OCR A Chemistry H432 · Year 13 · 5.2.3

Part 3: Feasibility, storage cells and fuel cells

All 3 parts available. Reviewed 6 October 2026.

Use supplied potentials to predict electron transfer, then explain why concentration, kinetics and practical engineering limit the prediction.

Choose the reduction and oxidation, then check the stated direction

For a proposed forward reaction, calculate E°cell using the half-cell reduced minus the listed reduction potential of the half-cell oxidised. A positive value supports thermodynamic feasibility under standard conditions. A negative value means the reverse direction is favoured under those conditions.

Original illustrative data: Ag⁺/Ag = +0.80 V and Fe³⁺/Fe²⁺ = +0.77 V. For Ag⁺ + Fe²⁺ → Ag + Fe³⁺, E°cell = +0.03 V. This is a small positive standard driving force, not proof of a fast or complete reaction in every mixture.

Identify the actual species before comparing. An oxidising agent is the electron-accepting species on the left of a reduction half-equation. Saying “the more positive metal reacts” can confuse a metal with its ions. Use “more positive” or “more negative” rather than an ambiguous “higher” when signs differ.

Standard predictions have two different limitations

Concentrations other than standard conditions change electrode potentials and can alter the favoured direction, especially where the standard voltage is small. For Fe³⁺ + e⁻ ⇌ Fe²⁺, increasing the oxidised-to-reduced concentration ratio favours reduction and raises the reduction potential qualitatively. A numerical Nernst-equation treatment is not required here.

Kinetic barriers can prevent an otherwise favourable reaction proceeding at a useful rate. Electrode surfaces, passivation and catalysts affect the pathway and rate. A positive E°cell therefore answers a thermodynamic question, while the observed current also depends on transport and kinetics.

In a multistep reduction, compare the reducing agent with each relevant couple separately. An agent capable of one reduction step may not reduce the next oxidation state. Sum balanced half-equations only after establishing the direction for the step being considered.

Apply the same principles to unfamiliar batteries

A storage cell supplies electrical energy through a spontaneous redox reaction. During discharge, oxidation is at the negative electrode and reduction at the positive electrode. In a rechargeable cell, an external power source drives the reverse chemistry during charging, provided the chemical changes are sufficiently reversible in practice.

OCR supplies the relevant storage-cell equations and data. Use them: balance electrons, add the discharge reactions, calculate the voltage and reverse the chemical changes for charging where appropriate. Do not replace an unfamiliar supplied system with a memorised lithium equation.

Practical advantages such as high energy density or rechargeability must be weighed against raw-material supply, degradation, recycling and hazards including fire or toxic components. These are context-dependent comparisons; a claim that every rechargeable battery is automatically environmentally harmless is unsupported.

Worked example: sum the electrode reactions

A fuel cell uses reaction of a continuously supplied fuel with oxygen to create a voltage. For an illustrative acidic hydrogen cell, oxidation is 2H₂ → 4H⁺ + 4e⁻ and reduction is O₂ + 4H⁺ + 4e⁻ → 2H₂O. Cancelling H⁺ and electrons gives 2H₂ + O₂ → 2H₂O.

For alkaline supplied half-equations, the corresponding forms can be 2H₂ + 4OH⁻ → 4H₂O + 4e⁻ and O₂ + 2H₂O + 4e⁻ → 4OH⁻. Cancelling common species gives the same overall reaction. Do not combine an acidic oxidation equation with an alkaline reduction equation without converting the chemistry consistently.

Hydrogen produces water at the point of use, but lifecycle emissions depend on its production and transport. Storage and distribution, catalyst cost and fuel supply can limit use. A fuel cell differs from a closed storage battery because reactants are supplied during operation. OCR does not require recall of particular fuel-cell equations; practise interpreting those provided.

Quick checks

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

Q1. For X²⁺/X = −0.20 V and Y²⁺/Y = +0.50 V, predict the spontaneous standard reaction.Show answer

Y²⁺ is reduced and X is oxidised: X + Y²⁺ → X²⁺ + Y. E°cell = 0.50 −(−0.20) = +0.70 V. Electrons move externally from X to the Y electrode.

Q2. A calculated E°cell is +0.02 V. Can you guarantee reaction direction in a very non-standard mixture?Show answer

No. The standard driving force is small, and concentration changes can alter the electrode potentials. The calculation also says nothing about the speed.

Q3. What happens chemically when an ideal rechargeable cell is charged?Show answer

An external electrical supply drives the reverse of the discharge reactions, regenerating active materials. Real reversibility and degradation limit how well this works.

Q4. Why must electrons cancel from a summed cell reaction?Show answer

Electrons released at oxidation are consumed at reduction; the cell transfers them internally through the external circuit. Multiply half-equations to equalise electron amounts before addition; no net electron remains in the overall chemical equation.

Q5. Give a balanced evaluation of “a hydrogen fuel cell is zero-carbon”.Show answer

Pure hydrogen oxidation produces no CO₂ at the point of use. However, making hydrogen, supplying energy and transporting it may produce greenhouse gases. State the lifecycle boundary before making the claim.

Sources

Sources and examiner guidance (reviewed 6 October 2026)

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