore the ratio of these two. From Gibbs free energy and spontaneity: ${\eta }_{\text{max}}=\frac{\Delta G}{\Delta H}$ Both are negative, so the ratio is positive: ${\eta }_{\text{max}}=\frac{-237}{-286}=\frac{237}{286}=0.829\dots \approx \boxed{83\%}$ The missing $17\%$ is the $T\Delta S$ term — energy that stays as heat/entropy and can never become electricity, no matter how good the engineering.

Which tool? We use $\text{energy}=\text{power}\times \text{time}$. Why this and not something fancier? Power (kW) is energy per unit time, so multiplying it back by the time restores the total energy — the units literally say $\text{kW}\times \text{h}=\text{kWh}$. Then a fixed conversion factor swaps units.

• (a) Time $=24\times 8=192\text{h}$. $E=1.5\text{kW}\times 192\text{h}=\boxed{288\text{kWh}}$
• (b) Convert with $1\text{kWh}=3.6\text{MJ}$: $288\times 3.6=\boxed{1036.8\text{MJ}}$

Which tool? Mass ratio — the reaction is a fixed recipe. From $2{\text{H}}_2+{\text{O}}_2\to 2{\text{H}}_2\text{O}$: for every $1\text{kg}$ of H₂, $9\text{kg}$ of water appears. $m_{{\text{H}}_2\text{O}}=25\times 9=\boxed{225\text{kg}}$ This matches the parent note's Apollo figure — far more water than the crew drinks, so the surplus cools equipment.

• (a) Voltage lost $=1.23-0.90=0.33\text{V}$. As a fraction: $\frac{0.33}{1.23}=0.268\dots \approx \boxed{27\%}$
• (b) The three losses (overpotentials) are:
  1. Activation overpotential — energy hump to start each electrode reaction. The Pt catalyst attacks this one by lowering the activation barrier (see Catalysis).
  2. Ohmic loss — plain resistance of electrodes, electrolyte, wiring ($V=IR$).
  3. Concentration overpotential — reactant runs low right at the electrode surface if gas supply lags.

Write both half-reactions balanced to $4e^{-}$: $\text{Anode: }2{\text{H}}_2+4{\text{OH}}^{-}\to 4{\text{H}}_2\text{O}+4e^{-}$ $\text{Cathode: }{\text{O}}_2+2{\text{H}}_2\text{O}+4e^{-}\to 4{\text{OH}}^{-}$ The anode consumes $4{\text{OH}}^{-}$; the cathode regenerates $4{\text{OH}}^{-}$. Net ${\text{OH}}^{-}$ change $=-4+4=0$. So ${\text{OH}}^{-}$ is merely a shuttle carrying charge from cathode to anode, not a reactant that depletes. Only H₂ and O₂ are truly consumed, and only water is truly made.

Which tool? $\eta =\frac{\text{useful out}}{\text{chemical in}}\Rightarrow \text{chemical in}=\frac{\text{useful out}}{\eta }$.

• Fuel cell: $\frac{288}{0.60}=480\text{kWh}$ of chemical energy.
• Combustion: $\frac{288}{0.25}=1152\text{kWh}$ of chemical energy.
• Ratio: $\frac{1152}{480}=\boxed{2.4}$ — the combustion engine must carry 2.4× the fuel. On a spacecraft, that extra mass is decisive.

Which tool? The bridge between electricity and thermodynamics is $\Delta G^{\circ }=-nF{E}_{\text{cell}}^{\circ }$ Here $n$ = moles of electrons transferred, $F$ = charge per mole of electrons (coulombs), $E_{\text{cell}}^{\circ }$ = volts (joules per coulomb). Multiplying charge $\times$ voltage gives energy in joules. The minus sign encodes that a positive cell voltage means a negative (spontaneous) $\Delta G$ — the link back to Gibbs free energy and spontaneity. $\Delta G^{\circ }=-(4)(96485)(1.23)=-474,706\text{J}\approx \boxed{-475\text{kJ}}$ This is for $2$ mol H₂. Per mole of H₂: $\frac{-475}{2}=\boxed{-237\text{kJ/mol}}$ Beautiful check: this matches the $\Delta G=-237\text{kJ/mol}$ used in the efficiency formula (Problem 2.1). Electrochemistry and thermodynamics agree.

