Q: What does H2 Chemistry Electrochemistry cover in SEAB 9476? A: Standard electrode potentials, Ecell calculations, galvanic vs electrolytic cells, selective discharge of ions, Faraday's laws, and real-world applications including batteries, fuel cells, and corrosion.
TL;DR Electrochemistry turns redox reactions into cell voltage or electrolysis products. Most marks come from half-equations, electrode roles, Ecell∘ signs, and electron-to-mole calculations.
Redox is electron transfer: Mark oxidation and reduction.
Galvanic cells produce voltage; electrolytic cells use voltage: Label anode, cathode, and electron flow.
The calculation follows from roles, signs, ions, and booklet values: Write half-equations before using E∘
Concrete example: In a galvanic cell, the half-cell with the more positive reduction potential is the cathode. The other half-cell runs oxidation. Do not multiply E∘ values by stoichiometric coefficients.
Electrochemistry connects redox chemistry to real-world energy devices and industrial processes. It is one of the more calculation-heavy sections of H2 Chemistry and regularly appears in Papers 2 and 3 as structured questions involving cell potential computation, electrolysis reasoning, or data-booklet lookups.
Status: SEAB's current H2 Chemistry (9476) syllabus PDF is labelled for 2026, and the current Chemistry Data Booklet is labelled 8873/9476/9813 for use from 2026 in all non-practical Chemistry papers. [1][2]
Quick revision box
What this topic tests: Standard electrode potentials, cell potential calculations, galvanic vs electrolytic cell identification, electrolysis product prediction, selective discharge, Faraday's law computations, and applications.
Top mistakes to avoid: Sign errors in Ecell∘; confusing anode polarity between galvanic and electrolytic cells; ignoring concentration effects on electrolysis products; multiplying E∘ by stoichiometric coefficients (never do this).
20-minute sprint plan: 5 min redox sign conventions and electrochemical series; 8 min cell potential and electrolysis product reasoning; 7 min Faraday's law drill.
Before calculating or predicting products, decide which cell type the question describes. The same word can carry different polarity depending on whether the cell is producing current or being driven by an external supply.
Question clue
Cell type
Anode role
Cathode role
First action
A spontaneous redox reaction produces a voltage.
Galvanic cell
Oxidation, negative electrode
Reduction, positive electrode
Pick the more positive reduction potential as the cathode.
An external power supply drives a non-spontaneous reaction.
Electrolytic cell
Oxidation, positive electrode
Reduction, negative electrode
List ions present before predicting discharge.
A question asks for Ecell∘.
Standard cell potential
Lower E∘ half-cell is oxidised
Higher E∘ half-cell is reduced
Use Ecathode∘−Eanode∘ without multiplying values by coefficients.
A question asks for mass deposited or gas volume.
Electrolysis calculation
Supplies electrons to the circuit
Receives electrons from the circuit
Convert charge to moles of electrons before using the half-equation ratio.
Common trap: anode always means oxidation and cathode always means reduction, but electrode signs change between galvanic and electrolytic cells.
1 Oxidation States and Redox Foundations
Before tackling electrochemistry, ensure you can assign oxidation states reliably and identify which species is oxidised or reduced. Electrochemistry builds directly on the redox concepts from earlier topics.
Key principles to recall:
Oxidation is loss of electrons; reduction is gain of electrons (OIL RIG).
An oxidising agent is itself reduced; a reducing agent is itself oxidised.
Balance half-equations for both atoms and charge before combining.
Balancing half-equations - step-by-step:
Write the unbalanced species.
Balance atoms other than O and H.
Balance oxygen by adding HX2O.
Balance hydrogen by adding HX+.
Balance charge by adding electrons.
Example: Balance the half-equation for MnOX4X− being reduced to MnX2+ in acidic solution.
MnOX4X−+8HX++5eX−MnX2++4HX2O
2 Standard Electrode Potentials
2.1 The standard hydrogen electrode (SHE)
All electrode potentials are measured relative to the SHE, which is assigned E∘=0.00V by convention. The SHE consists of a platinum electrode in contact with HX2 gas at 1bar and HX+ ions at 1.00moldm−3, at 298K.
