How does changing an ion concentration affect electrode potential in JC2 H2 Chemistry? Students often know the standard cell formula yet fail an unfamiliar question because they do not identify which species is oxidised and which is reduced. The chemical half-equation and the relative concentrations matter. A standard electrode potential belongs to defined conditions, not every mixture a student might prepare.
For H2 Chemistry syllabus 9476, the concentration treatment is qualitative. Students should predict how electrode potential varies with an aqueous ion’s concentration, understand standard cell potentials and connect E° to ΔG° through −nFE°. A correct explanation separates thermodynamic tendency, actual concentration and reaction rate rather than treating them as one quantity.
This guide offers original worked calculations and reasoning drills for Hougang JC2 families. eduKateSG tutorials are held near Sixth Avenue MRT at 8 Fourth Avenue, not in Hougang; parents should confirm current suitability, fees and travel arrangements.
Begin with the half-reaction
A standard electrode potential belongs to a specified reduction half-reaction, not merely to a metal’s name. For Cu²⁺(aq)+2e⁻⇌Cu(s), Cu²⁺ accepts electrons when reduction occurs. For Fe³⁺(aq)+e⁻⇌Fe²⁺(aq), Fe³⁺ is the oxidised form and Fe²⁺ is the reduced form. Writing these roles first is essential when students predict what a change in concentration will do.
Many wrong answers begin with an unexamined slogan: ‘more ions means a bigger potential’. That can be misleading when the additional ion is the reduced form rather than the oxidised form. The correct reasoning identifies the species being produced and consumed by the written reduction reaction, with other relevant conditions held fixed.
What the 2027 syllabus expects
The 2027 SEAB H2 Chemistry 9476 syllabus includes standard electrode potentials, standard cell potentials, redox feasibility, batteries, electrolysis and the relationship ΔG°=−nFE°. It also requires students to predict qualitatively how an electrode potential varies with the concentration of an aqueous ion. This stated learning outcome is qualitative, rather than a requirement to derive and apply the Nernst equation numerically.
A useful JC2 revision task therefore combines a balanced half-reaction, a prediction about concentration, and a correct interpretation of a standard cell calculation. Extra university-level equations can enrich the topic later, but they should not crowd out the precise H2 examination target. The student must learn what changes physically, not merely display a formula.
Higher Cu²⁺ concentration in the copper half-cell
For Cu²⁺(aq)+2e⁻⇌Cu(s), increasing the concentration of Cu²⁺ relative to the unchanged solid copper tends to increase the reduction potential under the simple model. A higher concentration of the oxidised ionic species favours the reduction direction relative to the reverse process. Reducing Cu²⁺ concentration tends to shift the reduction potential in the less positive direction.
The copper metal is a pure solid; adding extra solid metal is not equivalent to changing the aqueous Cu²⁺ activity. Students should distinguish the composition of solution from the physical area or mass of the electrode. Electrode area can affect kinetic behaviour or attainable current, but that is not the same quantity as the equilibrium reduction potential.
The Fe³⁺ and Fe²⁺ contrast
In Fe³⁺(aq)+e⁻⇌Fe²⁺(aq), both forms are dissolved ions. Increasing the ratio of Fe³⁺ to Fe²⁺, with other conditions comparable, tends to make the reduction potential more positive. Increasing the relative amount of Fe²⁺ instead tends to make the reduction potential less positive. The role of an ion in the half-reaction matters more than the mere fact that its concentration has increased.
This is an excellent diagnostic for students who have memorised a rule from the copper example. Ask which iron ion is oxidised and which is reduced, then reverse the concentration change and predict the new direction. If the learner repeats the same answer in both cases without explanation, the conceptual distinction needs rebuilding.
Standard conditions and actual conditions
Standard potentials are tabulated under defined standard-state assumptions and compared against a reference electrode. In real cells, aqueous-ion concentrations and other conditions may differ from those assumptions. A standard voltage is therefore not a guarantee of the exact voltage measured for every gas pressure, solution composition or temperature.
