Library › Redox Equilibria

Chemistry (IAL)

Redox Equilibria

Revise Redox Equilibria for Chemistry (IAL) — revision notes and instant AI marking. Free to start.

📖 Revision notes · preview

Edexcel IAL Chemistry — Unit 5

Redox Equilibria

Big idea: Every half-cell "wants" to gain or lose electrons to a different degree — we measure that "want" as a standard electrode potential (E°), and comparing two E° values tells us which way electrons will actually flow, whether a reaction is feasible, and how to calculate unknown concentrations by titration.

Chapter Summary

Oxidation and reduction can be defined three ways: gain/loss of oxygen, gain/loss of hydrogen, and gain/loss of electrons (the last one is the one that matters here) — remember with OIL RIG.
Standard electrode potential (E°) is measured against the Standard Hydrogen Electrode (SHE), which is fixed at 0.00 V, under standard conditions (1.00 mol dm⁻³, 298 K, 100 kPa).
Three types of half-cell exist: metal/metal-ion, non-metal/non-metal-ion, and ion/ion (different oxidation states) — the last two need an inert platinum electrode.
The salt bridge (usually KNO₃) completes the circuit by letting ions flow, balancing charge, without letting electrons pass through it.
E°_cell = E°_right − E°_left = E°_reduction − E°_oxidation. The more positive half-cell is reduced; the less positive one is oxidised.
Conventional cell diagrams use | for phase boundaries and , for species in the same phase, with the oxidation half on the left and reduction half on the right.
A reaction is thermodynamically feasible (spontaneous) when E°_cell is positive — but this says nothing about the rate of reaction.
E° is directly proportional to total entropy change (ΔS) and to ln K (the equilibrium constant).
Fuel cells (like hydrogen-oxygen) generate electricity directly from a chemical reaction without combustion — cleaner and more efficient, but hydrogen storage brings its own problems.
Redox titrations (manganate(VII) and iodine-thiosulfate) let us calculate unknown concentrations using stoichiometric ratios from balanced half-equations.

1. Reduction & Oxidation Recap

The Three Definitions

Chemists use three interchangeable definitions of oxidation and reduction, depending on what's easiest to spot in a given reaction. They all describe the same underlying process — electron transfer — just viewed from different angles.

Process	Oxidation	Reduction
Oxygen	Addition of oxygen 2Mg + O₂ → 2MgO	Loss of oxygen 2CuO + C → 2Cu + CO₂
Hydrogen	Loss of hydrogen	Addition of hydrogen
Electrons	Loss of electrons Al → Al³⁺ + 3e⁻	Gain of electrons F₂ + 2e⁻ → 2F⁻

🧠 Memory trick — OIL RIG Oxidation Is Loss (of electrons) | Reduction Is Gain (of electrons). This is the definition that matters for everything in this chapter — electrode potentials are entirely about electron transfer.

How different blocks of the periodic table behave

Because this chapter deals with ions constantly, it helps to know instinctively what charge an element's ion will carry:

s-block elements are oxidised, losing electrons to form 1+ or 2+ ions (e.g. Na → Na⁺ + e⁻, Ca → Ca²⁺ + 2e⁻).
p-block metals are also oxidised to positive ions — usually with a charge matching their group number (Al → Al³⁺ + 3e⁻), though some (like Sn → Sn²⁺ + 2e⁻) don't follow this rule exactly.
p-block non-metals are usually reduced, gaining electrons to form negative ions with charge = (group number − 8). So Group 7 → −1 (F⁻), Group 6 → −2 (O²⁻).
d-block elements are usually oxidised to positive ions, but because they have variable oxidation states they can form a whole range of charges — Cu²⁺, Cr³⁺, V⁵⁺, and so on. This variability is exactly why they're so useful in redox titrations later in this chapter.

Q1. Identify whether each of the following is an oxidation or reduction, and justify your answer using the electron-transfer definition:
(a) Fe²⁺ → Fe³⁺ + e⁻ (b) Cl₂ + 2e⁻ → 2Cl⁻

2. Standard Electrode Potential (E°)

Why we need a "standard"

Electrode potential isn't a fixed number for a given half-reaction — it changes with temperature, gas pressure, and concentration of reagents (Le Chatelier's principle at work). That's a problem if you want to compare how "keen" different species are to gain or lose electrons.

