Redox and electrode potentials — OCR A-Level Chemistry

This section heading introduces the redox content of 5.2.3, where you build on the oxidation-number and electron-transfer ideas met earlier in 2.1.5 Redox and apply them to electrode potentials. Read the specification statements under the heading as a connected sequence: first secure the language of oxidation and reduction in terms of electron loss and gain and changes in oxidation number, then use that language to identify oxidising and reducing agents, and then apply it to half-equations and to cells built from two half-cells. Treat the heading itself as an organiser rather than a separate examinable fact. When you revise, write each half-equation as a reduction with its electrode potential, check that oxidation numbers balance as well as atoms and charge, and practise deciding which species is oxidised and which is reduced in a full equation.

An oxidising agent is a species that oxidises another species and is itself reduced, so it gains electrons. A reducing agent is a species that reduces another species and is itself oxidised, so it loses electrons. The two definitions are mirror images, and the key is to track electron transfer: the species that accepts electrons is the oxidising agent, and the species that donates electrons is the reducing agent. For example, in the reaction between zinc and copper(II) ions, zinc atoms lose electrons and are oxidised, so zinc acts as the reducing agent, while copper(II) ions gain electrons and are reduced, so they act as the oxidising agent. Oxidation numbers can confirm this: the oxidising agent's oxidation number falls, and the reducing agent's oxidation number rises.

Redox equations balance both atoms and charge. Two routes are examined. Route one uses oxidation numbers: assign oxidation states to each atom, identify which elements change, and balance the change in oxidation number by equalising the total increase in oxidation number with the total decrease. Route two uses half-equations: write separate oxidation and reduction half-equations, balance atoms other than O and H, balance O by adding H₂O, balance H by adding H⁺ (in acidic conditions), balance charge by adding electrons, then multiply each half-equation so the electrons cancel before adding them. For example, MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O combined with Fe²⁺ → Fe³⁺ + e⁻ gives MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺. Check that atoms and total charge balance on both sides.

Electron transfer underlies redox. An oxidising agent accepts electrons and is itself reduced; a reducing agent donates electrons and is itself oxidised. To interpret or predict whether a reaction occurs, compare the standard electrode potentials of the two half-cells. The half-equation with the more positive E⦵ value proceeds as a reduction (forward), and the other is reversed as an oxidation. For example, a Cl₂/Cl⁻ half-cell (E⦵ = +1.36 V) will oxidise a Br⁻/Br₂ half-cell (E⦵ = +1.07 V), so chlorine displaces bromine. The cell emf is the difference between the two E⦵ values and is positive for a feasible reaction. Identify which species is oxidised and which is reduced, and track the electrons transferred.

This row is a course heading, not an assessed answer target. It signals that you must study the technique of redox titration as a practical method: a solution of known concentration (the titrant) is added from a burette to a measured volume of the analyte until the oxidising and reducing species have reacted in their exact stoichiometric ratio. You learn to choose an indicator or use self-indicating systems, read a burette to the bottom of the meniscus, and calculate an unknown concentration from moles. Work through the linked practical procedures, including the iron(II)/manganate(VII) and iodine/thiosulfate systems, and practise the mole calculations that follow each titration.

You must know the practical techniques for redox titrations, applied to two systems. In the Fe²⁺/MnO₄⁻ titration, acidified potassium manganate(VII) is added from a burette to a measured volume of iron(II) solution; MnO₄⁻ is self-indicating, turning from purple to colourless as it is reduced to Mn²⁺, with the first permanent pink marking the end point. In the I₂/S₂O₃²⁻ titration, iodine is titrated with sodium thiosulfate using starch indicator, which gives a blue-black colour that disappears at the end point. Both rely on accurate burette readings, concordant titres and mole-ratio calculations.