The correct exam language distinguishes E°, which is standard, from the potential under the actual stated conditions. If a question changes concentration, explain the qualitative potential change before assuming any exact value. This protects students from using a table entry as if it were a fixed physical property that never responds to the chemical environment.
The zinc–copper standard cell
For the standard zinc–copper galvanic cell, E°(Cu²⁺/Cu)=+0.34 V and E°(Zn²⁺/Zn)=−0.76 V. The cathode reduction is Cu²⁺+2e⁻→Cu, and zinc is oxidised at the anode. Calculate E°cell=E°cathode(reduction)−E°anode(reduction)=+0.34−(−0.76)=+1.10 V. The corresponding overall reaction is Zn(s)+Cu²⁺(aq)→Zn²⁺(aq)+Cu(s).
Do not multiply a standard potential by the number of electrons when balancing the overall reaction. Voltage is not a molar enthalpy. The electron count becomes relevant when calculating quantities of charge or Gibbs free energy, but the potential subtraction uses the listed reduction values and a consistent cathode-minus-anode rule.
Changes in Zn²⁺ and Cu²⁺ do not act alike
In the zinc–copper reaction, Cu²⁺ is consumed while Zn²⁺ is produced. Increasing the relative amount of Cu²⁺ under appropriate otherwise comparable conditions can favour the forward redox direction, whereas increasing Zn²⁺ relative to Cu²⁺ tends to diminish that forward driving force. These qualitative concentration effects cannot be understood by saying that all metal ions increase the voltage.
Have students write the overall balanced equation and put an arrow beside each ion indicating production or consumption. This makes the pressure of reactants and products on the reaction direction visible. The same method works for many unfamiliar aqueous redox systems where the names of the metals change.
A concentration-cell example
Imagine two copper electrodes in solutions with unequal Cu²⁺ concentration, joined by an external conducting path and a suitable ionic connection. The more concentrated copper-ion half-cell has the more positive reduction potential in the simple model and can act as the cathode. The dilute half-cell can act as an anode, with copper dissolving to form Cu²⁺.
The resulting chemical change tends to reduce the initial concentration difference. The driving force arises even though both electrodes contain copper. This is a useful conceptual demonstration that electrode potential depends on solution composition, not only the identity of the metal. The example is for reasoning; actual electrochemical experiments require suitable laboratory supervision.
Salt bridge and circuit reasoning
A salt bridge or suitable ionic connection helps maintain electrical neutrality while electrons travel through the external circuit of a galvanic cell. Ions move through the solutions or bridge; electrons do not simply travel through the salt bridge as though it were a metal wire. Failure to distinguish these charge carriers can cause confusion about why a cell stops supplying current.
This is separate from the calculation of the standard cell potential but belongs to the same scientific picture. A thermodynamic driving force alone is insufficient to maintain a useful current without a functioning ionic and electronic circuit. Students should learn the role of each part of the cell rather than memorise a drawing with unlabeled arrows.
Positive E°cell and ΔG°
The H2 syllabus states ΔG°=−nFE°. In the standard zinc–copper example, n=2 moles of electrons per mole of the balanced overall reaction. Taking F≈9.65×10⁴ C mol⁻¹ and E°=+1.10 V, ΔG°=−2×9.65×10⁴×1.10≈−2.12×10⁵ J mol⁻¹, or about −212 kJ mol⁻¹. The negative free-energy change corresponds to thermodynamic favourability under standard conditions.
The negative sign must not be forgotten. Another frequent mistake is inserting n into the voltage subtraction instead of only the energy relation. Once the cell equation is balanced and the correct E° established, the number of electrons determines the charge equivalent of one mole of the specified reaction. Show that distinction explicitly in written working.
Why positive E° does not mean fast
A positive standard cell potential addresses thermodynamic favourability, not the rate of electron transfer. Actual observed cell current can depend on resistance, electrode surface, concentration, reaction kinetics and other conditions. A student who writes that the redox reaction ‘happens quickly because the potential is positive’ is making a claim not supported by the potential alone.
