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Electrochemistry

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Electrochemistry

Measure and combine electrode potentials, predict cell reactions, and calculate what electrolysis produces.

A-Level 9476 (2026-2027)

01

Track electrons before calculating a voltage

Oxidation loses electrons and raises oxidation number; reduction does the reverse.

In a redox reaction, oxidation and reduction occur together. Oxidation is electron loss or an increase in oxidation number. Reduction is electron gain or a decrease in oxidation number. The oxidising agent accepts electrons and is itself reduced; the reducing agent supplies electrons and is itself oxidised.

For Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s), zinc changes from 0 to +2 and releases two electrons. Copper changes from +2 to 0 and accepts them. Separating the two half-reactions can make those electrons travel through an external wire, delivering electrical energy.

Balance a half-equation in acidic solution
  1. Balance the changing element

    Write its reactant and product species with the correct charge.

  2. Balance oxygen and hydrogen

    Add H2O to balance O, then H+ to balance H.

  3. Balance charge

    Add electrons to the more positive side; check both atoms and charge.

  4. Combine

    Multiply half-equations to cancel electrons, add them, and cancel common species.

Worked example

Combine an oxyanion reduction with peroxide oxidation

Construct the reaction of acidified MnO4- with H2O2, forming Mn2+ and O2.

  1. Reduction: MnO4- + 8H+ + 5e- → Mn2+ + 4H2O.
  2. Oxidation: H2O2 → O2 + 2H+ + 2e-. Oxygen rises from -1 to 0.
  3. Multiply the first by 2 and the second by 5. Cancel ten electrons and ten of the sixteen H+ ions.
Answer

2MnO4- + 6H+ + 5H2O2 → 2Mn2+ + 8H2O + 5O2. All species except water and oxygen are aqueous; water is liquid and oxygen is gas. Both sides have net charge +4.

For a half-equation required in alkaline solution, an acid-balanced equation can be converted by adding enough OH- to both sides to consume every H+, then cancelling water. Use the actual stated product: permanganate, for example, does not necessarily form Mn2+ outside acidic conditions.

02

Measure every half-cell against the same reference

An isolated electrode has no independently measurable absolute potential.

A standard electrode potential, E°, is the potential difference of a half-cell relative to the standard hydrogen electrode, with all species in their standard states and negligible current flowing. Values are quoted for the half-reaction written as a reduction. A more positive value indicates a stronger tendency for that reduction under the stated standard conditions.

For the usual school measurements, aqueous concentrations are 1 mol dm-3, gases have standard pressure 100 kPa, and pure solids or liquids are used in their standard states. Strictly, dissolved species have unit activity; concentration is the course approximation. Temperature must be stated: the Data Booklet values are at 298 K. Standard state is not another name for 273 K gas conditions.

The standard hydrogen electrode (SHE) uses an inert platinum electrode coated with finely divided platinum in contact with H2(g) at 100 kPa and aqueous H+ at 1 mol dm-3, at the stated temperature. Its reversible half-equation is 2H+(aq) + 2e- ⇌ H2(g), and its standard potential is assigned 0.00 V. Platinum provides electrical contact and a catalytic surface; it is not consumed in this equation.

Connect the test half-cell to the SHE using a salt bridge and high-resistance voltmeter
Type to measureTest half-cell arrangementReduction represented
Metal with aqueous ionsA clean copper strip in 1 mol dm-3 Cu2+(aq).Cu2+ + 2e- ⇌ Cu.
Non-metal with aqueous ionsInert platinum contacting Cl2(g) at 100 kPa and 1 mol dm-3 Cl-(aq).Cl2 + 2e- ⇌ 2Cl-.
Two oxidation states of the same elementInert platinum in a solution with both Fe3+ and Fe2+, each at 1 mol dm-3.Fe3+ + e- ⇌ Fe2+.

Keep both half-cells at 298 K when using the tabulated values. A suitable salt bridge, such as one containing KNO3, permits ion movement and prevents charge build-up without rapidly mixing the solutions. The voltmeter draws negligible current so the reading approximates the equilibrium electromotive force. Record which terminal is positive, not only the size of the reading: copper is +0.34 V relative to the SHE, while zinc is -0.76 V.

Check your understandingA standard test half-cell is connected to the SHE. Its terminal is negative and the potential difference is 0.25 V. What is its standard electrode potential?Think it through, then reveal the answer
E° = -0.25 V relative to the SHE. Reporting +0.25 V would reverse the reduction tendency; the magnitude alone does not determine the sign.
03

The spontaneous cell sends electrons from anode to cathode

Use reduction potentials consistently, then balance the chemical equation.