The fix: agree on a common set of conditions and a common reference point, so every measurement is directly comparable. That's what "standard" means here.

Standard Conditions Ion concentration = 1.00 mol dm⁻³ | Temperature = 298 K | Pressure = 100 kPa These three conditions must all be true before you can call a measured potential a "standard" electrode potential.

Definition: the standard electrode potential (E°) is the potential difference (voltage) produced when a standard half-cell is connected to a standard hydrogen electrode, under standard conditions, measured with a high-resistance voltmeter (so essentially no current flows — this matters because if current flowed, the concentrations would start changing and it wouldn't be "standard" anymore).

💡 Think of it like a tug-of-war scoreboard Imagine every half-reaction is a team pulling on a rope, trying to "win" electrons. The Standard Hydrogen Electrode is the fixed anchor point at 0.00 V. A very positive E° (like Br₂/Br⁻ at +1.09 V) means that half-reaction pulls electrons towards itself really strongly — it's a strong oxidising agent, easily reduced. A very negative E° (like Na⁺/Na at −2.71 V) means that half-reaction barely wants to hold onto electrons at all — sodium metal readily gives them away.

The Standard Hydrogen Electrode (SHE)

This is the reference point everything else is measured against, deliberately assigned a value of exactly 0.00 V. It's built from:

Hydrogen gas at 100 kPa, in equilibrium with H⁺ ions at 1.00 mol dm⁻³: 2H⁺(aq) + 2e⁻ ⇌ H₂(g)
An inert platinum electrode (often coated in finely-divided "platinum black" to increase surface area and speed up the equilibrium) in contact with both the gas and the ions.

When any other half-cell is connected to the SHE via a salt bridge and a high-resistance voltmeter, the reading you get off the voltmeter is that half-cell's E° value.

Q2. The half-equation Br₂(l) + 2e⁻ ⇌ 2Br⁻(aq) has E° = +1.09 V, while 2H⁺(aq) + 2e⁻ ⇌ H₂(g) has E° = 0.00 V. Which species is more likely to be reduced, and what does this tell you about bromine as an oxidising agent compared to H⁺?

3. Measuring Standard Electrode Potential

There are exactly three types of half-cell you need to be able to set up and analyse. In every single case, the half-cell is connected to a Standard Hydrogen Electrode via a salt bridge, and a high-resistance voltmeter reads the E° directly.

Type 1 — Metal / Metal-ion half-cell

The simplest case: a solid metal electrode dipped into a solution of its own ions. Example: Ag⁺/Ag half-cell.

Ag⁺(aq) + e⁻ ⇌ Ag(s) E° = +0.80 V

Connected to the SHE (E° = 0.00 V), the silver half-cell is more positive, so it's the positive pole, and E°_cell = (+0.80) − (0.00) = +0.80 V. Since Ag⁺ has the greater E°, it's more likely to be reduced — reduction happens at the positive electrode, oxidation at the negative electrode. This rule (reduction at positive, oxidation at negative) holds for every cell you'll meet in this chapter.

Type 2 — Non-metal / Non-metal-ion half-cell

Because a non-metal like Br₂ can't be shaped into a solid electrode, you need an inert platinum wire or foil dipped into the solution to make electrical contact — platinum doesn't react, it just carries electrons in and out. Example: Br₂/Br⁻ half-cell.

Br₂(aq) + 2e⁻ ⇌ 2Br⁻(aq) E° = +1.09 V

E°_cell = (+1.09) − (0.00) = +1.09 V. Br₂ is more positive than SHE, so it's the positive pole and gets reduced.

Type 3 — Ion / Ion half-cell

Here both species are in solution but in different oxidation states of the same element — again, a platinum electrode is needed purely to carry the electrons. Example: MnO₄⁻/Mn²⁺ half-cell (Mn goes from +7 down to +2).

MnO₄⁻(aq) + 8H⁺(aq) + 5e⁻ ⇌ Mn²⁺(aq) + 4H₂O(l) E° = +1.52 V Notice the H⁺ ions appear in the half-equation itself — they're needed to convert the oxygen in MnO₄⁻ into water, so the solution must be acidified.

