What you'll learn
Redox reactions — reactions involving the transfer of electrons — power batteries, rust iron, extract metals and keep water clean. Electrochemistry studies how these electron transfers can be harnessed to produce electricity, or driven by electricity to produce chemicals. This guide covers oxidation and reduction in terms of electrons and oxidation numbers, constructing and balancing half-equations and full redox equations, redox titrations with potassium manganate(VII) and sodium thiosulfate, standard electrode potentials (E°), the standard hydrogen electrode, electrochemical cells and cell e.m.f., predicting the feasibility of reactions, the effect of concentration on electrode potentials, and applications including fuel cells, batteries, corrosion and electrolysis. Caribbean examples include the corrosion of steel in salty coastal air and the electrolytic extraction of aluminium from alumina.
CAPE papers regularly ask you to balance redox equations, calculate cell e.m.f., predict whether reactions will occur, and perform redox titration calculations. Systematic methods make these reliable.
Key terms and definitions
Oxidation — Loss of electrons, or an increase in oxidation number.
Reduction — Gain of electrons, or a decrease in oxidation number.
Oxidising agent — A species that accepts electrons (and is reduced).
Reducing agent — A species that donates electrons (and is oxidised).
Oxidation number — The charge an atom would have if all its bonds were ionic.
Half-equation — An equation showing either the oxidation or the reduction part of a redox reaction, with electrons.
Standard electrode potential, E° — The potential of a half-cell measured relative to the standard hydrogen electrode under standard conditions (298 K, 1 mol dm⁻³ solutions, 1 atm gases).
Standard hydrogen electrode (SHE) — Hydrogen gas at 1 atm bubbled over platinum in 1 mol dm⁻³ H⁺ at 298 K; assigned E° = 0.00 V.
Electromotive force (e.m.f.), E°cell — The potential difference between two half-cells: E°cell = E°(positive electrode) − E°(negative electrode).
Salt bridge — Connects two half-cells, allowing ions to move to complete the circuit without mixing the solutions.
Electrolysis — Using electricity to drive a non-spontaneous redox reaction.
Core concepts
Oxidation numbers
Rules:
- Uncombined elements have oxidation number 0.
- In a simple ion, the oxidation number equals the charge.
- The sum of oxidation numbers in a compound is 0; in an ion it equals the ion's charge.
- Oxygen is usually −2 (−1 in peroxides); hydrogen is usually +1 (−1 in metal hydrides).
- Fluorine is always −1; Group I metals +1; Group II +2.
Example: in KMnO₄, Mn = +7; in Cr₂O₇²⁻, Cr = +6; in S₂O₃²⁻, S = +2 (average).
Balancing redox equations in acidic solution
- Write the skeleton half-equation.
- Balance atoms other than O and H.
- Balance O by adding H₂O.
- Balance H by adding H⁺.
- Balance charge by adding electrons.
- Multiply half-equations so electrons cancel, then add.
Manganate(VII): MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O Iron(II): Fe²⁺ → Fe³⁺ + e⁻ Overall: MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O
Dichromate(VI): Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O
Redox titrations
Potassium manganate(VII) titrations are self-indicating: purple MnO₄⁻ is decolorised as it is reduced to nearly colourless Mn²⁺, so the end point is the first permanent pale pink colour. The solution must be acidified with dilute sulfuric acid (not HCl, which would be oxidised).
Iodine–thiosulfate titrations: iodine is titrated with sodium thiosulfate: I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻ Starch is added near the end point; the blue-black colour disappears at the end point. These titrations are used to find the concentration of oxidising agents (which first liberate iodine from iodide), such as chlorine in bleach.
Standard electrode potentials
A half-cell consists of an element in contact with a solution of its ions (or two ions of the same element with a platinum electrode). Its E° is measured against the standard hydrogen electrode.
- A more positive E° means the species on the left of the half-equation is a stronger oxidising agent (more easily reduced).
- A more negative E° means the species on the right is a stronger reducing agent (more easily oxidised).
| Half-equation | E° / V |
|---|---|
| Zn²⁺ + 2e⁻ ⇌ Zn | −0.76 |
| Fe²⁺ + 2e⁻ ⇌ Fe | −0.44 |
| 2H⁺ + 2e⁻ ⇌ H₂ | 0.00 |
| Cu²⁺ + 2e⁻ ⇌ Cu | +0.34 |
| Ag⁺ + e⁻ ⇌ Ag | +0.80 |
| Cl₂ + 2e⁻ ⇌ 2Cl⁻ | +1.36 |
| MnO₄⁻ + 8H⁺ + 5e⁻ ⇌ Mn²⁺ + 4H₂O | +1.51 |
Electrochemical cells
In a cell made from two half-cells, the half-cell with the more positive E° is the positive electrode (reduction occurs there); the one with the more negative E° is the negative electrode (oxidation occurs). Electrons flow through the external wire from negative to positive.
