JC2 H2 Chemistry Tuition Hougang | Fe²⁺ Catalysis of Iodide and Peroxodisulfate

JC2 H2 Chemistry tuition for Hougang students can make a difficult catalysis question surprisingly logical: why can Fe²⁺ speed the reaction between iodide and peroxodisulfate without appearing in the overall equation? The starting reaction involves two negatively charged species, I⁻ and S₂O₈²⁻, whose direct electron transfer can be kinetically slow. An Fe²⁺/Fe³⁺ cycle provides an alternative pathway. The iron ions change oxidation state and then return to the starting catalyst form.

This H2 Chemistry 9476 guide explains the catalytic redox cycle involving Fe²⁺, I⁻ and S₂O₈²⁻, including balanced half-equations, standard-electrode-potential reasoning, homogeneous catalysis and the difference between rate and thermodynamic feasibility. It includes complete worked equations, stoichiometric calculations and appropriately bounded kinetic examples. eduKate Singapore’s premium three-student tutoring format lets each learner reconstruct the catalyst cycle independently and receive an exact correction when a charge, electron or sign is wrong. Families from Hougang can enquire at 8 Fourth Avenue, Singapore 268674, near Sixth Avenue MRT; there is no Hougang branch.

Arrange a JC Chemistry consultation · JC2 H2 Chemistry Tutor Hougang | Small Groups Tutorials · H2 Chemistry 9476 national study guide

The Immediate Answer: Fe²⁺ and Fe³⁺ Form a Catalytic Cycle

The uncatalysed overall reaction is S₂O₈²⁻ + 2I⁻ → 2SO₄²⁻ + I₂. In the presence of an appropriate iron catalyst, one schematic pathway separates the electron transfer into two redox stages: peroxodisulfate oxidises Fe²⁺ to Fe³⁺, and Fe³⁺ oxidises iodide to iodine while regenerating Fe²⁺.

The first stage can be written S₂O₈²⁻ + 2Fe²⁺ → 2SO₄²⁻ + 2Fe³⁺. The second is 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂. When those stages are added, the Fe²⁺ and Fe³⁺ terms cancel and the net redox reaction remains the one above.

The chemical value of the iron species is that they can participate repeatedly. It is not that the catalyst makes the overall reaction turn into a new product or that it changes the equilibrium constant at the same temperature.

Why the 2027 H2 Syllabus Names This Example

The official Singapore-Cambridge H2 Chemistry 9476 syllabus for 2027, Reaction Kinetics, explicitly requires students to outline the catalytic role of Fe²⁺ in the I⁻/S₂O₈²⁻ reaction. It also covers homogeneous and heterogeneous catalysis, rates and rate equations, catalysts as alternative pathways with lower activation energies, and standard redox potentials in its electrochemistry strand.

This page owns the named Fe²⁺/iodide/peroxodisulfate electron-transfer example, distinct from the general H2 catalysis guide and from broader redox balance explanations. The student should be able to show why the iron catalyst is regenerated, not merely repeat that catalysts are ‘not used up’.

Begin With Charge and Oxidation States

Iodide is I⁻ and is oxidised to iodine I₂. Each iodine atom in I₂ has oxidation state zero, up from −1 in iodide. The peroxodisulfate ion S₂O₈²⁻ is reduced to sulfate SO₄²⁻ in the overall reaction through cleavage of the peroxo linkage and electron uptake.

Fe²⁺ loses an electron to form Fe³⁺ in one stage and Fe³⁺ gains an electron to return to Fe²⁺ in the other. This makes iron an electron-transfer mediator in the simplified homogeneous aqueous catalytic model.

Fe²⁺ → Fe³⁺ + e⁻: iron(II) is oxidised.
Fe³⁺ + e⁻ → Fe²⁺: iron(III) is reduced.
2I⁻ → I₂ + 2e⁻: iodide is oxidised.
S₂O₈²⁻ + 2e⁻ → 2SO₄²⁻: peroxodisulfate is reduced.
Homogeneous catalysis: the catalyst and reactants are in the same liquid phase in the ordinary aqueous model.

One dangerous shortcut is to label Fe²⁺ as always a reducing agent and stop there. Its role depends on the chemical partner. Fe²⁺ acts as a reducing species when it is oxidised by peroxodisulfate; Fe³⁺ acts as an oxidising species when it oxidises iodide. Both forms are part of one catalyst cycle.