The standard cell potential E°cell is the cell electromotive force, or maximum potential difference measured at negligible current, when all components are in standard states at the stated temperature. For a chosen reaction, cell = E°cathode - E°anode, with both tabulated values kept as reduction potentials.

At the cathode, reduction consumes electrons. At the anode, oxidation releases them. In a spontaneous galvanic cell, electrons flow through the wire from the negative anode to the positive cathode. Ions carry charge through the electrolyte and salt bridge; electrons do not flow through the bridge.

A zinc-copper galvanic cell

Zinc is the negative anode and dissolves to zinc ions. Electrons move through the external wire to the positive copper cathode, where copper ions are reduced. Bridge anions enter the zinc side and bridge cations enter the copper side to maintain electrical neutrality.

For standard concentrations at 298 K, the Zn/Cu cell has E°cell = 1.10 V. The sketch shows discharge; an ideal potential measurement draws negligible current.

Worked example

Predict the direction and voltage

Use E°(Zn2+/Zn) = -0.76 V and E°(Cu2+/Cu) = +0.34 V.

  1. The copper reduction is more favourable, so Cu2+ is reduced at the cathode.
  2. Reverse the zinc half-equation: Zn → Zn2+ + 2e-. Zinc is the anode and reducing agent.
  3. cell = 0.34 - (-0.76) = +1.10 V.
  4. The half-equations already transfer two electrons each: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s).
Answer

Electrons flow from zinc to copper. The written reaction is thermodynamically spontaneous under standard conditions. Zinc mass falls and copper mass rises.

Interpret the sign for the reaction as written
cellStandard thermodynamic conclusion
PositiveForward reaction is favourable: ΔG° < 0.
NegativeForward reaction is unfavourable; the reverse is favoured under standard conditions.
ZeroΔG° = 0; there is no standard thermodynamic driving force in either direction.
04

A standard prediction is not a guarantee of a rapid real reaction

Concentration affects equilibrium; activation barriers affect the time taken.

Electrode potentials depend on the actual composition. Think of each half-equation as a reduction equilibrium: a change favouring reduction generally makes that electrode potential more positive. A change favouring the reverse makes it less positive. These are qualitative predictions at fixed temperature, with other conditions held constant.

Concentration changes with other conditions fixed
Half-equationChangeEffect on reduction potential
Cu2+ + 2e- ⇌ CuIncrease [Cu2+].More positive: reduction of Cu2+ is favoured.
Fe3+ + e- ⇌ Fe2+Increase [Fe3+], or decrease [Fe2+].More positive; the oxidised-to-reduced concentration ratio rises.
2H+ + 2e- ⇌ H2Decrease [H+] at fixed hydrogen pressure.Less positive: hydrogen-ion reduction is less favoured.

In the Zn/Cu cell, diluting Cu2+ lowers the copper reduction potential. Increasing Zn2+ raises the zinc reduction potential. Either change reduces Ecell = ECu - EZn. Far enough from standard conditions, an E° prediction can cease to describe the actual direction. Complex formation and precipitation matter because they change the free ion concentrations.

Even a positive actual cell potential says nothing about how quickly a reaction proceeds. Large activation barriers, slow electrode reactions or a protective surface film can make a thermodynamically favoured change difficult to observe. Different temperature, gas pressure, pH or chemical species can also invalidate a direct application of tabulated E° values. Use the exact relevant half-equation and the conditions, not only the names of the elements.

Check your understandingExcess ammonia greatly reduces free Cu2+ by forming a complex. What happens to the potential of the free Cu2+/Cu couple at fixed temperature?Think it through, then reveal the answer
It becomes less positive because the concentration of the ion being reduced has fallen. Total dissolved copper can remain large: the free-ion concentration is the relevant one. If using a tabulated ammine-complex half-equation instead, include its ligand species and conditions explicitly.
05

Add Gibbs energies when combining half-reactions

The electron count connects an intensive potential to an extensive energy change.

ΔG° = -nFE°cell. Here n is the number of moles of electrons transferred per mole of the cell reaction as written, and F is the charge per mole of electrons. With F in C mol-1 and E in V, ΔG° is in J mol-1 of reaction because 1 V = 1 J C-1.