The defining half-equation:
2HX+(aq)+2eX−HX2(g)E∘=0.00V
2.2 Interpreting the electrochemical series
Half-equations in the Data Booklet are written as reductions. A more positive E∘ means a stronger tendency to be reduced, so the species on the left is a stronger oxidising agent. A more negative E∘ means a stronger tendency for the reverse reaction (oxidation), so the reduced form is a stronger reducing agent.
The table below shows selected values from the SEAB Chemistry Data Booklet. Always use Data Booklet values in exams.
Reading the table: Species near the top of the list (most positive E∘) are the strongest oxidising agents. Species near the bottom (most negative E∘) have the strongest reducing forms.
2.3 Standard conditions
Standard electrode potentials apply at 298K, with all dissolved species at 1.00moldm−3 and all gases at 1bar. Deviations from these conditions shift the measured potential. The SEAB syllabus expects you to predict qualitatively how concentration changes affect electrode potential.
2.4 Predicting whether a reaction is feasible
A reaction is thermodynamically feasible under standard conditions if Ecell∘>0. The species with the higher (more positive) E∘ acts as the oxidising agent (cathode half-reaction); the species with the lower E∘ acts as the reducing agent (anode half-reaction).
Worked example: Will FeX3+ oxidise IX− to IX2?
FeX3++eX−FeX2+, E∘=+0.77 V
IX2+2eX−2IX−
Ecell∘=Ecathode∘−Eanode∘=+0.77−(+0.54)=+0.23 V
Since Ecell∘>0, the reaction is feasible. FeX3+ can oxidise IX− to IX2.
3 Electrochemical (Galvanic) Cells
3.1 How they work
A galvanic cell converts chemical energy into electrical energy through a spontaneous redox reaction. Two half-cells are connected by a salt bridge that maintains electrical neutrality and an external wire through which electrons flow.
Anode: Oxidation occurs here. In a galvanic cell, the anode carries a negative polarity.
Cathode: Reduction occurs here. In a galvanic cell, the cathode carries a positive polarity.
Salt bridge: Allows migration of ions to complete the circuit and prevent charge build-up. Typically contains an inert electrolyte such as KNOX3(aq) or KCl(aq).
Salt bridge charge-balance checkpoint
When explaining a galvanic cell, separate electron flow in the wire from ion flow in the salt bridge. The salt bridge does not carry electrons; it keeps each half-cell electrically neutral as redox occurs.
Cell region
What changes during operation
Charge problem without salt bridge
What the salt bridge supplies
Anode half-cell
Metal atoms are oxidised, forming more cations in solution.
Positive charge builds up and opposes further oxidation.
Anions from the salt bridge migrate into the anode half-cell.
Cathode half-cell
Cations are reduced and removed from solution.
The solution becomes relatively negative as cations are consumed.
Cations from the salt bridge migrate into the cathode half-cell.
External wire
Electrons flow from anode to cathode.
The circuit would not stay complete without matching ion movement in solution.
No electrons enter the salt bridge; ions complete the internal circuit.
Worked check: in a zinc-copper cell, Zn(s)ZnX2+(aq)+2eX− occurs at the zinc anode, so ZnX2+ builds up. Anions from the salt bridge move into the zinc half-cell. At the copper cathode, CuX2+ is reduced to Cu(s), so cations from the salt bridge move into the copper half-cell.
Misconception check: electrons move through the external wire. Ions move through the salt bridge and solutions. Do not say electrons travel through the salt bridge.
3.2 Cell notation
The conventional way to represent a galvanic cell:
Zn(s)∣ZnX2+(aq)∣∣CuX2+(aq)∣Cu(s)
Read left to right: anode | anode solution || cathode solution | cathode. The double vertical line represents the salt bridge. Electrons flow from left (anode, oxidation) to right (cathode, reduction) through the external circuit.
3.3 Calculating cell potential
Ecell∘=Ecathode∘−Eanode∘
For the Daniell cell:
Ecell∘=+0.34V−(−0.76V)=+1.10V
A positive Ecell∘ indicates the reaction is thermodynamically feasible under standard conditions.