E°_cell = (+1.52) − (0.00) = +1.52 V.

🌉 The Salt Bridge — what it actually does The salt bridge (commonly KNO₃ or KCl solution) completes the circuit by allowing ions — not electrons — to flow between the two half-cells, balancing the charge that builds up as electrons flow around the external wire. A metal wire can't be used as a salt bridge because metals only carry electrons, and a salt bridge specifically needs to carry ions. KCl and KNO₃ work well because chloride and nitrate ions are highly soluble, unreactive with the ions already in the half-cells, and unlikely to form precipitates that would block ion flow.

Q3. Explain why a platinum electrode is required for the MnO₄⁻/Mn²⁺ half-cell but not for the Ag⁺/Ag half-cell.

4. Conventional Cell Representation

Drawing out a full diagram of every electrochemical cell (beakers, wires, voltmeters and all) is slow. Instead, chemists use a compact shorthand called a cell diagram (or cell notation) that captures all the same information in one line.

Reading the notation Zn(s) | Zn²⁺(aq) ‖ Cu²⁺(aq) | Cu(s) | = phase boundary (solid ↔ solution)
, = same phase — two species that are both aqueous, or both gaseous, separated by a comma not a line
‖ = salt bridge
Left-hand side = oxidation (anode) | Right-hand side = reduction (cathode)

By convention, the half-cell with the more negative E° is always written on the left, and the more positive one on the right. This lines up neatly with the formula:

E°_cell = E°_right − E°_left Or equivalently, since oxidation is always on the left and reduction is always on the right: E°_cell = E°_reduction − E°_oxidation

Handling species that need a platinum electrode

When both species in a half-reaction are dissolved (no solid metal electrode), an inert platinum electrode must be shown, separated from the solution species by a phase-boundary line, e.g. for the Fe²⁺/Fe³⁺ half-cell:

Pt | Fe²⁺(aq), Fe³⁺(aq) Notice Fe²⁺ is written before Fe³⁺ — because on the left-hand (oxidation) side, Fe²⁺ is being oxidised into Fe³⁺, so the order follows the direction of the reaction: reactant, then product.

Combining this with a chlorine half-cell (Cl₂(g) + 2e⁻ ⇌ 2Cl⁻(aq)) gives the full cell diagram:

Pt | Fe²⁺(aq), Fe³⁺(aq) ‖ Cl₂(g) | 2Cl⁻(aq) | Pt

🦁 Examiner's mnemonic — "Lio the Lion goes ROOR!" Left Is Oxidation. And "ROOR" reminds you of the species order across a full cell diagram: Reduced / Oxidised (salt bridge) Oxidised / Reduced. Read left to right, the species shift from their reduced form to their oxidised form on the left, cross the salt bridge, then go from oxidised to reduced form on the right.

⚠️ Cell diagrams are NOT quantitative You'll often combine a one-electron half-reaction (like Fe²⁺/Fe³⁺) with a two-electron half-reaction (like Cl₂/Cl⁻) in the same cell diagram — and that's fine. The diagram just represents which materials and redox processes are involved; it isn't meant to show balanced electron numbers. Most textbooks balance them anyway out of habit, but don't panic if a question doesn't.

Q4. Given E°(Cu²⁺/Cu) = +0.34 V and E°(Zn²⁺/Zn) = −0.76 V, write the full conventional cell diagram and calculate E°_cell.

5. Thermodynamics & Electrode Potential

Feasibility — will the reaction actually go?

The E° values of two half-equations tell you, on paper, which way electrons "want" to flow. A reaction is described as thermodynamically feasible (or spontaneous) if, when you set it up as a cell, the overall E°_cell comes out positive.