Worked example

Convert a cell voltage into free energy

Find ΔG° for Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s), with E°cell = 1.10 V and F = 9.65 × 104 C mol-1.

  1. The balanced reaction transfers two electrons, so n = 2.
  2. ΔG° = -2(9.65 × 104)(1.10) = -212300 J mol-1.
  3. Divide by 1000 to express the result in kJ mol-1.
Answer

ΔG° = -212 kJ mol-1. Doubling the reaction would double n and ΔG°, but E°cell would remain 1.10 V.

When two reduction steps combine to make a new reduction half-equation, their E° values cannot simply be added. Convert each to its corresponding -nFE° energy contribution relative to the same reference, add those contributions with the required signs, then divide by -nF for the combined half-equation.

Worked example

Find the potential for a combined iron reduction

Given Fe3+ + e- → Fe2+, E°1 = +0.77 V, and Fe2+ + 2e- → Fe, E°2 = -0.44 V, find E° for Fe3+ + 3e- → Fe.

  1. The first step contributes -F(0.77); the second contributes -2F(-0.44).
  2. Add: ΔG°combined = -F[0.77 + 2(-0.44)] = +0.11F.
  3. The combined step transfers three electrons, so -3FE°combined = +0.11F.
  4. combined = [0.77 + 2(-0.44)]/3 = -0.0367 V from the rounded supplied data.
Answer

Approximately -0.04 V, consistent with the Data Booklet. The result is electron-weighted, not the sum +0.33 V or an unweighted average.

Check your understandingA cell reaction has E° = +0.50 V and transfers two electrons. What changes if the equation is reversed?Think it through, then reveal the answer
E° becomes -0.50 V and ΔG° changes sign. The magnitude of n remains 2. Reversing changes the direction; multiplying coefficients changes the energy and electron amount, not the potential magnitude.
06

Compare useful energy, size and mass

A higher voltage can help, but it does not measure capacity by itself.

A cell converts chemical free energy into electrical energy. A rechargeable battery stores reactants internally and uses an external supply to restore them during charging. A fuel cell can operate while fuel and oxidant are supplied continuously and products are removed. Hydrogen-oxygen fuel cells and improved batteries offer possible gains in compactness, lower mass and cell voltage; compare the complete system, not just an attractive electrode reaction.

Hydrogen-oxygen fuel cell in an acidic electrolyte
LocationHalf-equation and role
Anode2H2(g) → 4H+(aq) + 4e-. Hydrogen is oxidised.
CathodeO2(g) + 4H+(aq) + 4e- → 2H2O(l). Oxygen is reduced.
Overall2H2(g) + O2(g) → 2H2O(l). Electrons travel through the external circuit.

The ideal standard voltage for this acidic cell is 1.23 - 0.00 = 1.23 V at 298 K; doubling its chemical equation does not give twice the voltage. Real operating voltage is lower because current flow involves losses. Several cells in series add their voltages. The fuel cell produces water at use, but hydrogen production, storage and transport affect the wider environmental and mass comparison.

What an improved cell would offer
Design gainWhy it can matter in an electric vehicleWhat else must be checked
Higher energy per unit massLess battery mass for the same stored energy, or more range for a given mass.The full pack includes casing, cooling and control systems.
Higher energy per unit volumeA smaller pack can store the same energy.Smaller size alone does not establish useful capacity.
Higher cell voltageMore energy can be delivered for the same charge; fewer series cells may meet a target voltage.Capacity, power output, lifetime and practical operating voltage still matter.

Worked example

Compare two hypothetical cells on a fair basis

Cell A delivers 3.2 V with capacity 2.0 Ah and mass 50 g. Cell B delivers 3.6 V with the same capacity and mass 40 g. Compare approximate stored energy per mass, ignoring voltage variation during discharge.

  1. Energy in Wh is voltage × capacity in Ah. A: 3.2 × 2.0 = 6.4 Wh; B: 3.6 × 2.0 = 7.2 Wh.
  2. Convert mass to kg: A = 0.050 kg; B = 0.040 kg.
  3. A: 6.4/0.050 = 128 Wh kg-1. B: 7.2/0.040 = 180 Wh kg-1.
Answer

B has higher energy per mass in this supplied comparison. These are illustrative data, not specifications of a current commercial battery.

07

Choose the products from the species and conditions

An external supply drives reduction at the cathode and oxidation at the anode.