Important:E∘ values are intensive properties. Do not multiply them by stoichiometric coefficients when calculating cell potentials. The E∘ of the CuX2+/Cu half-cell remains +0.34 V regardless of whether the half-equation is written for 1 or 2 moles of Cu.
3.4 Worked example: Zinc-silver cell
Question: A galvanic cell uses ZnX2+/Zn and AgX+/Ag half-cells. Write the conventional cell notation, the overall cell reaction, and calculate Ecell∘.
Given: E∘(AgX+/Ag)=+0.80 V; E∘(ZnX2+/Zn)=−0.76 V.
Solution:
Identify anode (lower E∘) and cathode (higher E∘):
Anode: Zn(s)ZnX2+(aq)+2eX−
Cathode: AgX+(aq)+eX−Ag(s)
Cell notation:
Zn(s)∣ZnX2+(aq)∣∣AgX+(aq)∣Ag(s)
Overall reaction (multiply cathode by 2):
Zn(s)+2AgX+(aq)ZnX2+(aq)+2Ag(s)
Cell potential:
Ecell∘=+0.80−(−0.76)=+1.56 V
Note: the E∘ for AgX+/Ag is not multiplied by 2 even though the half-equation was multiplied by 2.
3.5 Worked example: Iron(III)/iron(II) and iodine/iodide cell
Question: Consider the cell Pt(s)∣IX2(s)∣IX−(aq)∣∣FeX3+(aq),FeX2+(aq)∣Pt(s). Identify the anode and cathode, write the half-equations, and calculate Ecell∘.
Given: E∘(FeX3+/FeX2+)=+0.77 V; E∘(IX2/IX−)=+0.54 V.
Solution:
By convention, left = anode, right = cathode.
Anode (oxidation): 2IX−(aq)IX2(s)+2eX−
Cathode (reduction): 2FeX3+(aq)+2eX−2FeX2+(aq)
Overall: 2FeX3+(aq)+2IX−(aq)2FeX2+(aq)+IX2(s)
Ecell∘=+0.77−(+0.54)=+0.23 V
The reaction is spontaneous under standard conditions. Note that a platinum inert electrode is used when neither species is a solid metal.
4 Electrolytic Cells
4.1 How they differ from galvanic cells
An electrolytic cell uses an external power source to drive a non-spontaneous redox reaction. The polarity of the electrodes is reversed compared to a galvanic cell.
Feature
Galvanic cell
Electrolytic cell
Energy conversion
Chemical to electrical
Electrical to chemical
Spontaneity
Spontaneous (Ecell>0)
Non-spontaneous; external voltage required
Anode polarity
Negative
Positive
Cathode polarity
Positive
Negative
Common use
Batteries, fuel cells
Electroplating, metal refining, industrial synthesis
The one constant: oxidation always occurs at the anode, and reduction always occurs at the cathode.
4.2 Selective discharge of ions
Three factors determine which species are preferentially discharged at each electrode:
Position in the electrochemical series: Species with more positive E∘ values are preferentially reduced at the cathode. At the anode, species with more negative E∘ for the reduction half-reaction are more easily oxidised.
Concentration of ions: A high concentration of a particular ion can shift the product. For example, concentrated NaCl(aq) favours ClX2 at the anode rather than OX2 from water.
Nature of electrodes: Inert electrodes (graphite, platinum) do not participate in electrode reactions. Reactive electrodes (e.g. copper) may dissolve at the anode, altering the products entirely.
Key default products from aqueous electrolysis with inert electrodes:
At cathode (reduction)
Condition
HX2 and OHX−
Dilute solutions of metal salts where E∘(metal)<E∘(HX+/HX2)
Metal deposited
Higher E∘ metals: CuX2+, AgX+
At anode (oxidation)
Condition
OX2
Dilute solutions or solutions without high halide concentration; water is oxidised
ClX2
Concentrated ClX− solutions
BrX2
Concentrated BrX−
4.3 Worked example: Electrolysis of dilute sulfuric acid
With inert electrodes:
Cathode:2HX+(aq)+2eX−HX2(g) - hydrogen ions are more easily reduced than water.
Anode:2HX2O(l)OX2(g)+4HX+(aq)+4eX−
Volume ratio of HX2:OX2 collected is 2:1.
