IB Chemistry HL Topic 4 — Energy from Fuels Paper 1 & 2 Core idea ~12 min read

Fuel Cells

Every page so far has burnt the fuel. A fuel cell does something cleverer: it separates the oxidation from the reduction, forces the electrons to travel round an external circuit, and takes the energy out as electricity directly. No flame, no pistons, and no losing most of the energy as heat on the way.

📘 What you need to know

What makes it different from burning

Burn hydrogen in air and the electrons pass straight from hydrogen to oxygen where the two molecules collide. You get the energy, but you get it as heat, and heat is an awkward form to work with.

A fuel cell keeps the two halves of that reaction physically apart. Hydrogen is oxidised at one electrode and releases electrons; oxygen is reduced at the other electrode and needs electrons. The only route between them is the wire you have connected, so the electrons have to travel through your circuit and do useful electrical work on the way.

This is the same trick as any voltaic cell you have met before. The novelty is only that the reactants are gases fed in continuously rather than solids and solutions sealed inside.
The hydrogen–oxygen fuel cell electric current e⁻ e⁻ H₂ in unused H₂ out O₂ (air) in water out OH⁻ ionsANODE oxidation, negative membrane CATHODE reduction, positiveElectrons take the long way round; ions cut through the electrolyte. Nothing is stored, so it works for as long as fuel and air are supplied.
The membrane matters more than it looks. If hydrogen and oxygen mixed inside the cell they would simply react together, releasing the energy as heat and producing no current at all.

The parts, and what each is for

The half-equations for the hydrogen cell

In an alkaline electrolyte, hydroxide ions are the species available, so they appear in both half-equations. At the anode, hydrogen is oxidised:

Anode — oxidation 2H2(g) + 4OH−(aq) → 4H2O(l) + 4e−   E° = −0.83 V

At the cathode, oxygen is reduced, and the hydroxide ions are regenerated:

Cathode — reduction O2(g) + 2H2O(l) + 4e− → 4OH−(aq)   E° = +0.40 V

Add them together and cancel what appears on both sides — the 4e−, the 4OH−, and two of the four waters:

Overall 2H2(g) + O2(g) → 2H2O(l)   E°cell = +1.23 V

Notice what the hydroxide is doing: it is consumed at the anode and regenerated at the cathode, so the electrolyte is not used up. It is carrying charge, not taking part in the net reaction.

WORKED EXAMPLE

Using the two half-cell potentials above, calculate E°cell for the alkaline hydrogen–oxygen fuel cell, and identify which electrode is the anode.

Step 1: The more positive potential is reduced, so it is the cathode +0.40 V (oxygen) is more positive than −0.83 V, so oxygen is reduced Step 2: The other half-cell must be the anode hydrogen is oxidised, so the hydrogen electrode is the anode − and it is negative Step 3: Apply E°cell = E°cathode − E°anode E°cell = (+0.40) − (−0.83) E°cell = +1.23 V, hydrogen electrode is the anode a positive E°cell confirms the reaction is spontaneous, as it must be

Change the electrolyte and the half-equations change

This catches people out badly. The same cell in an acidic electrolyte, such as a proton exchange membrane, has completely different-looking half-equations, because now it is H+ that is available rather than OH−:

Acidic electrolyte anode: 2H2(g) → 4H+(aq) + 4e−
cathode: O2(g) + 4H+(aq) + 4e− → 2H2O(l)

Add those together and you get… exactly the same overall reaction, 2H2 + O2 → 2H2O, and the same +1.23 V. The electrolyte changes the route, not the destination.

Same reaction, two different electrolytes ALKALINE (KOH) OH⁻ moves cathode → anode H₂ + 2OH⁻ → 2H₂O + 2e⁻ at the anode ACIDIC (PEM) H⁺ moves anode → cathode H₂ → 2H⁺ + 2e⁻ at the anodeoverall, either way 2H₂(g) + O₂(g) → 2H₂O(l)The half-equations look nothing alike; the overall reaction is identical.
In both cases the ion moves so as to cancel the charge building up: away from the electrode gaining positive charge, towards the one losing it.

🧩 Building a fuel-cell half-equation from scratch

  1. Decide which species is oxidised (the fuel, at the anode) and which is reduced (oxygen, at the cathode).
  2. Ask what ion the electrolyte provides: OH− if alkaline, H+ if acidic. That ion is allowed to appear.
  3. Balance the atoms other than O and H first.
  4. Balance oxygen using H2O, then hydrogen using H+ or OH−.
  5. Balance the charge by adding electrons to whichever side needs them.
  6. Scale the two half-equations so the electrons cancel, add them, and cancel anything appearing on both sides.

Why fuel cells are so efficient

This is the strongest argument for the technology, and it is a thermodynamic one rather than an engineering one.

A petrol engine converts chemical energy into heat, heat into mechanical work, and mechanical work into electricity. The middle step is the problem: converting heat into work has a hard theoretical ceiling, and real engines manage only 25–30% overall.

A fuel cell goes from chemical energy straight to electrical energy. There is no heat step to be limited by, so the ceiling is set by a different and much more generous quantity.