Electrolysis uses electrical energy to drive a reaction. The supply delivers electrons to the negative cathode and removes them from the positive anode. Cathode still means reduction and anode still means oxidation: those definitions do not change when the electrode signs differ from a galvanic cell.

Predict an electrolytic product
  1. List the actual species

    A molten salt contains its ions; an aqueous electrolyte also contains water and its acid-base species.

  2. Check the electrode

    An inert electrode provides a surface. An active metal anode may itself oxidise.

  3. Compare plausible half-reactions

    At the cathode compare reductions; at the anode compare reversed reduction equations. Use potentials under the relevant conditions.

  4. Account for concentration and kinetics

    Competing discharge depends on ion concentration, pH and electrode surface. Standard potentials alone can be insufficient.

Representative products with their conditions
Electrolyte and electrodesCathode reductionAnode oxidation
Molten NaCl; inert electrodesNa+(l) + e- → Na(l).2Cl-(l) → Cl2(g) + 2e-.
Aqueous CuSO4; inert electrodesCu2+(aq) + 2e- → Cu(s), while sufficient Cu2+ remains.2H2O(l) → O2(g) + 4H+(aq) + 4e-.
Dilute aqueous NaCl; suitable inert electrodes2H2O(l) + 2e- → H2(g) + 2OH-(aq).Oxygen is favoured in the usual dilute-solution model; chloride oxidation can compete as conditions change.
Concentrated aqueous NaCl; suitable inert electrodesWater gives H2; sodium metal is not deposited.2Cl-(aq) → Cl2(g) + 2e- is favoured under the usual concentrated-brine conditions.
Aqueous CuSO4; copper electrodesCu2+(aq) + 2e- → Cu(s).Cu(s) → Cu2+(aq) + 2e-; the anode dissolves.

In aqueous sodium chloride, water reduction is much easier than Na+ reduction, so hydrogen forms at the cathode. At an inert anode, water/oxygen and chloride/chlorine compete. Increasing chloride concentration favours chloride oxidation; oxygen formation also has a kinetic barrier at many electrode surfaces. This is why a simple comparison of the standard +1.23 V oxygen and +1.36 V chlorine reduction values cannot, by itself, correctly predict every brine experiment.

Electrolysis can change its own conditions. As Cu2+ is depleted near an inert cathode, hydrogen evolution can become important; changes in concentration and pH may alter the products. An electrolyte conducts because its ions move, not because electrons pass through the bulk liquid.

Check your understandingWhy does changing the anode from platinum to copper alter the products in aqueous CuSO4?Think it through, then reveal the answer
Platinum supplies an inert surface, so water is oxidised and oxygen forms under the usual conditions. Copper provides an additional, more readily oxidised reactant: Cu → Cu2+ + 2e-. The anode dissolves instead. The electrolyte name alone is therefore insufficient.
08

Turn charge into product through the half-equation

One mole of electrons carries one Faraday of charge.

The Faraday constant F is the charge per mole of electrons: F = Le, where L is the Avogadro constant and e is the magnitude of the charge on one electron. The Data Booklet uses L = 6.02 × 1023 mol-1, e = 1.60 × 10-19 C and F = 9.65 × 104 C mol-1. The small difference in multiplying the printed rounded constants comes from rounding.

For a constant current, Q = It, with current in amperes, time in seconds and charge in coulombs. Then n(e-) = Q/F. Divide by the number of electrons required per product particle in the half-equation. Finally use mass = amount × molar mass, or gas volume = amount × the stated molar volume. If current varies, Q is the area under the current-time graph.

The charge-to-product calculation

Current multiplied by time gives charge. Dividing by the Faraday constant gives moles of electrons. The balanced half-equation converts electron amount into product amount, which can then be converted to mass or gas volume.

Do not convert charge directly to moles of metal without the electron ratio: Ag+, Cu2+ and Al3+ require different electron amounts.

Worked example

One current, two electrode products

Pass 2.50 A for 1930 s through aqueous CuSO4 using inert electrodes. Find copper mass and oxygen volume at a molar gas volume of 24.0 dm3 mol-1. Use F = 9.65 × 104 C mol-1, M(Cu) = 63.5 g mol-1, sufficient Cu2+, and 100% current efficiency.

