Feasibility of redox reactionsEdexcel International A Level Chemistry: Revision notes

A redox reaction is thermodynamically feasible under standard conditions if its E°cell is positive. To test a reaction:

1. Split it into two half-equations and find their E° values.
2. The species being reduced has the right-hand (positive) half-cell; the species being oxidised has the left-hand half-cell.
3. E°cell = E°(reduced) − E°(oxidised); if positive, the reaction is feasible.

Example: Cl₂ + 2Br⁻ → 2Cl⁻ + Br₂. Cl₂ is reduced (+1.36 V) and Br⁻ is oxidised (+1.09 V): E°cell = +1.36 − 1.09 = +0.27 V, so it is feasible. The reverse reaction would have E°cell = −0.27 V and is not feasible.

A stronger oxidising agent (more positive E°) will oxidise the reduced form of any half-cell with a less positive E°.

Disproportionation is a reaction in which the same species is both oxidised and reduced.

Copper(I) in aqueous solution: 2Cu⁺(aq) → Cu²⁺(aq) + Cu(s). Cu goes from +1 to +2 (oxidised) and from +1 to 0 (reduced).

• Cu⁺ → Cu has E° = +0.52 V (this is the reduction)
• Cu²⁺ → Cu⁺ has E° = +0.15 V (the reverse is the oxidation)
• E°cell = (+0.52) − (+0.15) = +0.37 V, so it is feasible

An intermediate oxidation state disproportionates when the E° for it being reduced is more positive than the E° for it being oxidised. Other examples: hydrogen peroxide into water and oxygen, and chlorine in cold aqueous alkali.

E°cell is directly proportional to the total entropy change and to ln K:

• ΔS_total = nFE°cell / T
• ln K = nFE°cell / (RT)

where n is the number of electrons transferred, F = 96 500 C mol⁻¹ and R = 8.31 J K⁻¹ mol⁻¹.

Worked example: 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂, E°cell = +0.23 V, n = 2, T = 298 K.

ΔS_total = (2 × 96 500 × 0.23) / 298 = +149 J K⁻¹ mol⁻¹

ln K = (2 × 96 500 × 0.23) / (8.31 × 298) = 17.9, so K ≈ 6.1 × 10⁷.

A positive E°cell means ΔS_total is positive and K > 1; a negative E°cell means ΔS_total is negative and K < 1.

E°cell tells you whether a reaction is feasible, not whether it will occur at an observable rate. A reaction with a positive E°cell may have a very high activation energy, so very few molecules have enough energy and the reaction is negligibly slow. The species is then kinetically stable (inert).

Example: acidified MnO₄⁻ (+1.51 V) should oxidise water (O₂|H₂O +1.23 V; E°cell = +0.28 V), but manganate(VII) solutions can be stored for weeks because the reaction is extremely slow at room temperature.

E° values apply only at 298 K, 100 kPa and 1.00 mol dm⁻³. In other conditions the electrode equilibria shift and the real emf differs from E°cell. A small E°cell (about ±0.3 V or less) can change sign.

Example: MnO₂ + 4H⁺ + 2Cl⁻ → Mn²⁺ + Cl₂ + 2H₂O has E°cell = (+1.23) − (+1.36) = −0.13 V. In concentrated hydrochloric acid, the high [H⁺] shifts the MnO₂ equilibrium right (more positive) and the high [Cl⁻] shifts the Cl₂ equilibrium left (less positive), so the reaction occurs and chlorine is made.