• (a) Running the cell backward needs at least the same $\boxed{1.23\text{V}}$ — you must repay exactly the free energy the fuel cell would release.
• (b) Overpotential always opposes what you want:
  • In the fuel cell (getting energy out), overpotential subtracts → you see less than 1.23 V (e.g. 0.9 V).
  • In electrolysis (putting energy in), overpotential adds → you must supply more than 1.23 V (e.g. 1.8–2.0 V). In both directions the kinetic toll is a loss; it just shows up on opposite sides of the ideal.

Step (a) — charge. Which tool? We need the total charge $Q$, and voltage is energy per unit charge ($V=\text{J/C}$). Rearranging that definition gives $Q=E/V$ — divide energy by voltage to recover charge. This is why we use $Q=E/V$ and not, say, $Q=It$ (we were never given the current directly). Electrical energy $E=288\text{kWh}=288\times 3.6\times 10^6\text{J}=1.0368\times 10^9\text{J}$. $Q=\frac{1.0368\times 10^9}{0.90}=1.152\times 10^9\text{C}$

Step (b) — moles of electrons. Which tool? The Faraday constant $F$ is the charge of one mole of electrons, so dividing total charge by $F$ converts coulombs → moles of electrons. $n_{e^{-}}=\frac{Q}{F}=\frac{1.152\times 10^9}{96485}=1.1940\times 10^4\text{mol }{e}^{-}$

Step (c) — H₂ needed. Each mole of H₂ gives $2$ mol $e^{-}$: $n_{{\text{H}}_2}=\frac{n_{e^{-}}}{2}=5.970\times 10^3\text{mol}$ $m_{{\text{H}}_2}=5.970\times 10^3\times 2.0\text{g}=1.194\times 10^4\text{g}\approx \boxed{11.9\text{kg}}$

Step (d) — water. $2{\text{H}}_2\to 2{\text{H}}_2\text{O}$, so mol water = mol H₂; $M({\text{H}}_2\text{O})=18\text{g/mol}$: $m_{{\text{H}}_2\text{O}}=5.970\times 10^3\times 18\text{g}=1.075\times 10^5\text{g}\approx \boxed{107\text{kg}}$

Reality check vs. parent note: the note quotes ~25 kg H₂ because real efficiency is ~60 %, not 100 %. Our Faraday number (11.9 kg) is the ideal floor; dividing by the ~0.5 practical fuel-to-electricity fraction pushes it toward the ~25 kg figure. Both are consistent once you separate the ideal from the operational.

The alkaline cell's real weakness is CO₂: atmospheric CO₂ reacts with the KOH electrolyte to form carbonate, which clogs the cell. Apollo could tolerate this because it ran on pure stored H₂ and O₂ (no ambient air at all) and already carried CO₂ scrubbers for the crew's breathing, so contamination was fully managed; on top of that KOH was cheap, robust 1960s technology needing less platinum. A hydrogen car, by contrast, must cold-start in seconds and often breathes ambient air laden with CO₂, both of which cripple an alkaline cell. The PEM cell answers exactly these needs — its solid Nafion membrane runs cool (about $60$–$80{}^{\circ }\text{C}$, in the same modest band as a modern alkaline cell), starts fast, is compact, and shrugs off the driving environment — so the same electrochemistry ($2{\text{H}}_2+{\text{O}}_2\to 2{\text{H}}_2\text{O}$) ends up with two different electrolytes chosen entirely by operating constraints, not by the reaction itself.

Why does a fuel cell beat a combustion engine on efficiency? ::: It converts chemical energy directly to electricity (ceiling $\Delta G/\Delta H\approx 83\%$) instead of going chemical→heat→work, so it dodges the Carnot limit that pins engines near 25 %.