Worked Example 1: Write the Oxidation Half-Equation for Iodide

Begin with I⁻ as the aqueous reactant and I₂ as product.
Balance iodine atoms by writing 2I⁻ → I₂.
Balance net charge by showing two electrons released on the right.
Obtain 2I⁻ → I₂ + 2e⁻.
Explain that iodide loses electrons, so it is oxidised.
Do not write a positive iodide ion or create a hydrogen atom to balance this electron-only half-equation.

The half-equation is simple enough that a tutor should expect an independent answer before moving to the more complicated peroxo species. Errors in the electron direction should be repaired at this stage.

Worked Example 2: Reduce Peroxodisulfate to Sulfate

Write S₂O₈²⁻ → 2SO₄²⁻, keeping sulfur and oxygen atom counts equal.
Check total left charge: −2.
Check right ionic charge before electrons: 2(−2) = −4.
Add two electrons to the reactant side to make total left charge −4.
The balanced reduction is S₂O₈²⁻ + 2e⁻ → 2SO₄²⁻.
Recognise this as an electron-gain process.

Students sometimes add two electrons to the product side simply because there are two sulfate ions. That produces an incorrect charge balance. Charge, not the number of product species, determines electron placement.

Worked Example 3: Derive the Uncatalysed Net Equation

Write oxidation: 2I⁻ → I₂ + 2e⁻.
Write reduction: S₂O₈²⁻ + 2e⁻ → 2SO₄²⁻.
Add the half-equations and cancel two electrons from both sides.
Obtain S₂O₈²⁻ + 2I⁻ → 2SO₄²⁻ + I₂.
Check oxygen and sulfur atom counts: two sulfur, eight oxygen on each side.
Check charge: left −2 −2 = −4; right 2(−2) = −4.

This is the net equation with or without the catalyst. Iron participates in the mechanism, not the net atom balance when a full catalytic turnover is completed.

The Fe²⁺ Oxidation Stage

When peroxodisulfate oxidises Fe²⁺, each Fe²⁺ becomes Fe³⁺ by losing one electron. Because peroxodisulfate accepts two electrons in its reduction half-equation, two Fe²⁺ species participate in the balanced schematic catalytic step.

The combined stage S₂O₈²⁻ + 2Fe²⁺ → 2SO₄²⁻ + 2Fe³⁺ can be derived from the redox half-equations. The iron ions are not present in sulfate product as a permanent constituent; they are converted to Fe³⁺ and await regeneration.

Worked Example 4: Balance the First Catalytic Stage

Start from S₂O₈²⁻ + 2e⁻ → 2SO₄²⁻.
Use Fe²⁺ → Fe³⁺ + e⁻ to supply the electrons.
Multiply the iron oxidation half-equation by two.
Add to obtain S₂O₈²⁻ + 2Fe²⁺ → 2SO₄²⁻ + 2Fe³⁺.
Check left charge: −2 + 4 = +2.
Check right charge: −4 + 6 = +2, so charge balances.

This one charge check is a useful tutorial diagnostic. The correct species may all be present in a student’s answer even while the equation is electrically inconsistent.

The Fe³⁺ Regeneration Stage

Fe³⁺ can accept an electron from iodide and return to Fe²⁺. One Fe³⁺ accepts one electron, so two Fe³⁺ ions are required to account for the two electrons released when two iodides form one iodine molecule.

The stage is 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂. The iron(II) species regenerated here can participate again in the first stage, allowing a catalytic cycle rather than a single-use chemical reagent.

Worked Example 5: Balance the Regeneration Stage

Write iodide oxidation: 2I⁻ → I₂ + 2e⁻.
Write iron(III) reduction: Fe³⁺ + e⁻ → Fe²⁺.
Multiply the iron(III) reduction by two.
Add to give 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂.
Left charge is +6 −2 = +4.
Right charge is +4, so charge is balanced and Fe²⁺ has been regenerated.

An answer ending at Fe³⁺ would describe iron(II) being consumed as a reagent, not yet a complete catalyst cycle. The second stage is the evidence of regeneration.

Worked Example 6: Add the Two Catalytic Stages

Take the two stages above and add them. Fe²⁺ appears as a reactant in the first and as a product in the second. Fe³⁺ appears as a product in the first and as a reactant in the second.

Add S₂O₈²⁻ + 2Fe²⁺ → 2SO₄²⁻ + 2Fe³⁺.
Add 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂.
Cancel both occurrences of 2Fe²⁺.
Cancel both occurrences of 2Fe³⁺.
Obtain the original net equation S₂O₈²⁻ + 2I⁻ → 2SO₄²⁻ + I₂.
Conclude that iron mediates the electron transfer while not being permanently consumed in the net catalytic cycle.