Worked Example — Determining Feasibility

Two half-equations:

Cl₂(g) + 2e⁻ ⇌ 2Cl⁻(aq) E° = +1.36 V

Cu²⁺(aq) + 2e⁻ ⇌ Cu(s) E° = +0.34 V

Step 1 — Identify which is reduced: Cl₂ has the more positive E°, so Cl₂ gets reduced (forward direction): Cl₂(g) + 2e⁻ → 2Cl⁻(aq)

Step 2 — Identify which is oxidised: Cu²⁺/Cu has the less positive E°, so copper is oxidised (backward direction of its own equation): Cu(s) → Cu²⁺(aq) + 2e⁻

Step 3 — Combine, cancelling electrons: Cu(s) + Cl₂(g) → 2Cl⁻(aq) + Cu²⁺(aq), i.e. Cu(s) + Cl₂(g) → CuCl₂(s)

Step 4 — Calculate E°_cell: (+1.36) − (+0.34) = +1.02 V → feasible (spontaneous)

Step 5 — Check the reverse: The backward reaction would give (+0.34) − (+1.36) = −1.02 V → not feasible.

The Golden Rule A reaction is feasible when E°_cell > 0. The half-equation with the more positive E° proceeds forwards (as written, i.e. is reduced); the one with the less positive E° is reversed (is oxidised).

Entropy and the equilibrium constant, K

There's a deeper thermodynamic story here. A larger cell potential corresponds to a bigger overall change in entropy, and also to a larger equilibrium constant. You don't need to derive these, just know the relationships:

E° ∝ ΔS_total E° ∝ ln K These come from combining ΔG = −nFE°_cell with ΔG = −RT ln K. You are NOT expected to use the Faraday-based equation numerically (F isn't given in the Data Booklet) — just know the two proportionality relationships conceptually.

Why "feasible" doesn't always mean "it happens"

This is one of the most commonly-tested ideas in this whole topic, and it's genuinely important to understand deeply, not just memorise. E° values are a purely thermodynamic tool — they tell you about the energy relationship between reactants and products, nothing about the pathway between them.

Reaction rate / kinetics: a reaction might be feasible but proceed so slowly it looks like nothing is happening at all.
Activation energy: a reaction can have a large positive E°_cell (very thermodynamically favourable) but still not occur in practice because the activation energy barrier is too high. Classic example: Cu²⁺(aq) + H₂(g) → Cu(s) + 2H⁺(aq) has E°_cell = +0.34 V, which says "yes, feasible" — yet this reaction doesn't actually proceed at a noticeable rate because the reactants are kinetically stable.
Non-standard conditions: E° values are only valid at 1.00 mol dm⁻³, 298 K, 100 kPa. If concentrations change, Le Chatelier's principle shifts the equilibrium, which changes the actual electrode potential — see below.
Non-aqueous conditions: many real redox reactions don't happen in aqueous solution at all, so standard electrode potential data (which is inherently based on aqueous half-cells) simply doesn't apply.

🔬 Worked concentration example — V³⁺/V²⁺ system V³⁺(aq) + e⁻ ⇌ V²⁺(aq) E° = +0.26 V

If [V³⁺] is increased above the standard 1.0 mol dm⁻³, Le Chatelier shifts the equilibrium to the right, removing electrons from the system — this makes the electrode potential less negative (more positive).

If [V²⁺] is increased above 1.0 mol dm⁻³ instead, the equilibrium shifts to the left, adding electrons back to the system — this makes the electrode potential more negative.

Rule of thumb: increasing the concentration of the species on the LEFT of a reduction half-equation makes E more positive; increasing the concentration of the species on the RIGHT makes E more negative.

Q5. A reaction has E°_cell = +0.80 V, calculated from standard half-equations, yet when the two chemicals are mixed at room temperature in the lab, nothing visibly happens even after an hour. Explain this observation.

6. Fuel Cells

Electricity without combustion

A fuel cell is an electrochemical cell in which a fuel is continuously supplied and donates electrons at one electrode (getting oxidised), while oxygen is continuously supplied and gains electrons at the other electrode (getting reduced). As long as fuel and oxygen keep flowing in, the cell keeps generating a potential difference — unlike a normal battery, it doesn't "run down" because the reactants aren't stored inside the cell, they're fed in from outside.