4.4 Worked example: Electrolysis of concentrated NaCl(aq)
Cathode:2HX2O(l)+2eX−HX2(g)+2OHX−(aq) - water is reduced because NaX+ has a very negative E∘ (−2.71 V).
Anode:2ClX−(aq)ClX2(g)+2eX−
The solution becomes alkaline near the cathode as OHX− accumulates. This is the basis of the chlor-alkali industry, which produces NaOH, ClX2, and HX2 from brine.
4.5 Worked example: Electrolysis of molten NaBr
Molten salts contain only the ions of the salt - no water is present.
Cathode:NaX+(l)+eX−Na(l) - the only cation present is NaX+.
Anode:2BrX−(l)BrX2(g)+2eX−
Unlike aqueous solutions, there is no competition from water.
4.6 Worked example: Electrolysis of CuSO4(aq) with copper electrodes
When the anode is made of copper (reactive):
Cathode:CuX2+(aq)+2eX−Cu(s) - copper is deposited.
Anode:Cu(s)CuX2+(aq)+2eX−
The concentration of CuX2+ in solution remains approximately constant, and the mass of the anode decreases as mass of the cathode increases. This is the basis of copper refining (electrorefining), which purifies impure copper to over 99.99%.
5 Faraday's Laws
5.1 Core relationships
The amount of substance deposited or liberated at an electrode is proportional to the charge passed:
Q=I×t
where Q is charge in coulombs (C), I is current in amperes (A), and t is time in seconds (s).
The number of moles of electrons transferred:
ne=FQ
where the Faraday constant F=96485Cmol−1 (often approximated as 96500Cmol−1 in calculations).
5.2 Calculation procedure
Calculate the total charge: Q=It (ensure t is in seconds).
Find moles of electrons: ne=Q/F.
Use the stoichiometry of the half-equation to find moles of substance produced.
Convert to mass, volume at RTP/STP, or other required unit.
Faraday ratio checkpoint
After finding moles of electrons, pause before calculating mass or gas volume. The half-equation tells you how many moles of electrons are needed for one mole of product.
Product wanted
Half-equation ratio
Conversion from ne
Common trap
Cu(s) from CuX2+
CuX2++2eX−Cu
nCu=ne/2
Using ne directly as moles of copper.
Al(s) from AlX3+
AlX3++3eX−Al
HX2(g) from aqueous acid or water
2HX++2eX−HX2
OX2(g) from water at the anode
2HX2OOX2+4HX++4eX−
Worked check: if 0.0400mol of electrons pass through CuSOX4(aq), the maximum copper deposited is 0.0400/2=0.0200mol. Multiply this by the molar mass only after the electron-to-product ratio is settled.
Misconception check: charge gives moles of electrons, not moles of product. The electrode half-equation is the bridge between the two.
5.3 Worked example: Copper deposition
Question: A current of 2.50A is passed through CuSOX4(aq) for 30.0min using inert electrodes. Calculate the mass of copper deposited.
Solution:
Q=2.50×(30.0×60)=4500C
ne=4500/96485=0.04664mol
From CuX2++2eX−Cu
Mass =0.02332×63.5=1.48g
5.4 Worked example: Aluminium production (Hall-Heroult process)
Question: In the Hall-Heroult process, aluminium is produced by electrolysis of molten AlX2OX3. A current of 1.50×104A is applied for 2.00h. Calculate the mass of aluminium produced.
Given: Mr(Al)=27.0gmol−1.
Solution:
t=2.00×3600=7200s
Q=1.50×104×7200=1.08×108C
ne=(1.08×108)/96485=1.119×103mol
From AlX3++3eX−Al
Mass =373×27.0=1.01×104g (approximately 10.1 kg)
5.5 Worked example: Volume of gas evolved
Question:HX2SOX4(aq) is electrolysed with inert electrodes. A current of 0.800A is applied for 45.0min. Calculate the volume of oxygen gas produced at the anode at RTP (1bar, 298K).
The SEAB syllabus requires you to predict qualitatively how changes in ion concentration shift electrode potentials and cell potentials. When the concentration of the oxidised form increases (or the reduced form decreases), the reduction potential becomes more positive, favouring reduction.