The cancellation is not an arbitrary algebra trick. It encodes the fact that all catalytic species return to their starting total chemical form after a complete turnover in the simplified reaction scheme.

Why This Is Homogeneous Catalysis

Homogeneous catalysis occurs when the catalyst is in the same phase as the reactants in the relevant chemical system. Here Fe²⁺ and Fe³⁺ are aqueous species, and the reacting iodide and peroxodisulfate are also aqueous ions. That makes the ordinary model homogeneous.

This differs from the familiar Haber process in which gas reactants interact with the surface of a solid iron-containing catalyst. Both are catalysis, but the detailed ways they provide an alternative pathway are not the same.

A tutor should make students identify the catalyst’s phase from the actual chemical system rather than conclude that all iron catalysis must be heterogeneous simply because iron metal is a solid.

The Electrochemical Evidence: A Pair of Favourable Steps

Standard reduction potentials provide one way to understand the thermodynamic tendency of these redox stages under standard conditions. Use representative E° values S₂O₈²⁻/SO₄²⁻ ≈ +2.01 V, Fe³⁺/Fe²⁺ ≈ +0.77 V and I₂/I⁻ ≈ +0.54 V, with the numerical values taken as illustrative rounded data for the calculation.

The potential comparison does not measure reaction rate directly. It tells whether the specified redox direction is thermodynamically favourable under the reference conditions. A favourable uncatalysed reaction can still be slow because of its kinetic barrier.

Worked Example 7: E°cell for Peroxodisulfate Oxidising Fe²⁺

Reduction at the cathode: S₂O₈²⁻ + 2e⁻ → 2SO₄²⁻, E° ≈ +2.01 V.
Oxidation at the anode: Fe²⁺ → Fe³⁺ + e⁻, but the table supplies the Fe³⁺/Fe²⁺ reduction E° ≈ +0.77 V.
Use E°cell = E°red(cathode) − E°red(anode).
Calculate +2.01 − (+0.77) = +1.24 V.
The positive value supports standard-state feasibility of this catalytic step as written.
Do not multiply the Fe³⁺/Fe²⁺ potential when doubling the electron equation.

Fe²⁺ is oxidised, so it acts as the reducing species of the two redox partners in this stage. The potential calculation is consistent with the electron direction.

Worked Example 8: E°cell for Fe³⁺ Oxidising Iodide

Reduction: Fe³⁺ + e⁻ → Fe²⁺, E° ≈ +0.77 V.
Oxidation: 2I⁻ → I₂ + 2e⁻, reversing the tabulated I₂/2I⁻ reduction E° ≈ +0.54 V.
Use the listed reduction potentials without scaling by electron coefficients.
Calculate E°cell = +0.77 − (+0.54) = +0.23 V.
Conclude that this regeneration step is also thermodynamically favourable under the stated standard conditions.

The catalyst has opposite redox roles across the cycle. Fe²⁺ is oxidised in one stage; Fe³⁺ is reduced in another. That alternation is the precise chemical reason the iron couple can mediate the electron transfer.

Worked Example 9: E°cell for the Overall Reaction

Overall cathode reduction is S₂O₈²⁻/2SO₄²⁻ with E° ≈ +2.01 V.
Overall anode oxidation is iodide to iodine, using tabulated reduction E° ≈ +0.54 V.
Calculate E°cell = 2.01 − 0.54 = +1.47 V.
Recognise that the overall reaction is favourable under standard-state conditions.
Explain why its direct reaction can nevertheless be kinetically slower than a catalysed pathway.

A positive overall potential is not evidence that adding Fe²⁺ must change the overall E° to +1.47 plus another number. A catalyst changes kinetics, not the net standard free-energy difference for the same chemical states.

Why the Catalysed Rate Can Be Faster

A catalyst changes the mechanism and can provide an alternative pathway with a lower effective activation-energy barrier. For the named redox process, iron ions mediate electron transfer in two steps rather than requiring the same direct encounter between iodide and peroxodisulfate species.

Both main reactant ions are negatively charged in the overall process, which can contribute to unfavourable direct approach and electron transfer in an aqueous environment. The catalyst’s differently charged redox partners can offer an alternative route. But a fully quantitative rate prediction requires actual kinetic data, not charge signs alone.

The catalyst does not move the products to a different final energy level or change the standard cell potential of the net reaction at the same temperature.