Alkaline hydrogen-oxygen fuel cell

Negative electrode (anode): H₂(g) + 2OH⁻(aq) → 2H₂O(l) + 2e⁻ E° = −0.83 V
Positive electrode (cathode): O₂(g) + 2H₂O + 4e⁻ → 4OH⁻(aq) E° = +0.40 V
Overall: 2H₂(g) + O₂(g) → 2H₂O(l) E° = +1.23 V

Acidic hydrogen-oxygen fuel cell

Negative electrode (anode): H₂(g) → 2H⁺(aq) + 2e⁻ E° = 0.00 V
Positive electrode (cathode): O₂(g) + 4H⁺(aq) + 4e⁻ → 2H₂O(l) E° = +1.23 V
Overall: 2H₂(g) + O₂(g) → 2H₂O(l) E° = +1.23 V

Notice both types of fuel cell give the same overall equation and same overall E° — the electrolyte (acidic vs alkaline) only changes the mechanism (which ions carry the charge through the electrolyte), not the net chemistry.

✅ Benefits	⚠️ Risks / Problems
Water is the only product — no CO₂, no harmful emissions No combustion means all the bond energy converts directly to electrical energy instead of being lost as heat/light No high-temperature combustion means no harmful nitrogen oxides form Used on spacecraft — the water product doubles as drinking water for astronauts	Hydrogen is highly flammable — production and storage carry real safety hazards Needs very thick-walled cylinders/pipes to store hydrogen safely (economic cost) Hydrogen production is currently a by-product of the crude oil industry — a non-renewable, finite resource High energy density per gram, but low energy density per unit volume (it's a gas) — needs bulky containers vs. liquid fuels

✅ Benefits

⚠️ Risks / Problems

Water is the only product — no CO₂, no harmful emissions

No combustion means all the bond energy converts directly to electrical energy instead of being lost as heat/light

No high-temperature combustion means no harmful nitrogen oxides form

Used on spacecraft — the water product doubles as drinking water for astronauts

Hydrogen is highly flammable — production and storage carry real safety hazards

Needs very thick-walled cylinders/pipes to store hydrogen safely (economic cost)

Hydrogen production is currently a by-product of the crude oil industry — a non-renewable, finite resource

High energy density per gram, but low energy density per unit volume (it's a gas) — needs bulky containers vs. liquid fuels

🔑 Key difference from other electrochemical cells A fuel cell runs continuously as long as fuel and oxygen are supplied — energy is not stored inside the cell (unlike a rechargeable battery). Also remember: in a fuel cell, the anode is negative and the cathode is positive — the opposite convention feels intuitive to some students, so double-check it in exam answers.

Q6. Explain why hydrogen fuel cells are described as more environmentally friendly than petrol combustion engines, but also explain one significant practical limitation of hydrogen as a fuel.

7. Redox Titration Calculations

Using redox reactions to find unknown concentrations

In a redox titration, an oxidising agent is titrated against a reducing agent, with electrons transferred from one species to the other. Because many transition metal ions change colour naturally as their oxidation state changes, you often don't even need a separate indicator — the colour change itself signals the endpoint.

Potassium Manganate(VII) Titrations

MnO₄⁻ is a powerful oxidising agent (deep purple in solution) that gets reduced to pale pink/colourless Mn²⁺. It's commonly titrated against Fe²⁺, which is oxidised to Fe³⁺. Because the reaction mixture must be acidic (H⁺ is needed in the half-equation itself), the acid used must be dilute sulfuric acid — not hydrochloric acid (Cl⁻ can be oxidised by MnO₄⁻, contaminating the results) and not nitric acid (it's itself an oxidising agent that could interfere).

Worked Example — Balancing the Half-Equations

Half-equations given:

MnO₄⁻(aq) + 5e⁻ + 8H⁺(aq) → Mn²⁺(aq) + 4H₂O(l)

Fe²⁺(aq) → Fe³⁺(aq) + e⁻

Step 1 — Balance electrons: MnO₄⁻ needs 5e⁻, Fe²⁺ only loses 1e⁻, so multiply the iron equation by 5: 5Fe²⁺(aq) → 5Fe³⁺(aq) + 5e⁻

Step 2 — Add together, cancelling electrons: MnO₄⁻(aq) + 8H⁺(aq) + 5Fe²⁺(aq) → Mn²⁺(aq) + 4H₂O(l) + 5Fe³⁺(aq)

Result — mole ratio: 1 mole MnO₄⁻ : 5 moles Fe²⁺

Worked Example — Percentage of Iron in a Tablet

An iron tablet weighing 0.960 g was dissolved in dilute sulfuric acid. An average titre of 28.50 cm³ of 0.0180 mol dm⁻³ KMnO₄ solution was needed to reach the endpoint. Find the percentage by mass of iron in the tablet.