  1. Q = 2.50 × 1930 = 4825 C. Hence n(e-) = 4825/96500 = 0.0500 mol.
  2. Cu2+ + 2e- → Cu, so n(Cu) = 0.0500/2 = 0.0250 mol.
  3. m(Cu) = 0.0250 × 63.5 = 1.5875 g.
  4. 2H2O → O2 + 4H+ + 4e-, so n(O2) = 0.0500/4 = 0.0125 mol.
  5. V(O2) = 0.0125 × 24.0 = 0.300 dm3.
Answer

Copper mass = 1.59 g; oxygen volume = 0.300 dm3 = 300 cm3. The same charge passes both electrodes, but the product amounts differ because their electron ratios differ.

Check your understandingThe same charge deposits 0.0100 mol of silver from Ag+ and copper from Cu2+. How many moles of copper form?Think it through, then reveal the answer
Silver needs one electron per atom, so the charge represents 0.0100 mol of electrons. Copper needs two electrons per atom, giving 0.00500 mol Cu. Equal charge means equal electron amount, not equal moles or equal masses of metal.
09

Use electrode reactions to explain the industrial purpose

Copper purification transfers metal; anodising grows an oxide layer.

Electrolytic purification of copper
PartWhat happensWhy it purifies
Positive anode: impure copperCu(s) → Cu2+(aq) + 2e-.Copper leaves the impure material as dissolved ions.
Negative cathode: pure copper sheetCu2+(aq) + 2e- → Cu(s).Copper deposits on the pure sheet.
Electrolyte: acidified copper(II) sulfateCopper ions are replaced at the anode as they are removed at the cathode.The copper-ion concentration is approximately maintained during normal operation.
ImpuritiesLess readily oxidised metals such as Ag and Au collect as anode sludge; more readily oxidised metals can dissolve.Under the chosen conditions, dissolved impurities such as Zn2+ are not preferentially deposited with copper.

The net intended process is transfer of copper from the impure anode to the pure cathode. The power supply controls this electrolytic process. Swapping the electrodes would defeat the purpose; simply filtering the original copper cannot remove metallic impurities within it.

In anodising aluminium, the aluminium object is connected as the positive anode in a suitable acidic electrolyte, such as dilute sulfuric acid. Oxidation builds a thicker protective aluminium oxide layer on its surface. An idealised oxide-forming anode equation is 2Al(s) + 3H2O(l) → Al2O3(s) + 6H+(aq) + 6e-.

At the cathode, hydrogen ions can be reduced: 6H+(aq) + 6e- → 3H2(g). The oxide layer adheres to and protects the object; anodising is not deposition of aluminium metal onto a cathode. Detailed plant design and operating parameters are outside this syllabus.

Check your understandingA learner connects the aluminium object to the negative terminal to anodise it. Explain the error.Think it through, then reveal the answer
The negative electrode is the cathode, where reduction occurs. Anodising needs oxidation of aluminium to form its oxide surface, so the object must be the positive anode. The purpose determines which electrode reaction is required.

Quick revision

Revisit the essentials, then return to an explanation when you need it.

The relationships and their conditions
TaskUseEssential check
Predict a standard cellcell = E°cathode - E°anode.Both values are reduction potentials; positive means favourable for the written reaction under standard conditions.
Relate voltage to energyΔG° = -nFE°cell.n belongs to the balanced reaction; volts with coulombs give joules.
Combine half-reactionsAdd their Gibbs-energy contributions, then convert back to E°.Potentials are not directly additive or multiplied by stoichiometric coefficients.
Calculate electrolysis yieldQ = It; n(e-) = Q/F; use the half-equation.Time in seconds, correct electron ratio, current efficiency and gas conditions.
Explain an industrial cellWrite both electrode reactions first.Purification: impure Cu anode/pure Cu cathode. Anodising: aluminium object is the anode.

Always: oxidation at the anode, reduction at the cathode. Galvanic discharge: anode negative, cathode positive. Electrolysis: anode positive, cathode negative. Standard-potential tables predict thermodynamic direction; composition and kinetic barriers explain why a real observation can differ.

Scope and references

Learning outcomes and sources

12. Electrochemistry. Use the outcome map to find the explanation for a particular syllabus requirement.

See the learning outcome map
  1. 12(a) Explain redox using electrons and oxidation numbers.

    • Electron loss/gain
    • Increase/decrease of oxidation number
    • Oxidising and reducing agent roles

    Track electrons before calculating a voltage

  2. 12(b) Define standard electrode and standard cell potentials.

    • (i) Standard reduction potential relative to SHE
    • (ii) Standard cell electromotive force
    • Standard states, negligible-current measurement and stated temperature

    Measure every half-cell against the same referenceThe spontaneous cell sends electrons from anode to cathode

  3. 12(c) Describe the standard hydrogen electrode.

