Commercial electrochemical cellsAQA A-Level Chemistry: Revision notes
Section 1
Cells as a commercial source of energy
An electrochemical cell converts chemical energy into electrical energy. Commercial cells fall into three types:
- Non-rechargeable (irreversible) cells: the cell reaction runs one way until a reactant is used up, then the cell is discarded (e.g. zinc–silver oxide button cells).
- Rechargeable cells: the electrode reactions are reversible, so an external voltage can drive them backwards (e.g. lead–acid, nickel–cadmium, lithium-ion).
- Fuel cells: reactants are supplied continuously from outside, so the cell does not need electrical recharging and gives current for as long as fuel is fed in.
The cell EMF is the potential difference when no current flows: EMF = E(positive electrode) − E(negative electrode).
Section 2
Using electrode data to deduce cell reactions
Given standard electrode potentials, the half-cell with the more negative E is the negative electrode. When the cell supplies current, it goes in the oxidation direction (electrons released), and the other half-cell goes in the reduction direction at the positive electrode.
Worked example (Ni–Cd): NiO(OH) + H₂O + e⁻ ⇌ Ni(OH)₂ + OH⁻, +0.52 V; Cd(OH)₂ + 2e⁻ ⇌ Cd + 2OH⁻, −0.88 V.
- Negative: Cd + 2OH⁻ → Cd(OH)₂ + 2e⁻
- Positive: NiO(OH) + H₂O + e⁻ → Ni(OH)₂ + OH⁻
- EMF = +0.52 − (−0.88) = +1.40 V
Do not multiply E values by the number of electrons. Electrons flow through the external circuit from negative to positive, and this flow is the current. In a rechargeable cell the solid products stay on the electrodes, so applying an external voltage larger than the EMF reverses both reactions.
Do not double an E value to balance the electrons. E is not changed by the stoichiometry, so scale the equations only after deciding which way each half-cell goes.
Say that the cell reaction is driven backwards by an external voltage greater than the EMF when explaining recharging.
Section 3
Lithium cells
In a lithium cell the simplified electrode reactions on discharge are:
- Negative electrode: Li → Li⁺ + e⁻
- Positive electrode: Li⁺ + CoO₂ + e⁻ → Li⁺[CoO₂]⁻
Lithium is oxidised and cobalt is reduced from +4 to +3. The electrons travel through the external circuit and Li⁺ ions move through the electrolyte, completing the circuit. Lithium has a very negative electrode potential and a low density, so lithium cells give a high EMF and a lot of energy for their mass, which suits phones and laptops. Lithium-ion cells are rechargeable: applying a voltage reverses both reactions.
Section 4
Alkaline hydrogen–oxygen fuel cells
Hydrogen and oxygen are fed to separate electrodes in contact with an alkaline electrolyte (aqueous KOH).
- Negative: H₂ + 2OH⁻ → 2H₂O + 2e⁻ (E = −0.83 V as the reverse)
- Positive: O₂ + 2H₂O + 4e⁻ → 4OH⁻ (E = +0.40 V)
- Overall: 2H₂ + O₂ → 2H₂O
- EMF = +0.40 − (−0.83) = +1.23 V
The only product is water. A fuel cell does not run down like a normal cell, because fresh fuel is supplied, so it needs refuelling rather than recharging.
In an alkaline fuel cell write OH⁻ in the equations, not H⁺. Check that H₂O and OH⁻ balance for atoms and charge.
Section 5
Benefits and risks to society
Benefits: portable power with no mains supply; rechargeable cells reduce waste because they are reused; fuel cells give no CO₂ or pollutants at the point of use and refuel quickly.
Risks and limitations: non-rechargeable cells add to waste and use up metals such as silver; cobalt, lithium and cadmium are scarce or toxic and must be recycled safely; lithium cells can overheat and ignite if damaged; hydrogen is flammable and hard to store, and is often made from fossil fuels or from electricity, so the benefit depends on the source.
When asked to evaluate, give both sides and a reasoned conclusion.
Must Know
- Non-rechargeable, rechargeable and fuel cells differ in whether the reaction is irreversible, reversible or fed continuously.
- More negative E means negative electrode; EMF = E(positive) − E(negative).
- Lithium cell: Li → Li⁺ + e⁻ and Li⁺ + CoO₂ + e⁻ → Li⁺[CoO₂]⁻.
- Alkaline H₂/O₂ fuel cell: EMF +1.23 V; product water.
- Evaluate benefits and risks with a conclusion.
That's the notes covered.
Carry on to the next subtopic.
Exam questions on Commercial electrochemical cells
- A manufacturer makes rechargeable nickel–cadmium cells for cordless power tools. The cell is based on two half-equations with standard electrode potentials: NiO(OH) + H₂O + e⁻ ⇌ Ni(OH)₂ + OH⁻, E = +0.52 V; and Cd(OH)₂ + 2e⁻ ⇌ Cd + 2OH⁻, E = −0.88 V. The electrolyte is aqueous alkali.Explain why this cell can be recharged.2 marks
- A lithium cell in a mobile phone uses a lithium-containing negative electrode and a cobalt(IV) oxide positive electrode. When the cell supplies current, the simplified electrode reactions are: negative electrode, Li → Li⁺ + e⁻; positive electrode, Li⁺ + CoO₂ + e⁻ → Li⁺[CoO₂]⁻.Explain how the electrode reactions in this cell generate an electric current.2 marks
- A bus company is trialling hydrogen fuel-cell buses that use an alkaline hydrogen–oxygen fuel cell. Hydrogen and oxygen are fed continuously to separate porous electrodes in contact with aqueous potassium hydroxide. The relevant electrode potentials are: O₂ + 2H₂O + 4e⁻ ⇌ 4OH⁻, E = +0.40 V; and 2H₂O + 2e⁻ ⇌ H₂ + 2OH⁻, E = −0.83 V.Deduce the equation for the reaction at each electrode when the cell is supplying current, and calculate the EMF of the cell.3 marks
Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).