Step 1: moles of MnO₄⁻ = 0.0180 × (28.50/1000) = 5.13 × 10⁻⁴ mol

Step 2: ratio is 1 : 5 (MnO₄⁻ : Fe²⁺), so moles of Fe²⁺ = 5 × 5.13 × 10⁻⁴ = 2.565 × 10⁻³ mol

Step 3: mass of iron = 55.8 × 2.565 × 10⁻³ = 0.143127 g

Step 4: % by mass = (0.143127 / 0.960) × 100 = 14.9%

Iodine-Thiosulfate Titrations

This method determines the concentration of an oxidising agent indirectly. The oxidising agent first reacts with excess iodide ions to produce iodine, and then the iodine produced is titrated against sodium thiosulfate of known concentration:

2S₂O₃²⁻(aq) + I₂(aq) → 2I⁻(aq) + S₄O₆²⁻(aq) The brown/yellow iodine colour fades as it's converted to colourless iodide. When the solution turns pale straw-yellow, starch indicator is added — this makes the solution turn blue-black, and the endpoint is reached the instant that blue-black colour disappears completely.

Worked Example — Chlorate(I) in Household Bleach

ClO⁻(aq) + 2I⁻(aq) + 2H⁺(aq) → Cl⁻(aq) + I₂(aq) + H₂O(l). 10.0 cm³ bleach was made up to 250.0 cm³. A 25.0 cm³ portion was mixed with excess KI and acidified, then titrated against 0.05 mol dm⁻³ sodium thiosulfate — average titre 25.20 cm³.

Step 1: moles of S₂O₃²⁻ = 0.05 × (25.20/1000) = 1.26 × 10⁻³ mol

Step 2: ratio S₂O₃²⁻ : I₂ is 2:1, so moles I₂ (= moles ClO⁻) in the 25.0 cm³ sample = 1.26×10⁻³ / 2 = 6.30 × 10⁻⁴ mol

Step 3: scale up to the full 250.0 cm³: 6.30 × 10⁻⁴ × 10 = 6.30 × 10⁻³ mol ClO⁻

Step 4: this all came from 10.0 cm³ of original bleach, so 1.0 dm³ (1000 cm³) contains 6.30 × 10⁻³ × 100 = 0.630 mol dm⁻³ ClO⁻

📋 The General Method — 7 Steps for Any Redox Titration 1. Write the half-equations for oxidant and reductant
2. Deduce the overall balanced equation (balance electrons first!)
3. Calculate moles of manganate(VII)/dichromate(VI)/thiosulfate used in the titration
4. Use the mole ratio from the overall equation
5. Calculate moles of the reductant in the sample solution
6. Scale up to moles in the original solution (watch for dilution factors!)
7. Find the final concentration or percentage purity asked for

Q7. In a titration, 25.0 cm³ of Fe²⁺ solution required 22.40 cm³ of 0.0200 mol dm⁻³ KMnO₄ to reach the endpoint (ratio MnO₄⁻:Fe²⁺ = 1:5). Calculate the concentration of the original Fe²⁺ solution.

8. Redox Titration — Uncertainty

How confident can we be in a titration result?

Percentage uncertainty compares the significance of an instrument's absolute uncertainty relative to the size of the measurement taken — it is not the same thing as percentage error, which compares a result to a known literature value.

Percentage Uncertainty Formula Percentage uncertainty = (uncertainty ÷ measured value) × 100

Adding or subtracting measurements

Any time you read an instrument twice to get a single quantity — like reading a burette's initial and final volume to get a titre, or a balance's initial and final mass — you must add the two absolute uncertainties together. This is because each reading could independently be "out" by the stated uncertainty, so the combined figure could be out by up to double.