Redox titration calculations turn burette readings and solution concentrations into an amount of substance. Work in moles throughout: calculate the moles of the standard solution from concentration × volume in dm³, use the balanced redox equation to find the mole ratio to the analyte, then convert to concentration, mass or percentage purity as asked. A structured calculation is broken into labelled steps; a non-structured one gives only the raw data, so you must plan the route yourself. Example: 24.20 cm³ of 0.0200 mol dm⁻³ KMnO₄ reacts with 25.00 cm³ of Fe²⁺ solution. Moles MnO₄⁻ = 0.0200 × 0.02420 = 4.84 × 10⁻⁴ mol. The equation MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O gives Fe²⁺ = 5 × 4.84 × 10⁻⁴ = 2.42 × 10⁻³ mol, so [Fe²⁺] = 2.42 × 10⁻³ ÷ 0.02500 = 0.0968 mol dm⁻³.

These two redox systems are the standard titrations of the topic. In Fe²⁺/MnO₄⁻, acidified manganate(VII) is the self-indicating titrant: MnO₄⁻ is purple and is reduced to almost colourless Mn²⁺, so the end-point is the first permanent pink tinge as excess MnO₄⁻ appears. The equation is MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O. In I₂/S₂O₃²⁻, iodine is titrated with thiosulfate: I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻. Iodine solution is brown; starch indicator gives a blue-black complex, and the end-point is when this blue-black colour disappears as iodine is used up. Recognise the colour changes, the mole ratios and the role of each reagent.

Non-familiar redox systems test whether you can transfer the standard electrode-potential method to a cell you have not met in the specification. Read the description of the two half-cells, identify which species is oxidised and which is reduced, and write each half-equation with the correct number of electrons and charge balance. Then combine them so electrons cancel, giving the overall cell equation. Use E° values to decide the spontaneous direction: the half-cell with the more positive E° is reduced, and the other is oxidised. Calculate E°cell = E°(reduction) − E°(oxidation), keeping signs. Check that the cell diagram, ion charges and state symbols are consistent, and that any unfamiliar ion, such as a complex or an oxoanion, is balanced for atoms, charge and electrons before you choose an answer.

This guided reading covers the electrode-potential content of section 5.2.3. Work through your textbook or course notes on standard electrode potentials, focusing on how a half-cell is set up under standard conditions (298 K, 1 mol dm⁻³ solutions, 100 kPa gases), how the standard hydrogen electrode is defined with E° = 0.00 V, and how E° values are measured and tabulated as reduction potentials. Note the convention that the more positive E° half-cell is reduced and the more negative is oxidised, and how E°cell = E°(reduction) − E°(oxidation) follows. Read worked examples of unfamiliar redox systems and practise writing half-equations and combining them. Use the reading to build a method you can apply to any cell, not to memorise a fixed answer.

A standard electrode (redox) potential, E⦵, measures the tendency of a half-cell to gain electrons relative to the standard hydrogen electrode (SHE), which is assigned E⦵ = 0.00 V. Standard conditions are 298 K, a solution concentration of 1 mol dm⁻³ for each aqueous species, and a gas pressure of 100 kPa (1 bar) for any gaseous species; a solid electrode such as platinum is used where the half-cell has no metal of its own. To measure E⦵, connect the half-cell to the SHE with a high-resistance voltmeter and a salt bridge, keeping the half-cell under those standard conditions. The sign of E⦵ shows the direction of electron flow: a positive value means the half-cell is reduced and the SHE is oxidised.

Measuring a cell potential means building a cell from two half-cells, each dipping into its own solution, joined by a salt bridge (often filter paper soaked in potassium nitrate) that completes the circuit without mixing the solutions. Clean metal electrodes are dipped into solutions of their own ions at known concentration; where a half-cell has no metal, platinum is used. A high-resistance voltmeter connects the two electrodes and reads the e.m.f. under near-zero current so the cell is close to equilibrium. Standard conditions (298 K, 1 mol dm⁻³ solutions, 100 kPa gases) give E⦵cell. The reading is the difference between the two electrode potentials, with the more positive couple on the right.

This clause describes the first common type of half-cell: a metal, or a non-metal, dipping into a solution containing its own ions. A metal electrode such as a copper rod in copper(II) sulfate solution, or a zinc rod in zinc sulfate solution, sets up an equilibrium between the solid element and its aqueous ions, for example Cu²⁺(aq) + 2e⁻ ⇌ Cu(s). A non-metal example is a platinum electrode passing chlorine gas into a solution of chloride ions, where Cl₂(g) + 2e⁻ ⇌ 2Cl⁻(aq) applies. The platinum is inert and only conducts; the reacting species are the gas and its ions. In each case the half-cell provides a potential because the forward and reverse reactions reach a dynamic equilibrium at the electrode surface. Recognise that the element and its ions must be the same substance, and that the aqueous ions are supplied by a solution of a soluble salt of that element.