    • Platinum with catalytic platinum surface
    • Hydrogen at standard pressure
    • Hydrogen-ion standard activity/concentration convention
    • Reference potential and reversible half-equation

    Measure every half-cell against the same reference

  4. 12(d) Describe standard electrode-potential measurements.

    • (i) Metal/aqueous-ion and non-metal/aqueous-ion arrangements
    • (ii) Same-element ions in different oxidation states
    • SHE, salt bridge, high-resistance voltmeter, sign and standard conditions

    Measure every half-cell against the same reference

  5. 12(e) Calculate standard cell potentials.

    • Cathode reduction potential minus anode reduction potential
    • Do not scale potentials with half-equation coefficients

    The spontaneous cell sends electrons from anode to cathode

  6. 12(f) Use cell potentials to predict electron flow and spontaneity.

    • (i) Simple-cell electron direction
    • (ii) Thermodynamic spontaneity for the reaction as written
    • Electrode roles and signs in a galvanic cell

    The spontaneous cell sends electrons from anode to cathode

  7. 12(g) Recognise limits of standard-potential predictions.

    • Non-standard composition, temperature, pressure and pH
    • Kinetic barriers and passivation
    • Complexation and precipitation affecting free ions

    A standard prediction is not a guarantee of a rapid real reactionChoose the products from the species and conditions

  8. 12(h) Construct overall redox equations from half-equations.

    • Atom and charge balances
    • Acidic and alkaline balancing method
    • Electron cancellation
    • Cross-reference Topic 13 redox systems

    Track electrons before calculating a voltage

  9. 12(i) Apply the Gibbs energy and potential relationship.

    • DeltaG standard = -nFE standard
    • Balanced-reaction electron count and units
    • Combining half-reactions through additive Gibbs energies
    • Electron-weighted combined potential

    Add Gibbs energies when combining half-reactions

  10. 12(j) Predict concentration effects on electrode potentials.

    • Metal-ion concentration
    • Oxidised/reduced ion ratio
    • Hydrogen-ion concentration
    • Qualitative treatment at fixed other conditions

    A standard prediction is not a guarantee of a rapid real reaction

  11. 12(k) Discuss possible advantages of improved cells.

    • Hydrogen/oxygen fuel cell
    • Improved batteries including electric vehicles
    • Smaller size, lower mass and higher voltage
    • Whole-system and capacity comparison

    Compare useful energy, size and mass

  12. 12(l) Relate Faraday, Avogadro and electron-charge constants.

    • F = Le
    • Charge per mole of electrons
    • Units and rounded Data Booklet values

    Turn charge into product through the half-equation

  13. 12(m) Predict electrolytic products.

    • Molten versus aqueous electrolyte
    • Redox-series/electrode-potential reasoning
    • Ion concentration
    • Competing water reactions and electrode material

    Choose the products from the species and conditions

  14. 12(n) Calculate charge and electrolysis yield.

    • (i) Quantity of charge passed
    • (ii) Mass and gas volume liberated
    • Half-equation electron ratio
    • Current efficiency and gas conditions

    Turn charge into product through the half-equation

  15. 12(o) Explain industrial electrolysis through electrode reactions.

    • (i) Aluminium anodising
    • (ii) Electrolytic copper purification
    • Electrode identities, products and purpose
    • Technical plant details not required

    Use electrode reactions to explain the industrial purpose

  • SEAB H2 Chemistry 9476, examination 2026

    Topic 12, printed pages 33-34. All 15 lettered groups and nested measurement, Gibbs-energy, cell-development and industrial requirements inspected.

  • SEAB H2 Chemistry 9476, examination 2027

    Topic 12, printed pages 33-34. Full scope compared with 2026 and found to agree; quantitative Nernst treatment and technical industrial details are not added as required outcomes.

  • SEAB Chemistry Data Booklet, for use from 2026

    Printed page 3 for constants and gas conditions; pages 8-12 for standard reduction potentials at 298 K. Current link verified from the SEAB syllabus page. Worked calculations use the supplied rounded values.

  • IUPAC Green Book, fourth edition abridged

    Printed page 53, section 4.11.1(v), recommends 100 kPa standard pressure; printed page 62, standard electrode potential, applies 10^5 Pa to the hydrogen-electrode reference. The SEAB Data Booklet separately specifies its electrode tables at 298 K.