E°cell = E°(positive) − E°(negative)
For a zinc–copper (Daniell) cell: E°cell = +0.34 − (−0.76) = +1.10 V.
Predicting feasibility
A reaction is feasible (thermodynamically possible) if E°cell is positive. However, E° predictions do not indicate rate: a feasible reaction may be too slow to observe because of a high activation energy. Predictions also apply only under standard conditions.
Effect of concentration
Changing concentration shifts the half-cell equilibrium. Increasing [Cu²⁺] in Cu²⁺ + 2e⁻ ⇌ Cu shifts the equilibrium to the right, making E more positive. This follows Le Chatelier's principle.
Batteries and fuel cells
Batteries are portable electrochemical cells: alkaline zinc–manganese cells are non-rechargeable; lithium-ion and lead–acid batteries are rechargeable because their reactions can be reversed.
Hydrogen–oxygen fuel cells react hydrogen and oxygen to produce electricity, with only water as a product. In alkaline conditions:
- Negative electrode: H₂ + 2OH⁻ → 2H₂O + 2e⁻
- Positive electrode: O₂ + 2H₂O + 4e⁻ → 4OH⁻
Advantages: no carbon dioxide at the point of use, high efficiency. Disadvantages: hydrogen is difficult to store and is often produced from fossil fuels.
Corrosion of iron
Rusting is an electrochemical process requiring oxygen and water:
- Fe → Fe²⁺ + 2e⁻ (oxidation at anodic areas)
- O₂ + 2H₂O + 4e⁻ → 4OH⁻ (reduction at cathodic areas)
- Fe²⁺ and OH⁻ form Fe(OH)₂, which is oxidised further to hydrated iron(III) oxide (rust).
Salt water speeds rusting because dissolved ions increase conductivity — a major problem for vehicles, bridges and boats in the Caribbean's coastal environment. Prevention includes painting, oiling, galvanising (zinc coating), and sacrificial protection, where a more reactive metal such as zinc or magnesium (more negative E°) is attached and corrodes instead of the iron.
Electrolysis
In electrolysis, electrical energy drives a non-spontaneous reaction. Cations move to the cathode (reduction) and anions to the anode (oxidation). Aluminium is extracted by electrolysis of alumina dissolved in molten cryolite: at the cathode, Al³⁺ + 3e⁻ → Al; at the anode, oxide ions are oxidised to oxygen, which reacts with the carbon anodes.
Worked examples
Example 1: Oxidation numbers
Find the oxidation number of chromium in K₂Cr₂O₇.
2(+1) + 2(Cr) + 7(−2) = 0 → 2 + 2Cr − 14 = 0 → Cr = +6.
Example 2: Cell e.m.f. and feasibility
Will silver ions oxidise copper metal? Calculate E°cell.
Ag⁺/Ag (+0.80) is more positive than Cu²⁺/Cu (+0.34), so Ag⁺ is reduced and Cu oxidised: 2Ag⁺ + Cu → 2Ag + Cu²⁺. E°cell = 0.80 − 0.34 = +0.46 V, positive, so the reaction is feasible.
Example 3: Manganate(VII) titration
25.0 cm³ of an Fe²⁺ solution requires 20.0 cm³ of 0.0200 mol dm⁻³ KMnO₄. Find [Fe²⁺].
Moles MnO₄⁻ = 0.0200 × 0.0200 = 4.00 × 10⁻⁴ mol. Ratio MnO₄⁻ : Fe²⁺ = 1 : 5, so moles Fe²⁺ = 2.00 × 10⁻³ mol. [Fe²⁺] = 2.00 × 10⁻³ ÷ 0.0250 = 0.0800 mol dm⁻³.
Example 4: Sacrificial protection
Explain why zinc blocks attached to a steel ship's hull reduce rusting.
Zinc has a more negative E° (−0.76 V) than iron (−0.44 V), so zinc is oxidised in preference to iron. Electrons from the zinc flow to the iron, preventing iron from being oxidised; the zinc corrodes instead and is replaced periodically.
Example 5: Available chlorine in bleach
10.0 cm³ of household bleach was added to excess acidified potassium iodide. The iodine released required 18.0 cm³ of 0.100 mol dm⁻³ sodium thiosulfate. Calculate the concentration of chlorate(I) ions in the bleach.