Worked Example — Why Bigger Titres Reduce Uncertainty

Case A: burette has uncertainty ±0.1 cm³ (per reading). Initial reading 0.0 cm³, final reading 5.0 cm³. Titre = 5.0 cm³.

Combined absolute uncertainty = 0.1 + 0.1 = 0.2 cm³

Percentage uncertainty = (0.2 / 5.0) × 100 = 4%

Case B: same burette, but the solution is diluted so more is needed. Initial reading 0.0 cm³, final reading 30.0 cm³. Titre = 30.0 cm³.

Percentage uncertainty = (0.2 / 30.0) × 100 = 0.67%

✅ Practical takeaway To reduce uncertainty in a titration, dilute the chemical in the burette so a larger volume (titre) is needed to reach the endpoint. The absolute uncertainty of the burette itself doesn't change — but because it's now a smaller fraction of a bigger number, the percentage uncertainty drops significantly. This is a genuinely common exam question — "suggest how the percentage uncertainty in this titration could be reduced" — and the answer is almost always some version of "use a more dilute solution so a larger titre volume is required."

Q8. A student uses a burette (uncertainty ±0.05 cm³ per reading) and records an initial reading of 0.20 cm³ and a final reading of 24.60 cm³. Calculate the percentage uncertainty in this titre.

What to Memorise

Standard Electrode Potential (E°)

The potential difference produced when a standard half-cell is connected to a standard hydrogen electrode, under standard conditions (1.00 mol dm⁻³, 298 K, 100 kPa).

Standard Hydrogen Electrode (SHE)

2H⁺(aq) + 2e⁻ ⇌ H₂(g), fixed at E° = 0.00 V, using an inert platinum electrode with H⁺ at 1.00 mol dm⁻³ and H₂ gas at 100 kPa.

E°_cell formula

E°_cell = E°_right − E°_left = E°_reduction − E°_oxidation. More positive half-cell = positive pole = gets reduced.

Feasibility rule

A reaction is thermodynamically feasible (spontaneous) only when E°_cell is positive. Says nothing about reaction rate.

Cell diagram notation

| = phase boundary, comma = same phase, ‖ = salt bridge. Oxidation always written on the left, reduction always on the right.

Salt bridge

Usually KNO₃(aq) or KCl(aq) — completes the circuit by allowing ion flow (not electron flow) and balances charge build-up.

Proportionality relationships

E° ∝ ΔS_total and E° ∝ ln K (derived from ΔG = −nFE°_cell and ΔG = −RT ln K).

Hydrogen-oxygen fuel cell overall equation

2H₂(g) + O₂(g) → 2H₂O(l), E° = +1.23 V — true for both acidic and alkaline electrolyte versions.

Manganate(VII) titration acid

Always dilute sulfuric acid — never HCl (oxidised, contaminates result) or HNO₃ (itself an oxidising agent).

Manganate(VII) balanced half-equation

MnO₄⁻(aq) + 8H⁺(aq) + 5e⁻ ⇌ Mn²⁺(aq) + 4H₂O(l). Purple → colourless/pale pink acts as its own indicator.

Iodine-thiosulfate equation

2S₂O₃²⁻(aq) + I₂(aq) → 2I⁻(aq) + S₄O₆²⁻(aq). Starch added near the endpoint (pale straw colour) — blue-black disappears at the true endpoint.

Percentage uncertainty formula

(uncertainty ÷ measured value) × 100. Add absolute uncertainties when a measurement involves two readings (e.g. burette initial + final).