Energy Profile: Same Endpoints, Different Route

An uncatalysed energy profile can have a high activation barrier even if the products lie at a lower Gibbs free-energy level than the reactants. A catalysed multi-step route may show more than one barrier separated by intermediate states, with a lower controlling barrier under the relevant conditions.

The overall reaction remains S₂O₈²⁻ + 2I⁻ → 2SO₄²⁻ + I₂. The net enthalpy and free-energy changes are properties of the reactant and product states under the specified conditions; a catalyst does not rewrite them.

Worked Example 10: Distinguish Barrier and Overall Energy

A hypothetical diagram gives the uncatalysed pathway a largest activation barrier of 80 kJ mol⁻¹ and the catalysed alternative a largest relevant barrier of 50 kJ mol⁻¹, with the same reactant and product energy levels. These numbers are invented solely for the exercise.

Identify the two routes as having the same net chemical reactants and products.
Compare the kinetic barriers: 50 is lower than 80 kJ mol⁻¹.
Conclude that the catalysed pathway can proceed faster under the relevant conditions.
Do not claim the net ΔH or ΔG has been changed by 30 kJ mol⁻¹.
Explain that a multi-step pathway’s detailed rate also depends on intermediate concentrations and the specific barriers.

A student who says ‘the catalyst raises the energy of Fe²⁺ until it reacts’ has not described the actual catalytic-cycle chemistry. The important model is a different available route with regeneration.

A Catalyst Is Not a Reactant Used in Stoichiometric Amount

Suppose an idealised experiment uses a small amount of Fe²⁺ catalyst and a substantially larger amount of peroxodisulfate. If the redox cycle turns over repeatedly, the catalyst can assist conversion of much more reactant than its original mole amount. There is no one-to-one rule that one mole Fe²⁺ yields exactly one mole iodine in the total product.

Catalyst deactivation, side reactions or changes in oxidation state can affect real systems. The idealised textbook picture assumes catalyst regeneration, not the assertion that a real catalyst remains infinitely active under every imaginable condition.

Worked Example 11: Theoretical Iodine from Peroxodisulfate

An illustrative mixture contains 0.0200 mol S₂O₈²⁻ and excess I⁻, with a suitable iron catalyst present. Assume the net reaction is completed with no interfering side reactions.

Write S₂O₈²⁻ + 2I⁻ → 2SO₄²⁻ + I₂.
One mole S₂O₈²⁻ produces one mole I₂ under the net stoichiometry.
Theoretical iodine amount is 0.0200 mol.
Using M(I₂) ≈ 253.8 g mol⁻¹, the theoretical iodine mass is 5.08 g.
Do not multiply this amount by moles of Fe²⁺ catalyst because Fe²⁺ does not appear in the net balanced reaction.

This is an atom and electron accounting result, not an instruction to prepare or isolate iodine outside a professional laboratory. Peroxodisulfate oxidants and iodine-containing reagents require appropriate safety controls.

Worked Example 12: Iodide as the Limiting Reactant

A second conceptual mixture contains 0.0200 mol S₂O₈²⁻ and only 0.0250 mol I⁻. Which reactant limits theoretical iodine production?

From the net equation, two moles I⁻ are required per mole S₂O₈²⁻.
0.0200 mol peroxodisulfate would require 0.0400 mol iodide.
Only 0.0250 mol iodide is present, so iodide is limiting.
The theoretical iodine amount is 0.0250/2 = 0.0125 mol I₂.
Peroxodisulfate consumed is 0.0125 mol and 0.00750 mol remains in the idealised complete-consumption model.

The catalyst may speed the reaction but does not change which reactant is limiting. That depends on net stoichiometry and the starting amounts.

Rates Data: Catalyst Present Versus Absent

To test whether a catalyst increases reaction speed, experiments can monitor the formation of iodine under controlled temperature and reactant concentrations. A suitable optical observation may be used if its connection to iodine concentration is calibrated and the measurement method has adequate selectivity.

A rate difference between two otherwise comparable trials supports a catalytic effect, but one numerical comparison does not establish the full detailed mechanism or rate law. The chemical half-equations and separate kinetic evidence support the proposed pathway together.

Worked Example 13: A Rate Comparison

Two hypothetical trials use the same initial concentrations and temperature. Trial A without Fe²⁺ has an initial iodine-production concentration rate of 2.0 × 10⁻⁶ mol dm⁻³ s⁻¹. Trial B with a suitable small Fe²⁺ amount has 1.0 × 10⁻⁵ mol dm⁻³ s⁻¹.

