Part 3: Storage cells and hydrogen–oxygen fuel cells

Follow oxidation and reduction during discharge and charging, then compare a stored supply of reactants with continuous fuel-cell operation.

What limits the energy supply?

A primary cell contains reactants that are consumed during discharge and is not designed for safe, efficient regeneration. A secondary or storage cell has a sufficiently reversible chemical system: an external power supply drives the discharge reaction backwards during charging. Rechargeability requires suitable chemistry and an electrode structure that survives cycling; writing a reversible arrow does not make an arbitrary cell rechargeable.

A fuel cell receives fuel and oxidant continuously and removes products. It converts part of the chemical energy of reaction into electrical work through separated electrode reactions. It can operate while the supply and operating conditions are maintained. A nearly steady supply can support a nearly steady voltage, but current draw, temperature, fuel concentration, catalyst condition and internal resistance still affect its actual voltage.

In discharge, anode means oxidation and cathode means reduction. During charging a secondary cell’s processes reverse. Keep the process definitions rather than memorising that a named physical electrode must always be the anode. A charging voltage must oppose the discharge emf and in practice also overcome losses.

Interpret a rechargeable cell from supplied data

A lead–acid cell provides a useful worked model. Suppose the supplied reduction potentials are PbSO₄ + 2e⁻ ⇌ Pb + SO₄²⁻, E° = −0.36 V, and PbO₂ + SO₄²⁻ + 4H⁺ + 2e⁻ ⇌ PbSO₄ + 2H₂O, E° = +1.69 V. During discharge the more positive system is reduced and lead is oxidised by reversing the first half-equation. E°cell = 1.69 − (−0.36) = +2.05 V for these rounded data.

Adding the half-equations cancels two electrons. Pb(0) becomes Pb(II), while Pb(IV) in PbO₂ becomes Pb(II). Sulfate and acid are consumed and water forms, so electrolyte composition changes during discharge. Charging reverses the overall reaction and regenerates Pb and PbO₂, using electrical energy. Do not multiply the potential when multiplying the equation, and do not assume the working cell meets standard conditions.

In unfamiliar battery questions, identify oxidation states, reverse the oxidation half-equation, equalise electrons, add and cancel common species. For evaluation, consider energy per mass, recharge cycles, charging time, cost, toxic materials and recycling using the supplied evidence. A large voltage alone does not imply a large total energy capacity: capacity also depends on the amount of reacting material.

Hydrogen–oxygen fuel cell with an acidic electrolyte

Hydrogen enters the negative anode and loses electrons. Oxygen enters the positive cathode and gains electrons, combining with H⁺ to produce water. Electrons travel through the external load; H⁺ is transported through the electrolyte or proton-conducting membrane. The membrane does not provide the electron path. Porous catalytic electrodes give gases access to the reaction sites.

Use equal electron totals before adding the half-equations. Twice the hydrogen equation supplies the four electrons needed by one oxygen molecule. H⁺ then cancels from the overall equation. The standard potentials +1.23 V for O₂/H₂O and 0.00 V for H⁺/H₂ give E°cell = +1.23 V at the usual standard conditions. Real operating voltages are lower under load because of kinetic and resistive losses.

Hydrogen–oxygen fuel cell with an alkaline electrolyte

In alkaline conditions write half-equations using OH⁻ and water instead of H⁺. Oxygen reduction makes OH⁻ at the cathode; hydrogen oxidation consumes OH⁻ at the anode. Hydroxide transport therefore runs from the oxygen side towards the hydrogen side, opposite to proton transport in an acidic cell. External electron flow is still from hydrogen to oxygen.

The tabulated reduction potentials are approximately −0.83 V for 2H₂O + 2e⁻ ⇌ H₂ + 2OH⁻ and +0.40 V for O₂ + 2H₂O + 4e⁻ ⇌ 4OH⁻. Their difference is +1.23 V. Add the discharge equations and cancel both OH⁻ and the water present on both sides; the overall hydrogen–oxygen reaction matches the acidic cell. Never combine an acidic half-equation with an alkaline one without converting the medium.

Hydrogen-rich fuels and an evidence-based comparison

Methanol and other hydrogen-rich fuels can be used in fuel cells, either directly or through processing that supplies hydrogen. In an idealised acidic direct-methanol cell the anode oxidation is CH₃OH + H₂O → CO₂ + 6H⁺ + 6e⁻. Combining with oxygen reduction gives CH₃OH + 1½O₂ → CO₂ + 2H₂O. Unlike pure hydrogen use, a carbon-containing fuel produces carbon dioxide at the point of use.

A hydrogen cell produces water locally and can avoid combustion pollutants such as carbon monoxide. It can also convert chemical energy efficiently without a combustion-engine cycle. Its whole-life impact depends on how hydrogen and electricity were produced, compression or liquefaction energy, catalyst manufacture and infrastructure. Hydrogen is an energy carrier: it is not automatically renewable.

Storage is challenging because hydrogen has low density. Compressed gas needs pressure-resistant vessels; liquid hydrogen requires very low temperatures; solid storage materials add mass and may release hydrogen slowly. Hydrogen’s flammability and the costs of catalysts and replacement cells are practical constraints. Compare a specified option using evidence rather than claiming all fuel cells have zero emissions or unlimited efficiency.

Quick checks

Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official Edexcel mark allocations.

Sources