Practise paired questions: one about whether the reaction direction is thermodynamically feasible under standard conditions, another about why an experimental cell might deliver little current. Their answers should involve different physical reasoning. This habit prevents electrochemistry from becoming a single memorised subtraction formula.
Do not confuse galvanic cells with electrolysis
Oxidation occurs at the anode and reduction at the cathode in both galvanic and electrolytic contexts. The polarity of those electrodes relative to the external circuit can differ between a spontaneous galvanic cell and a driven electrolytic setup. Students who simply memorise ‘anode is positive’ without the cell context can draw incorrect arrows or assign half-reactions wrongly.
Begin instead with electron transfer. Identify the species losing electrons, the species gaining them, and which process is oxidation or reduction. Then consider the cell type and the external energy source. This chemical foundation works more reliably than isolated signs, especially when the question includes an unfamiliar diagram.
Why electrode concentration and current are different
Changing aqueous-ion concentration can alter an electrode’s thermodynamic potential and can also affect current under particular experimental conditions, but those are distinct statements. Potential concerns the energetic tendency of a half-reaction; current is the rate of electrical charge flow. Neither can be predicted completely from a single concentration value without further context.
Students should avoid overextending a correct potential prediction into an unsupported numerical statement about current. If a paper asks only which way potential shifts, answer that question and give the relevant half-reaction reasoning. If it asks about an actual cell’s performance, look for the additional measured or supplied information.
The Nernst equation boundary
University-level electrochemistry uses the Nernst equation to quantify the dependence of electrode potentials on reaction activities. The 2027 H2 syllabus instead frames the concentration effect as a qualitative prediction. This is a valuable boundary: students need to explain the direction of change in terms of the oxidised and reduced forms, not derive a logarithmic equation as a prerequisite for every question.
A capable JC2 student can say what happens when Cu²⁺ is increased, what happens when Fe²⁺ is increased in its redox couple, and why those cases differ. The ability to interpret a chemically altered situation is more robust than copying an advanced formula without understanding its assumptions.
A three-part answer method
Write the half-equation as a reduction, identify the ion whose concentration is changing, then state whether the written reduction is relatively favoured or disfavoured under the changed conditions. The final sentence should indicate whether the reduction potential shifts more positive or more negative. If the question is about a complete cell, identify both half-cells and the resultant redox direction.
Students often skip the explanatory middle step and write an answer they cannot defend. In a small-group lesson, require them to say why their prediction follows from the species ratio. Then reverse the concentration change in a second question. That changed task reveals whether the learner owns the method.
A week’s focused revision
Session one covers oxidation, reduction and half-equations. Session two practises standard cell potentials. Session three adds ΔG°=−nFE°. Session four uses Cu²⁺/Cu concentration changes; session five uses Fe³⁺/Fe²⁺ ratios. Session six introduces a conceptual concentration cell. Session seven contains a mixed unseen task asking for both a standard numerical calculation and a qualitative non-standard prediction.
Retest the two most frequent errors after a delay. An incorrect oxidation direction needs different repair from a correct half-reaction with a missed negative sign. A tutor should measure independent performance on altered redox couples rather than repeatedly demonstrate the same copper–zinc example.
Questions parents can ask
Parents need not perform electrochemical calculations themselves. Ask the student to show an earlier mistaken answer and a changed question completed without guidance. Can they explain which species gains electrons, why an ion concentration changes the potential, and why a standard voltage is not necessarily the actual voltage? Those explanations provide evidence of progress.
Grade changes depend on many factors, and no fixed number of tutorials guarantees an A-Level outcome. A strong teaching arrangement keeps the student engaged with school laboratory work, other JC2 subjects and adequate rest. The aim is durable scientific understanding rather than an ever-growing folder of copied solutions.
Frequently asked questions
Does more Cu²⁺ make Cu²⁺/Cu reduction potential more positive? Yes in the simple comparable-conditions model. Does every increase in ion concentration have that effect? No; the ion may be the reduced rather than oxidised form. Does positive E°cell mean fast? No. Do we multiply electrode potentials by electron count? No. What does n mean in ΔG°=−nFE°? Moles of electrons transferred per mole of the balanced reaction.