Concepts Checklist

I can define oxidation and reduction using all three definitions (oxygen, hydrogen, electrons) and use OIL RIG confidently.
I can explain why standard conditions are needed to compare electrode potentials fairly.
I know the exact standard conditions: 1.00 mol dm⁻³, 298 K, 100 kPa.
I can describe the Standard Hydrogen Electrode and explain why it's assigned E° = 0.00 V.
I can identify and set up all three types of half-cell: metal/metal-ion, non-metal/non-metal-ion, and ion/ion.
I understand why a platinum electrode is needed when no solid metal is present.
I can explain the role of the salt bridge and why a metal wire can't be used instead.
I can calculate E°_cell using E°_right − E°_left and correctly identify the positive/negative poles.
I can write and interpret conventional cell diagrams, including phase boundaries and salt bridges.
I can determine whether a redox reaction is thermodynamically feasible using E°_cell.
I understand why a positive E°_cell does not guarantee a reaction will actually happen at a noticeable rate.
I can explain how changing concentration (non-standard conditions) shifts electrode potential via Le Chatelier's principle.
I know E° ∝ ΔS_total and E° ∝ ln K as proportionality relationships (not the numerical Faraday equation).
I can write the half-equations and overall equation for both acidic and alkaline hydrogen-oxygen fuel cells.
I can discuss the benefits and risks/problems of hydrogen fuel cells.
I can balance manganate(VII) half-equations and calculate percentage purity/concentration from titration data.
I understand why sulfuric acid (not HCl or HNO₃) is used to acidify manganate(VII) titrations.
I can perform iodine-thiosulfate titration calculations, including working out when to add starch indicator.
I can calculate percentage uncertainty and explain how to reduce it in a titration.
I know to add absolute uncertainties whenever two instrument readings are combined into one measurement.

Exam Tips & Common Mistakes

❌ Mistake 1 — Mixing up which pole is oxidised vs reduced Students often get confused about which electrode is positive and which reaction direction it represents. Fix it firmly: the more positive E° value means that half-reaction gets reduced, and reduction happens at the positive electrode in a voltaic cell. The less positive (or more negative) E° means that species gets oxidised, at the negative electrode. (Note: this flips for fuel cells and electrolytic cells — but for the standard voltaic cells in this chapter, positive = reduction.)

❌ Mistake 2 — Confusing "feasible" with "will definitely happen" A positive E°_cell only tells you a reaction is thermodynamically possible — examiners love to test whether you understand that kinetics (activation energy, rate) can prevent a feasible reaction from actually occurring. Always mention kinetic stability / high activation energy as the explanation when a question describes a reaction that "should" happen (positive E°) but doesn't.

❌ Mistake 3 — Wrong acid for manganate(VII) titrations Never suggest HCl (Cl⁻ ions get oxidised by MnO₄⁻, wasting some of the oxidant and giving an inaccurate titre) or HNO₃ (nitric acid is itself an oxidising agent and would interfere with the reaction). The safe, standard answer is always dilute sulfuric acid.

❌ Mistake 4 — Forgetting to balance electrons before adding half-equations Before combining two half-equations into an overall redox equation, always check the number of electrons matches on both sides — multiply one or both half-equations by the smallest whole number needed so the electrons cancel exactly. This is the single most common arithmetic slip in this whole topic.

❌ Mistake 5 — Not doubling uncertainty for burette readings Because a titre comes from two burette readings (initial and final), students often forget to add the uncertainties together and just quote the single-reading uncertainty. Always double it (or add the two absolute uncertainties) when a measurement is derived by subtraction of two readings.

What examiners specifically look for

State symbols in every half-equation and cell diagram — marks are frequently lost for missing (aq), (s), (g), (l).
Correct direction of arrows — half-equations are written with ⇌ (equilibrium) when quoted as standard E° data, but with a single → when describing the actual direction a specific reaction proceeds.
Full, precise definitions — for E°, examiners want "standard half-cell," "standard hydrogen electrode," AND "standard conditions" all mentioned, not just a vague "voltage produced by two half-cells."
Explaining, not just stating — if asked "explain" why a value changes, restating the value isn't enough; you must reference Le Chatelier's principle, electron transfer, or kinetics/activation energy as appropriate.
Significant figures in titration answers — match your final answer's precision to the least precise data given (usually 3 significant figures for typical titration data).

🔓 Read the full Redox Equilibria note — free You're seeing the preview · free account, no card needed

Also in the full note

1. Reduction & Oxidation Recap
5. Thermodynamics & Electrode Potential
Exam Tips & Common Mistakes

What's inside

📖 Revision notes 🎯 Learn mode ✦ AI flashcards ✓ Instant AI marking 🧊 3D explorers 🧪 Experiments & simulations 📈 Progress tracking

Read the full Redox Equilibria notes free

That's the preview — create a free account to read the rest, plus flashcards and practice questions with instant AI marking. No credit card.

Unlock the full notes free →