Compute the ratio 1.0 × 10⁻⁵/(2.0 × 10⁻⁶).
The catalysed trial’s measured initial rate is five times the uncatalysed trial’s in this invented dataset.
The result supports an increase in observed rate under the compared conditions.
Do not infer that the equilibrium constant has multiplied by five.
Do not conclude that all Fe²⁺ amounts or temperatures cause exactly the same fivefold enhancement.

The numbers are model data, not results of an actual experiment. The purpose is to practise distinguishing a rate comparison from an equilibrium or catalyst-mass claim.

Does Adding More Catalyst Always Make Rate Linear?

The rate may depend on the concentration of the active catalyst species, but detailed dependence can be affected by multiple steps, competing processes or catalyst speciation. Doubling the total iron salt added does not necessarily double the observed initial rate under every set of conditions.

A student should determine reaction orders from controlled initial-rate comparisons if the question supplies data. The overall net reaction’s stoichiometry does not by itself reveal the rate law of the two-step catalytic mechanism.

Worked Example 14: A Proposed Catalytic Rate Law Needs Evidence

An experiment reports that doubling total initial catalyst concentration doubles the initial product-formation rate under one controlled range. That supports an apparent first-order dependence on the varied catalyst amount in that range, provided other variables are controlled.

It does not prove that the elementary redox steps both have that same rate law or that the effect remains first order at high catalyst concentrations. A responsible answer distinguishes empirical observation from a microscopic mechanism.

Why Iron(III) May Also Lead Into the Cycle

The catalyst couple cycles between Fe²⁺ and Fe³⁺. If iron(III) is supplied initially, Fe³⁺ can participate in oxidising iodide and generating Fe²⁺ under appropriate conditions, after which the other catalytic stage can regenerate Fe³⁺. That helps explain why a coupled redox cycle cannot always be described solely by the oxidation state of the initially added iron salt.

The exact time development and catalyst species distribution can differ. H2 students need the named regenerative mechanism rather than a promise that any iron species in any solution behaves identically.

The Direct Reaction Is Not ‘Chemically Impossible’

Catalysis does not mean the uncatalysed reaction has no thermodynamic driving force. The standard potential comparison can favour the overall redox transfer even when the uncatalysed rate is slow. Fe²⁺ increases the accessibility of a productive path by mediating the transfer.

This is a useful general distinction: a reaction can be feasible but very slow, and a catalyst can make it useful in a practical timescale without making the products more stable than they already were.

Common H2 Catalysis Mistakes

Writing Fe²⁺ in the net stoichiometric equation as a permanently consumed reactant.
Showing Fe²⁺ oxidised to Fe³⁺ but omitting the step regenerating Fe²⁺.
Putting iodide oxidation electrons on the reactant side.
Forgetting to balance ionic charges and sulfate coefficients.
Calling the aqueous iron ion a heterogeneous solid catalyst merely because the element iron is a metal.
Claiming a catalyst changes overall E°cell or K at fixed temperature.
Using total added catalyst mass to infer theoretical iodine yield.
Assuming favourable E°cell means an uncatalysed reaction must be rapid.
Infering the order of a whole process from net coefficients rather than measured kinetics.

The First-Principles Fencing Method

Begin with iodide oxidation and peroxodisulfate reduction as separate half-equations.
Combine them into the uncatalysed net equation with correct charge.
Introduce Fe²⁺ → Fe³⁺ and its reverse as a redox couple.
Construct the first peroxodisulfate–iron stage and check its charge.
Construct the Fe³⁺–iodide regeneration stage and check its charge.
Add the stages and cancel the iron species to recover the net reaction.
Interpret rounded reduction-potential data for each stage and the overall process.
Finish with an unfamiliar mediator and ask for a valid regenerative cycle, not merely a catalyst name.

This sequence separates electron balance, thermodynamics and kinetics. Students who mix those three too early often memorise the arrow chain but cannot explain why it works.

Why a Premium Three-Student Tutorial Can Help

One learner may balance iodide correctly but write an incorrect peroxodisulfate charge. Another may obtain the two stages but fail to cancel the Fe species. A third may derive the net equation yet claim the catalyst makes the standard cell potential larger.

These misconceptions require different corrections. A tutor in a three-student group can inspect the independent first attempt, discuss the shared chemical principle and give each student a changed redox mediator question to test genuine transfer.