Is the Nernst equation required quantitatively for the specified concentration outcome? The H2 9476 syllabus states qualitative prediction. Can a cell use the same metal in both half-cells? A concentration cell can. Where should I check the official scope? The SEAB syllabus and corresponding examination-year resources.
An oxidation potential is a reversed half-reaction
Standard reference tables normally list reduction potentials. If a question asks about oxidation instead, reverse the chemical half-reaction and interpret the direction consistently. The numerical oxidation potential for the exactly reversed half-reaction may be expressed with the opposite sign, but you must not reverse the sign and also perform a subtraction that reverses it a second time. Such double reversals are a common source of incorrect cell voltages.
I teach one unambiguous classroom method: write both half-cells as reductions using tabulated E° values, identify the cathode and anode from the intended cell reaction, then use E°cell=E°cathode−E°anode. Balance electrons independently. If a student prefers oxidation-plus-reduction potentials, they must use that convention consistently throughout rather than mixing formulas.
What a higher oxidised-to-reduced ratio means
Consider a simple half-reaction Ox+ne⁻⇌Red. Increasing the relative chemical activity of Ox compared with Red favours the written reduction, making its reduction potential more positive under otherwise comparable conditions. Increasing Red relative to Ox tends to favour the reverse direction, making the reduction potential less positive. This is a qualitative relation rather than a numerical equation in the specified H2 learning outcome.
The distinction is clearest for Fe³⁺/Fe²⁺ because both species are dissolved. Increasing Fe³⁺ is not the same as increasing Fe²⁺. A student who sees only ‘iron concentration increased’ without knowing which oxidation state has changed lacks the information needed for a defensible prediction. Ask them to label the oxidised form before deciding.
The metal-ion case is simpler but can be misleading
For Cu²⁺(aq)+2e⁻⇌Cu(s), the pure solid copper has fixed activity in the ordinary model. Increasing Cu²⁺ raises the relative importance of the oxidised form, so the reduction potential moves in the more positive direction. This example is easy to remember because only one aqueous species appears in the half-equation.
However, transferring the simplistic phrase ‘more ions increases the potential’ to a pair of dissolved redox forms is incorrect. That is why students should practise one metal/ion couple and one ion/ion couple side by side. The underlying reasoning is the same, but the quantity being compared is the relative activity of the oxidised and reduced forms.
What happens when Cu²⁺ becomes very dilute?
If Cu²⁺ becomes less concentrated, the simple Cu²⁺/Cu reduction potential shifts less positive relative to its value at a higher concentration under comparable conditions. It would be careless to keep calling the electrode +0.34 V merely because that number appears in a Data Booklet. +0.34 V is the standard reference under its specified conditions, not a universal value for every solution.
This does not mean that dilution necessarily reverses the direction of every cell reaction. The complete cell includes the other half-cell and its own actual conditions. A qualitative comparison of two changed electrode potentials may show a strengthened or weakened driving force without demonstrating a reversal. State only the conclusion supported by the data.
Why the whole cell requires two half-cells
An isolated reduction half-equation helps define the tendency of an electrode, but a complete electrochemical cell requires a complementary oxidation. The measured cell potential is related to the difference in electrode potentials. If one half-cell’s concentration changes while the other is held fixed, the change in the complete cell potential follows from the change in their relative values and the selected reaction direction.
Before answering, identify which electrode is reduced and which oxidised under the proposed operating conditions. Do not assume the roles cannot change if the chemical conditions change sufficiently. In simple JC2 problems the intended direction is often clear, but the reasoning should come from the two reactions rather than a permanently memorised label.