ClO⁻ + 2I⁻ + 2H⁺ → Cl⁻ + I₂ + H₂O I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻
Moles S₂O₃²⁻ = 0.100 × 0.0180 = 1.80 × 10⁻³ mol. Moles I₂ = 1.80 × 10⁻³ ÷ 2 = 9.0 × 10⁻⁴ mol = moles ClO⁻. [ClO⁻] = 9.0 × 10⁻⁴ ÷ 0.0100 = 0.090 mol dm⁻³.
Example 6: Writing a half-equation for dichromate
Construct the half-equation for the reduction of Cr₂O₇²⁻ to Cr³⁺ in acid.
- Cr₂O₇²⁻ → 2Cr³⁺ (balance Cr).
- Cr₂O₇²⁻ → 2Cr³⁺ + 7H₂O (balance O).
- Cr₂O₇²⁻ + 14H⁺ → 2Cr³⁺ + 7H₂O (balance H).
- Charge on left: −2 + 14 = +12; on right: +6. Add 6e⁻ to the left: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O
Example 7: Predicting electrolysis products
Predict the products at each electrode when aqueous copper(II) sulfate is electrolysed with inert electrodes.
At the cathode, Cu²⁺ is reduced more readily than water (Cu²⁺/Cu has a more positive E° than H₂O/H₂): Cu²⁺ + 2e⁻ → Cu, so copper is deposited. At the anode, sulfate ions are not oxidised; water is oxidised instead: 2H₂O → O₂ + 4H⁺ + 4e⁻, so oxygen is given off and the solution becomes acidic. The blue colour fades as Cu²⁺ ions are removed.
Example 8: A feasible reaction that does not happen
E° values predict that hydrogen should react with oxygen at room temperature, yet a mixture of the two gases can be stored safely. Explain.
The positive E°cell shows the reaction is thermodynamically feasible, but the activation energy is very high, so the rate at room temperature is negligible. A spark or catalyst is needed to start the reaction — E° values say nothing about the rate of a reaction.
Electrochemistry at a glance
| Feature | Electrochemical cell | Electrolysis |
|---|---|---|
| Energy change | chemical → electrical | electrical → chemical |
| Reaction | spontaneous (E°cell positive) | non-spontaneous, driven by power supply |
| Negative electrode | anode (oxidation) | cathode (reduction) |
| Positive electrode | cathode (reduction) | anode (oxidation) |
Remember: oxidation always happens at the anode, and reduction always at the cathode, whatever the type of cell.
Common mistakes and how to avoid them
- Mixing up oxidation and reduction — remember OIL RIG (Oxidation Is Loss, Reduction Is Gain of electrons).
- Not balancing charge with electrons in half-equations.
- Using HCl to acidify manganate(VII) titrations.
- Subtracting electrode potentials the wrong way round.
- Thinking a positive E°cell guarantees a fast reaction.
- Multiplying E° values when balancing equations — E° values are not multiplied.
How this links to your Internal Assessment
Redox titrations are ideal SBA practicals: determining the iron content of iron tablets with potassium manganate(VII), the concentration of hydrogen peroxide, or the available chlorine in bleach using iodine–thiosulfate titrations. You could also build simple electrochemical cells from different metals and measure their voltages, or investigate factors affecting rusting of iron nails in salt water. Strong reports include accurate titres, clear calculations using mole ratios from balanced equations, and evaluation of sources of error such as end-point judgement and air oxidation of Fe²⁺.
Exam technique for Redox reactions and electrochemistry
- Assign oxidation numbers systematically and show your working.
- Balance half-equations step by step: atoms, O (H₂O), H (H⁺), charge (e⁻).
- Write E°cell = E°(positive) − E°(negative) and state whether the reaction is feasible.
- In titration calculations, state the mole ratio from the balanced equation.
- Mention both thermodynamic feasibility and kinetic limitations when asked to evaluate predictions.
Quick revision summary
- Oxidation = electron loss / oxidation number increase; reduction = gain / decrease.
- Half-equations balanced with H₂O, H⁺ and e⁻; combine so electrons cancel.
- KMnO₄ titrations are self-indicating (acidify with H₂SO₄); iodine–thiosulfate uses starch.
- E° measured against SHE (0.00 V); more positive = stronger oxidising agent.
- E°cell = E°(positive) − E°(negative); positive means feasible (but says nothing about rate).
- Fuel cells, batteries, rusting (needs O₂ and water; salt accelerates), sacrificial protection and electrolysis of alumina.