Part 2: Feasibility, conditions and disproportionation

Use electrode potentials to predict a direction, then state the conditions and kinetic limits of that prediction. Connect cell emf to entropy and equilibrium.

Read the electrochemical series as reduction tendencies

The electrochemical series lists reduction half-equations and their standard potentials. At the more positive end, the species on the left is more readily reduced and is a stronger oxidising agent under standard conditions. At the more negative end, the species on the right is more readily oxidised and is a stronger reducing agent. Always name the actual species: E°(Fe³⁺/Fe²⁺) describes Fe³⁺ accepting an electron, not metallic iron.

To test a proposed reaction, write the required oxidation and reduction, use the corresponding reduction-potential values, then calculate E°reduction − E°oxidation. A positive result predicts thermodynamic feasibility for that stated direction under standard conditions. A negative value predicts the reverse is favoured under those conditions. Do not choose a positive number first and then accidentally discuss the reverse chemical change.

Work from species to half-equations to a conclusion

Can chlorine oxidise Fe²⁺ to Fe³⁺? Chlorine must be reduced: Cl₂ + 2e⁻ → 2Cl⁻. Iron(II) must be oxidised: Fe²⁺ → Fe³⁺ + e⁻. Double the iron half-equation, cancel electrons and calculate 1.36 − 0.77 = +0.59 V. Therefore the proposed oxidation is thermodynamically feasible under standard conditions.

If given several successive oxidation states, test each reduction step. A reagent may reduce an element part of the way down a sequence but fail to reduce the next state. This is why naming the strongest-looking oxidant or quoting one positive emf cannot establish the final product. Topic 15 applies this to vanadium and chromium.

Positive emf connects to total entropy and equilibrium

For a specified balanced reaction at fixed temperature, E°cell is directly proportional to the standard total entropy change and to ln K. A positive E°cell corresponds to positive ΔStotal° and K greater than 1; a negative value corresponds to negative ΔStotal° and K less than 1. A larger positive value indicates a stronger equilibrium preference for products for a fixed electron-transfer number.

The quantitative bridge is ΔG° = −nFE°cell, ΔStotal° = nFE°cell/T and ln K = nFE°cell/(RT), where n is the electron transfer in the balanced reaction and F is the Faraday constant. These equations explain the proportionalities; detailed electrochemical calculations using them are an extension unless a question provides the needed relationship. ΔStotal includes surroundings as well as reacting substances, so it is not the same as ΔSsystem.

E°cell = 0 means K = 1, not equal concentrations of every species. As a real cell discharges, its reaction quotient changes until actual Ecell approaches zero at equilibrium. A large K makes products strongly favoured; it does not establish how fast they appear.

Non-standard conditions and kinetic barriers are different limitations

Changing concentration, gas pressure or temperature can change actual electrode potentials. For Cu²⁺ + 2e⁻ ⇌ Cu, increasing [Cu²⁺] increases the tendency for reduction and makes this electrode potential more positive. In Zn + Cu²⁺ → Zn²⁺ + Cu, increasing [Zn²⁺] or lowering [Cu²⁺] reduces the forward driving force and emf. Pure solid amounts do not enter this concentration argument provided those phases remain present.

Acid concentration matters when H⁺ occurs in a half-equation. Increasing [H⁺] favours the reduction side of Cr₂O₇²⁻ + 14H⁺ + 6e⁻ ⇌ 2Cr³⁺ + 7H₂O. Near-zero E°cell predictions are particularly sensitive to changed conditions. Temperature effects require the thermodynamic data: do not assert that every exothermic cell has a lower voltage at higher temperature solely from Le Chatelier’s principle. The temperature dependence of E° involves the reaction’s entropy change.

Kinetic inhibition means a favourable reaction has a high activation-energy barrier and proceeds too slowly to observe. A catalyst can accelerate it but does not change E°cell or K. Non-standard conditions instead alter the thermodynamic driving force itself. Passivating oxide layers can also prevent an expected metal reaction by blocking contact; a bare-metal prediction does not describe an inaccessible surface.

Use one oxidation state in both roles

In disproportionation, the same element in one initial oxidation state is both oxidised and reduced. The starting species therefore appears as the reduced member of one pair and the oxidised member of another. Use the reduction that consumes the starting state, and reverse the other half-equation to oxidise it.

For aqueous Cu⁺, Cu⁺ + e⁻ ⇌ Cu has E° = +0.52 V and Cu²⁺ + e⁻ ⇌ Cu⁺ has E° = +0.15 V. The first runs as reduction and the second in reverse: E°cell = 0.52 − 0.15 = +0.37 V. Adding gives 2Cu⁺ → Cu²⁺ + Cu. Copper changes from +1 to both +2 and 0.

This predicts the behaviour of the specified aqueous species under the stated conditions. It does not prove all copper(I) compounds are unstable. Precipitation, lattice energies and ligand binding can stabilise copper(I), so a sparingly soluble Cu(I) salt cannot automatically be treated as a standard 1 M Cu⁺ solution. Similarly, a colour change alone cannot establish disproportionation: identify both oxidation-state changes and balance atoms and charge.

Quick checks

Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official Edexcel mark allocations.

Sources