A stronger zinc–copper prediction
The cell reaction Zn(s)+Cu²⁺(aq)→Zn²⁺(aq)+Cu(s) consumes Cu²⁺ and produces Zn²⁺. If the initial Cu²⁺ concentration is increased while Zn²⁺ and other conditions remain fixed, the forward reaction’s driving force can become stronger. If the Zn²⁺ product concentration is increased substantially relative to Cu²⁺, the forward driving force can weaken.
This is a reaction-quotient idea expressed qualitatively. We are considering how the actual mixture differs from standard conditions. Students do not need to calculate a non-standard voltage by an unrequired advanced equation in order to make the directional prediction. They do need to recognise which ions are on the reactant and product sides of the balanced redox reaction.
The role of electrolyte concentration in an experiment
A practical cell’s behaviour depends on more than equilibrium potential. Solution conductivity, electrode surface processes and internal resistance can affect the observed current and terminal voltage under load. Changing electrolyte concentration may therefore have several effects at once. It is important not to attribute every observed voltage change uniquely to the equilibrium reduction potential without controlling other factors.
In a theory question focused on half-cell concentration, use the stated thermodynamic model. In a practical evaluation, explain what additional experimental controls or measurements would be needed to separate equilibrium potential from resistance and kinetics. This difference is a good example of the 9476 syllabus’s emphasis on applying and evaluating evidence.
Concentration cells teach the direction of chemical change
In a cell with copper electrodes at two different Cu²⁺ concentrations, the more concentrated side is more favourable for Cu²⁺ reduction under the simplified model. The dilute half-cell can supply oxidation, producing Cu²⁺ there. The resulting reactions reduce the difference in solution composition. Electrical energy can be obtained from a chemical-potential difference even though the metal is identical in both half-cells.
Ask students to state what happens to the solutions rather than merely memorise which side is called the cathode. The concentrated side loses Cu²⁺ through deposition while the dilute side gains Cu²⁺ through oxidation, until the driving force diminishes under suitable conditions. This connects the cell diagram to mass balance.
Why temperature is not a hidden constant in every question
Electrode potentials and equilibria depend on temperature, so a problem that changes temperature at the same time as ion concentration cannot normally attribute its observed result only to the concentration change. Read the conditions. The H2 qualitative outcome is best answered by explicitly saying ‘with temperature and other relevant conditions unchanged’ where the question intends such a comparison.
Students sometimes quote a standard E° value at one condition and compare it directly with a non-standard experimental measurement. That can mix effects. In a practical scientific argument, clarify which variables are controlled, which changed and what source supplied the reference potential.
A gas electrode introduces another composition variable
The hydrogen electrode illustrates that gases can participate in redox equilibria along with aqueous ions. Pressure of the gas and relevant aqueous-ion activity can affect non-standard electrode behaviour. The exact treatment depends on the half-reaction and conditions. A student should not assume every half-cell is a metal immersed in its own ion solution.
For the H2 syllabus, first understand the standard hydrogen electrode as the reference for tabulated potentials, then apply the same principle of carefully identifying reactants and products. Avoid introducing an advanced quantitative gas-electrode formula unless the task provides the necessary data and requires it. The important conceptual lesson is that reference conditions matter.
What a colour change cannot tell you alone
Some redox couples involve coloured ions, such as Fe³⁺ and Fe²⁺ in certain chemical environments or transition-metal species with characteristic hues. A visible change can suggest altered composition, but colour is not a direct meter reading of electrode potential. Speciation, complex formation and solution conditions can affect observed colour, so a potential prediction should be grounded in the half-equation and relevant activities.
An exam may combine an observation with E° data. Use the observation to identify a possible redox change, then check whether the potentials support the proposed direction under the supplied assumptions. Do not report a precise potential merely because one solution looks darker than another.
What ΔG°=−nFE° means dimensionally
Faraday’s constant F has units of coulombs per mole of electrons, and a volt is a joule per coulomb. Multiplying n mol of electrons per mole of the specified overall reaction by F and E° therefore gives joules per mole of reaction. That unit relationship is why the electrochemical potential can be converted into a thermodynamic free-energy change.