Why a fuel cell beats an engineFUEL CELL chemical energy electrical energy up to ~83% in theoryPETROL ENGINE chemical heat mechanical electricalheat is lost at every one of these steps typically only 25−30% efficientThe heat step is the bottleneck, and a fuel cell simply skips it.
Real hydrogen fuel cells reach 50–60% rather than 83%, because of resistance in the electrolyte and slow reaction at the electrodes. Even so, that is roughly double a petrol engine.
WORKED EXAMPLE

1.00 g of hydrogen is consumed by a fuel cell operating at 1.23 V. Calculate the charge that passes and the electrical energy produced. Compare it with the 141.8 kJ of enthalpy available from burning the same mass (ΔHc = −285.8 kJ mol−1). Take F = 96 500 C mol−1.

Step 1: Moles of hydrogen, then moles of electrons n(H₂) = 1.00 ÷ 2.016 = 0.496 mol each H₂ releases 2 electrons at the anode n(e⁻) = 2 × 0.496 = 0.992 mol Step 2: Charge, using Q = nF Q = 0.992 × 96 500 = 9.57 × 10⁴ C Step 3: Electrical energy, using W = QV W = 9.57 × 10⁴ × 1.23 = 1.18 × 10⁵ J Q = 9.57 × 10⁴ C, W = 118 kJ Step 4: Compare with the enthalpy 118 ÷ 141.8 × 100 = 83.0% so 83% of the chemical energy is available as electricity − the theoretical ceiling

Hydrogen: the honest assessment

AdvantagesDisadvantages
Water is the only product, so there are no harmful emissions at the point of useHydrogen is highly flammable, so storage and handling carry real safety risks
Operates at or near room temperature, avoiding combustion entirelyNeeds heavy, thick-walled pressurised tanks, adding cost, mass and complexity
No nitrogen oxides are formed, unlike in high-temperature combustionMost hydrogen is currently made from fossil fuels, so it is not renewable in practice
Energy goes directly to electricity, so efficiency is roughly double an engine’sLow energy density by volume, so containers must be large even though the mass is small
Used on spacecraft, where the water produced becomes drinking water for the crewUntil cheap, sustainable hydrogen production exists, wide use stays limited
Hydrogen is a carrier, not a source. There are no hydrogen mines. Every kilogram has to be made using energy from somewhere else — usually from natural gas at present. So a hydrogen car is exactly as clean as the electricity or fuel used to make its hydrogen. This is the same life-cycle argument that qualifies the carbon-neutral claim for biofuels.

The methanol fuel cell

Methanol solves hydrogen’s most annoying practical problem. It is a liquid at room temperature, so it can be stored and transported in an ordinary tank with no pressurisation at all. The cell works the same way, but the fuel supplying the hydrogen ions is methanol rather than hydrogen gas, and the electrolyte is acidic — typically a proton exchange membrane.

Anode — oxidation CH3OH(aq) + H2O(l) → CO2(g) + 6H+(aq) + 6e−
Cathode — reduction O2(g) + 4H+(aq) + 4e− → 2H2O(l)

The electrons do not cancel as written — six against four — so the cathode equation has to be scaled by 1.5 first.

WORKED EXAMPLE

Combine the two half-equations above to give the overall equation for the methanol fuel cell, using whole-number coefficients.

Step 1: Match the electrons anode gives 6e⁻, cathode takes 4e⁻, so multiply the cathode by 1.5 1.5O₂ + 6H⁺ + 6e⁻ → 3H₂O Step 2: Add the two half-equations CH₃OH + H₂O + 1.5O₂ + 6H⁺ + 6e⁻ → CO₂ + 6H⁺ + 6e⁻ + 3H₂O Step 3: Cancel what appears on both sides the 6H⁺ and 6e⁻ cancel; 3H₂O − 1H₂O leaves 2H₂O CH₃OH(aq) + 1.5O₂(g) → CO₂(g) + 2H₂O(l) Step 4: Double for whole numbers 2CH₃OH(aq) + 3O₂(g) → 2CO₂(g) + 4H₂O(l) check: C 2 = 2, H 8 = 8, O 2 + 6 = 8 = 4 + 4 ✓
Methanol vs hydrogenBetterWorse
Storage and transportA liquid at room temperature, so no pressurised tank; higher energy density by volumeToxic as well as flammable, so spills are more hazardous to people
Operating conditionsRuns at low pressure and temperature; the aqueous conditions give the membrane a longer lifespanLower voltage and power output per unit mass
Efficiency—Methanol crossover: the small polar molecule diffuses through the membrane and is oxidised at the cathode, wasting fuel and dropping the voltage
EmissionsCan be made from renewable biomass, so lower life-cycle emissions than hydrogen from fossil fuelsProduces carbon dioxide at the point of use; hydrogen produces only water
CostFuel is cheap and easy to distribute using existing infrastructureNeeds expensive catalysts such as ruthenium and palladium
Worth noticing: the methanol cell still releases CO2, so it is not a zero-carbon technology. What it avoids is combustion — so no nitrogen oxides, no carbon monoxide and no soot, and roughly double the efficiency of burning the same methanol in an engine.

💡 Exam tip

⚠ Common mix-up

Up next: Entropy and Spontaneity — enthalpy is only half the story. Some endothermic reactions happen anyway, and to explain why we need a second quantity.

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