Qualitative reasoning examples:
Increasing [CuX2+] at the cathode makes the cathode potential more positive, increasing Ecell.
Increasing [ZnX2+] at the anode makes the anode potential less negative (more positive), which decreases Ecell.
Increasing [HX+] for a half-reaction involving HX+ makes that electrode potential more positive.
Enrichment - the Nernst equation:
For quantitative treatment (beyond syllabus requirements, but useful for deeper understanding):
E=E∘−nFRTlnQ
At 298K, this simplifies to:
E=E∘−n0.0592log10Q
This equation is not required for exam calculations, but understanding the direction of the shift is examinable. [1]
7 Linking Electrochemistry to Other Topics
Chemical Energetics (Topic 7):ΔG∘=−nFEcell∘ connects cell potential to Gibbs free energy. A positive Ecell∘ corresponds to a negative ΔG∘, confirming spontaneity. At 298K, ΔG∘=−nFEcell∘ in joules (use F=96485Cmol−1).
Chemical Equilibria (Topic 9): At equilibrium, Ecell=0 and Q=K. This relationship links the equilibrium constant to the standard cell potential via ΔG∘=−RTlnK=−nFEcell∘
Transition Elements (Topic 13): Many electrochemical series entries involve transition metal ions; variable oxidation states are central to electrochemistry.
8 Applications of Electrochemistry
8.1 Primary and secondary cells (batteries)
Primary cells are non-rechargeable. The spontaneous redox reaction runs to completion and the cell cannot be regenerated.
Secondary cells (rechargeable batteries) can be recharged by reversing the electrode reactions using an external power source.
Lithium-ion battery: Used in mobile devices. LiX+ ions migrate through the electrolyte between electrodes. High energy density; no memory effect.
8.2 Fuel cells
A fuel cell is an electrochemical device that converts chemical energy from a fuel (such as hydrogen) directly into electrical energy without combustion. Unlike a battery, it does not run down as long as fuel is supplied.
The only product is water. Fuel cells are more efficient than combustion engines because they bypass the thermodynamic limitations of heat engines (Carnot efficiency).
Advantages over conventional combustion:
Higher energy conversion efficiency.
No COX2 or NOx emissions (when hydrogen is the fuel).
Operates continuously as long as fuel is supplied.
8.3 Corrosion
Corrosion is the electrochemical oxidation of metals in the presence of moisture and oxygen (and often electrolytes).
Iron rusting as an electrochemical process:
Anodic region (oxidation):Fe(s)FeX2+(aq)+2eX−
Cathodic region (reduction):OX2(g)+2HX2O(l)+4eX−4OHX−(aq)
FeX2+ is subsequently oxidised to FeX3+ and forms hydrated iron(III) oxide (rust): FeX2OX3⋅xHX2O
The driving force is the positive Ecell∘ for the overall reaction. Salt water accelerates corrosion by increasing the conductivity of the electrolyte.
Methods to prevent corrosion:
Sacrificial protection: Attach a more reactive metal (e.g. zinc or magnesium) to the iron. The reactive metal oxidises preferentially at the anode (E∘ is more negative), protecting the iron. Used in galvanised iron and marine structures.
Cathodic protection: Apply an external current to make the iron structure the cathode (negative electrode), preventing it from oxidising.
Protective coating: Paint, grease, or electroplating forms a barrier between the metal and the environment.
9 Common Exam Pitfalls
Sign errors: Always apply Ecell∘=Ecathode∘−Eanode∘. Never change the sign of E∘ values from the Data Booklet when using this formula.
Multiplying E∘ by coefficients: Electrode potentials are not multiplied by stoichiometric coefficients. They are intensive properties.
Confusing anode polarity: The anode is negative in a galvanic cell but positive in an electrolytic cell. Oxidation occurs at the anode in both.
Predicting Na metal from aqueous electrolysis:NaX+ has a very negative E∘; water is always reduced first in aqueous solution.
Ignoring electrode material: Reactive electrodes (e.g. copper anode in CuSOX4 electrolysis) dissolve, altering the products.