A Suggested 90-Minute JC2 H2 Chemistry Lesson

0–10 minutes: retrieve oxidation number, half-equations and charge conservation.
10–25 minutes: balance the uncatalysed I⁻/S₂O₈²⁻ reaction.
25–40 minutes: derive the Fe²⁺ oxidation and Fe³⁺ regeneration stages.
40–55 minutes: cancel the iron species and explain catalyst turnover.
55–68 minutes: calculate standard E° for both stages with supplied values.
68–80 minutes: distinguish rate evidence, overall equilibrium and theoretical iodine amount.
80–90 minutes: solve a changed catalytic cycle and record one exact error.

The lesson structure is illustrative. Actual teaching should fit the learner’s current college syllabus, prior redox skills and availability of a compatible premium group.

Three Learner Pathways

Repair: oxidation and reduction are confused

Begin with the two one-electron changes Fe²⁺ ⇌ Fe³⁺ and the iodide oxidation half-equation. Require charge-balance explanations before writing a two-step cycle.

Stabilisation: equations are right but catalyst reasoning is weak

Use a table showing iron consumed or produced at each stage, then ask the learner to cancel species and explain why iron does not determine the net product stoichiometry.

Extension: connect kinetics and standard potentials

Provide an unfamiliar mediator redox couple with two listed reduction potentials. Ask whether both mediated steps are thermodynamically feasible, then what separate experimental information would be needed to compare rates.

A Four-Week Retrieval Plan

Week 1: independently write iodide and peroxodisulfate half-equations.
Week 2: reconstruct and balance the Fe²⁺/Fe³⁺ two-step cycle.
Week 3: use standard reduction potentials to justify the direction of each stage.
Week 4: solve a mixed stoichiometry and kinetics evidence question with a changed mediator.

Students need not spend four full weeks on the same diagram if the explanation is already secure. The aim is to retain the causal model and transfer it to a new catalytic or redox dataset.

What Parents Can Look For

A learner making progress can explain why Fe²⁺ becomes Fe³⁺ in one step, why Fe³⁺ becomes Fe²⁺ in the next, and why both cancel from the overall equation. They can also explain why a favourable standard potential does not automatically mean a reaction proceeds quickly without catalyst.

These are independently testable chemical skills. The effect on examination performance still depends on broader Chemistry mastery, school work and sustained revision, and no tutor should guarantee a fixed grade.

Class Details and Travel From Hougang

Subject: JC2 H2 Chemistry 9476. Focus: named Fe²⁺ catalyst cycle in I⁻/S₂O₈²⁻ redox, balanced equations and kinetics. Tutorial approach: premium three-student small-group teaching with individual feedback and worked problems. Weekly placement: enquire about a suitable 90-minute group and current availability. Venue: eduKate Singapore, 8 Fourth Avenue, Singapore 268674, near Sixth Avenue MRT.

Hougang is the family search locality rather than the classroom address. Confirm fees, schedule, current group places and practical travel directly. The oxidants and iodine compounds in these examples are for supervised educational contexts only.

Frequently Asked Questions

What is the overall reaction between iodide and peroxodisulfate?

S₂O₈²⁻ + 2I⁻ → 2SO₄²⁻ + I₂, balanced for atoms and charge.

Why is Fe²⁺ a catalyst?

Fe²⁺ is oxidised to Fe³⁺ in one mediated stage, while Fe³⁺ is reduced back to Fe²⁺ in another, allowing a cycle.

Why is this homogeneous catalysis?

The iron mediator and reacting ions are aqueous species in the same liquid phase in the familiar syllabus example.

Does Fe²⁺ change the standard cell potential of the overall reaction?

No. The catalyst changes the pathway and kinetics, not the net thermodynamic state difference for the same reactants and products.

Is the uncatalysed iodide/peroxodisulfate reaction impossible?

No. A favourable redox potential can coexist with slow kinetics. A catalyst provides an alternative route.

Does the amount of Fe²⁺ determine iodine’s theoretical yield?

Not by a one-to-one permanent stoichiometric amount in a complete catalytic cycle. The net reactants and limiting-reagent amounts set the theoretical product.

Is the tutor centre in Hougang?

The centre described is Fourth Avenue near Sixth Avenue MRT. Families from Hougang can enquire about compatible current groups.

A Catalyst Is Defined by Its Complete Cycle

The strongest H2 answer balances the electrons, follows Fe²⁺ and Fe³⁺ through both stages, cancels the regenerated mediator and then separates kinetic improvement from standard-state thermodynamics. That explains why iron can repeatedly help the same overall reaction without becoming a permanent ingredient of the product.

Arrange a parent–student consultation with eduKate Singapore to review the student’s current JC2 redox and kinetics questions and compatible three-student group availability.