The negative sign links a positive standard cell potential to negative standard Gibbs free energy for the reaction as written. Students should show the unit conversion from J to kJ when needed. If a calculated free-energy change has units of coulombs instead of joules per mole, the formula or units have not been applied correctly.
Counting n correctly from half-equations
The electron count n belongs to the fully balanced overall redox reaction. For Zn(s)+Cu²⁺(aq)→Zn²⁺(aq)+Cu(s), two electrons are transferred per mole of that written reaction. If the reaction coefficients are scaled, the associated molar Gibbs change scales appropriately, while the cell potential itself does not simply multiply by the scaling factor.
This makes an excellent short classroom contrast. Give students the same cell written with all coefficients doubled. Ask how n, ΔG° for the newly specified reaction amount and E°cell change. The first two scale with the written reaction; the cell voltage remains an intensive driving-force measure under the same conditions. That distinction reveals whether formulas are understood physically.
When a positive E°cell conclusion is limited
A positive E°cell supports standard-state thermodynamic favourability in the written forward direction. It does not ensure reaction rate, quantitative conversion at arbitrary concentrations, or a particular industrial current under load. In a non-standard mixture, the actual driving force may be different. Distinguishing these limits is explicitly consistent with the syllabus’s treatment of electrochemistry and scientific models.
Students should answer the stated standard-state question before adding a limitation. A clear response might say that the calculated positive standard cell potential corresponds to negative ΔG° under standard conditions, but kinetics and changed concentrations can influence real behaviour. This is more useful than a vague warning that ‘the experiment may be different’.
A safe Paper 4-style reasoning exercise
Give the learner a diagram of two half-cells, the two ion concentrations and a set of fictional potential readings. Ask them to identify which solution is more concentrated, which half-reaction is expected to be reduced, and what observations would help check the claim. This can be done entirely on paper while the student learns data interpretation.
Actual galvanic-cell construction requires appropriate laboratory equipment, electrode preparation, electrolyte handling and supervision. A tuition worksheet can strengthen reasoning about the results but should not be represented as replacing hands-on school practical skills. The distinction matters when parents compare courses advertising ‘practical preparation’.
An error ledger specific to electrochemistry
Use five categories: half-equation direction, cathode/anode identification, standard potential subtraction, concentration-effect interpretation, and thermodynamics-versus-kinetics conclusion. Record the first wrong move from each marked question. A learner who consistently changes the sign of a reduction potential twice needs a different correction from someone who ignores the Fe³⁺/Fe²⁺ ratio.
After identifying the error, set one altered redox couple and retest after a delay. This is much more efficient than asking students to copy a perfect Daniell-cell solution repeatedly. Scientific understanding is visible when the method survives a change of ions, concentrations and question wording.
A fictional tutoring example
Imagine Aisha calculates the correct standard zinc–copper cell voltage but writes that increasing Zn²⁺ concentration strengthens the forward reaction because ‘there are more ions’. Ben correctly sees that increasing Zn²⁺ can diminish the forward driving force but mistakenly changes the standard E° values themselves rather than distinguishing actual conditions. Their final answers may both be wrong for different reasons.
A tutor should ask Aisha to label products and reactants and ask Ben to distinguish tabulated standard potentials from qualitative non-standard behaviour. The next tasks should then change the species and ask for independent explanations. This is how three-student tutorial time can be more diagnostic than a large lecture with one common model answer.
How parents can understand progress
Ask your JC2 student to explain what happens if the concentration of Cu²⁺ is raised and then ask whether the same rule applies when Fe²⁺ is raised in the Fe³⁺/Fe²⁺ couple. A learner who knows the difference and can point to the balanced half-equations has gained a usable concept rather than memorising a slogan.
Parents need not master electrochemical calculations. Look for a changed written question solved without hints, clear units and a sensible distinction between standard potential, actual concentration and reaction speed. Keep the study routine realistic alongside other A-Level subjects and adequate rest.