Forgetting units in Faraday calculations: Keep t in seconds, I in amperes, and state the Faraday constant with units.
Confusing dilute vs concentrated NaCl at anode: Dilute NaCl(aq) gives OX2; concentrated NaCl(aq)
Forgetting to write the overall equation: In structured questions asking for both half-equations and an overall equation, failing to combine them loses marks. Check that electrons cancel.
10 Quick Retrieval Check
What is the standard hydrogen electrode, and why is it assigned E∘=0.00V?
In the cell Mg(s)∣MgX2+(aq)∣∣FeX2+(aq)∣Fe(s), which electrode is the cathode? Calculate Ecell∘. (E∘(MgX2+/Mg)=−2.37 V; E∘(FeX2+/Fe)=−0.44 V)
Explain why concentrated NaCl(aq) produces ClX2 at the anode, but dilute NaCl(aq)
A current of 1.50A flows for 20.0min through molten NaCl. Calculate the mass of sodium deposited. (Mr(Na)=23.0
Will BrX2 oxidise FeX2+ to FeX3+
State two methods to prevent corrosion of iron and explain the electrochemical principle behind each.
Struggling with Electrochemistry? Our H2 Chemistry tuition programme covers this topic with structured practice, Paper 4 practical drills, and worked exam solutions.
FAQ
How do I calculate Ecell? Use Ecell∘=Ecathode∘−Eanode∘. The cathode is the electrode where reduction occurs (higher E∘). The anode is where oxidation occurs (lower E∘). Read both values directly from the SEAB Data Booklet as reduction potentials - do not flip the sign of the anode value before subtracting.
What is selective discharge in electrolysis? Selective discharge refers to the preferential oxidation or reduction of one ion over another when a mixture of ions is present. At the cathode, the ion with the more positive (or least negative) E∘ is preferentially reduced. At the anode, the ion that is most easily oxidised (corresponding to the lowest E∘ for the reduction half-reaction) is preferentially oxidised. High concentration of a particular ion can override the order predicted by E∘ alone - this explains why concentrated ClX− gives ClX2 at the anode despite OX2/HX2O having a lower electrode potential.
What is the difference between an electrolytic cell and a galvanic cell? A galvanic (voltaic) cell converts chemical energy into electrical energy through a spontaneous redox reaction (Ecell∘>0). An electrolytic cell uses an external power source to drive a non-spontaneous reaction (Ecell∘<0 without external voltage). In a galvanic cell the anode is negative; in an electrolytic cell the anode is positive (connected to the positive terminal of the power supply). In both cases, oxidation occurs at the anode and reduction occurs at the cathode.
How do I use Faraday's laws to calculate mass deposited? Follow four steps: (1) calculate total charge Q=It in coulombs (t must be in seconds); (2) calculate moles of electrons ne=Q/F using F=96485Cmol−1; (3) use the stoichiometry of the electrode half-equation to find moles of the product; (4) multiply by molar mass to get mass. For gas products, multiply moles by the molar volume at the stated conditions.
Is the Nernst equation examinable for SEAB 9476? The SEAB 9476 syllabus expects qualitative predictions of how concentration changes affect electrode potential. The Nernst equation itself is enrichment material - it is not required for exam calculations but understanding the direction of the shift (more concentrated oxidised form - more positive potential) is examinable. [1]
How does electrochemistry link to the practical paper? Paper 4 may include electrochemical cell setups, voltmeter readings, or electrolysis measurements. Familiarity with cell diagrams, salt bridge construction, inert vs reactive electrode choice, and data recording from potentiometric setups is helpful. See the Electrochemistry Instrumental Guide for Paper 4 practical coverage.
How does sacrificial protection work electrochemically? Attaching a more reactive metal (more negative E∘) to iron creates a galvanic cell where the reactive metal acts as the anode and preferentially oxidises, protecting the iron (cathode). The iron is protected as long as the sacrificial metal is present and electrically connected. Common examples include zinc blocks on ships and galvanised iron.
Where can I find the full H2 Chemistry Notes series? Start at the H2 Chemistry Notes hub, then follow the topic sequence from Atomic Structure through to Electrochemistry.