This clause describes a redox half-cell in which two ions of the same element, differing in oxidation state, are both present in solution and contact an inert platinum electrode. A typical example is a mixture of iron(II) and iron(III) ions, where Fe³⁺(aq) + e⁻ ⇌ Fe²⁺(aq) sets the potential. Another is a mixture of tin(II) and tin(IV) ions. The platinum does not react; it provides a conducting surface where electron transfer between the two oxidation states occurs and equilibrium is established. Both ions must be in the same solution so they can interconvert at the electrode. Recognise that the two species are the same element, that they differ in oxidation state, and that the electrode is inert platinum. The potential depends on the ratio of the two ion concentrations.

A standard cell potential is found by combining two standard electrode potentials, each measured under standard conditions against the standard hydrogen electrode. Identify which half-cell is the positive electrode (the one with the more positive E⦵ value) and which is the negative electrode. Then apply E⦵cell = E⦵(positive electrode) − E⦵(negative electrode). For example, with E⦵(Cu²⁺/Cu) = +0.34 V and E⦵(Zn²⁺/Zn) = −0.76 V, the cell potential is +0.34 − (−0.76) = +1.10 V. Keep the sign of each value, subtract the more negative from the more positive, and give the answer in volts with the correct sign.

A reaction is predicted to be feasible when the standard cell potential is positive, because a positive E⦵cell corresponds to a negative standard Gibbs energy change. Calculate E⦵cell from the two relevant standard electrode potentials and check its sign. A positive value predicts the reaction is thermodynamically feasible under standard conditions. The prediction has limitations: a positive E⦵cell says nothing about rate, so a reaction may be feasible but kinetically very slow (high activation energy), and changing concentrations from standard values alters the actual cell potential, so feasibility under non-standard conditions may differ.

This row is a course-guidance heading within redox and electrode potentials, not an assessed answer target. It signals that you should read the storage-cell and fuel-cell material alongside the electrode-potential work you have already met: standard electrode potentials, cell diagrams and the link between E°cell and spontaneity. As you read, connect each device to the underlying redox change. In a storage cell such as a lithium-ion or lead–acid battery the electrode reactions are reversible, so the cell can be recharged by driving the reverse reaction with an external supply. In a fuel cell, reactants are supplied continuously and the overall reaction is not reversed for storage. Note which electrode is oxidised and which is reduced, and how the sign of the electrode potential identifies the negative electrode.

This criterion asks you to apply electrode-potential principles to modern storage cells such as lithium-ion batteries. For a cell, write the two half-equations, identify which electrode is oxidised and which is reduced, and combine the standard electrode potentials to give E°cell. The more positive (less negative) electrode potential marks the positive electrode where reduction occurs; the more negative marks the negative electrode where oxidation occurs. A positive E°cell indicates the discharge reaction is feasible under standard conditions. In a rechargeable storage cell the electrode reactions are reversible, so an external supply drives the reverse reaction to recharge it. Questions may ask you to select the correct half-equation, the correct electrode polarity, or the sign of E°cell.

A fuel cell converts chemical energy from the reaction of a fuel with oxygen directly into electrical energy, producing a voltage across two electrodes without combustion. In a hydrogen–oxygen fuel cell with an alkaline electrolyte, hydrogen is oxidised at the negative electrode (anode): H₂ + 2OH⁻ → 2H₂O + 2e⁻. Oxygen is reduced at the positive electrode (cathode): O₂ + 2H₂O + 4e⁻ → 4OH⁻. Electrons flow through the external circuit from anode to cathode, doing electrical work, while ions move through the electrolyte to complete the circuit. The overall reaction is 2H₂ + O₂ → 2H₂O. The cell keeps producing a voltage while fuel and oxygen are supplied, unlike a secondary cell that must be recharged.