A self-check with three unfamiliar situations
Situation one: a copper half-cell becomes more concentrated in Cu²⁺ while other conditions remain comparable. Predict the reduction-potential direction. Situation two: a Fe³⁺/Fe²⁺ half-cell receives additional Fe²⁺ relative to Fe³⁺. Predict the direction again. Situation three: the standard zinc–copper cell has E°=+1.10 V; state what the sign means and whether it proves a rapid reaction.
The expected qualitative answers are more positive in situation one, less positive in situation two, and thermodynamic favourability under standard conditions without a kinetic guarantee in situation three. A correct response gives the chemical reason, not only those labels. Repeat the exercise with another redox couple to verify transfer.
Final four-question electrode-potential mastery check
Before putting away the electrochemistry notes, test four different decisions. First, predict the effect of raising Cu²⁺ concentration on the Cu²⁺/Cu reduction potential while other relevant conditions remain fixed. Second, predict the effect of raising Fe²⁺ relative to Fe³⁺ on Fe³⁺/Fe²⁺. Third, explain why a Cu²⁺/Cu standard potential cannot automatically be assigned to every non-standard copper solution. Fourth, state why a positive standard cell potential does not guarantee a rapid reaction.
Expected reasoning: a greater Cu²⁺ concentration tends to make reduction more favourable and the potential more positive; more Fe²⁺ relative to Fe³⁺ tends to make the iron reduction potential less positive. The standard potential is defined for reference conditions, and kinetic barriers affect the observed rate separately from thermodynamic favourability. The learner should say which half-reaction and ion role support each conclusion, not merely reproduce the directions.
After answering, change the ion ratio again or substitute another unfamiliar half-reaction and require the student to predict independently. This is a small but important transfer test: a student who can reason from chemical species and electron flow has a stronger foundation for A-Level examination work and future scientific study.
A complete cell-energy calculation with ΔG° = −nFE°
Consider the familiar standard zinc–copper cell: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s). The balanced reaction transfers two moles of electrons for every mole of reaction as written. With a standard cell potential of approximately +1.10 V and Faraday’s constant about 9.65 × 10⁴ C mol⁻¹, the standard Gibbs free energy change is ΔG° = −nFE° = −2 × (9.65 × 10⁴) × 1.10 = −2.12 × 10⁵ J mol⁻¹, or approximately −212 kJ mol⁻¹. The negative sign agrees with thermodynamic favourability under standard conditions.
The electron count n comes from the balanced overall redox equation; it is not merely the charge number on whichever ion appears first. The potential is an intensive quantity, so multiplying the overall reaction equation by two does not double E°. The corresponding ΔG° for twice the written amount of reaction does double, because the amount of chemical change has doubled. That distinction between an intensive voltage and an extensive energy change is one of the most useful consistency checks in H2 electrochemistry.
The meaning of a positive potential under non-standard concentrations
When ion concentrations differ from their standard values, the actual electrode potentials and cell potential can differ from standard values. For a reduction involving a metal ion becoming solid metal, an increased relevant metal-ion activity generally makes reduction more favourable relative to the same electrode at a lower metal-ion activity, at constant temperature and with other variables fixed. For a redox couple involving both aqueous oxidised and reduced forms, their relative activities matter. State the comparison carefully rather than treating the sign of the standard potential as a permanent label for every possible solution.
The revised H2 9476 syllabus asks for qualitative predictions of concentration effects; it does not require a numerical Nernst-equation calculation. An exam-ready response should identify the half-reaction as a reduction, name which relevant species concentration changes and explain the direction of the resulting potential change. If the question asks about the whole cell, perform that reasoning on both electrodes before predicting the effect on the measured cell potential.
Related Chemistry guides and tuition enquiries
Continue with JC2 H2 electrochemistry and redox calculations, positive cell potential versus reaction speed, JC2 Gibbs free energy and Data Booklet practice. The official SEAB H2 9476 syllabus provides the learning outcomes.
For tuition, contact eduKate Singapore or enquire about current JC2 places. Confirm the actual fees, timetable, group arrangement and location. Worked examples are teaching illustrations, not official SEAB examination questions.