Redox and Electrode Potentials

A-Level Chemistry · Redox Chemistry and Electrochemistry

Redox reactions

Redox = reduction and oxidation happening together.

Oxidation = loss of electrons (oxidation number increases).
Reduction = gain of electrons (oxidation number decreases).
OIL RIG: Oxidation Is Loss, Reduction Is Gain (of electrons).
An oxidising agent accepts electrons (is itself reduced); a reducing agent donates electrons (is itself oxidised).

Oxidation numbers (rules)

Uncombined elements = 0; simple ions = the charge.
O is usually −2 (−1 in peroxides); H is usually +1 (−1 in metal hydrides).
Oxidation numbers in a neutral compound sum to 0; in an ion, to the ion's charge.

Half-equations

Split a redox reaction into oxidation and reduction half-equations, balancing atoms then electrons. Combine so the electrons cancel.

Example: Zn → Zn²⁺ + 2e⁻ (oxidation) and Cu²⁺ + 2e⁻ → Cu (reduction).

Electrode potentials

When a metal is placed in a solution of its ions, an equilibrium sets up and a potential develops. This is measured relative to the standard hydrogen electrode (defined as 0.00 V) under standard conditions.

The more negative E⦵, the more easily the species is oxidised (better reducing agent).
The more positive E⦵, the more easily it is reduced (better oxidising agent).

Electrochemical cells and EMF

Connecting two half-cells makes a cell with an EMF:

E⦵cell = E⦵(positive electrode) − E⦵(negative electrode)
       = E⦵(reduction) − E⦵(oxidation)

A positive E⦵cell means the reaction is feasible (spontaneous).
Electrons flow from the more negative electrode (oxidation) to the more positive (reduction).

Predicting feasibility (and its limits)

Compare E⦵ values: the species with the more positive E⦵ is reduced. But E⦵ predictions assume standard conditions and say nothing about rate — a feasible reaction may still be too slow (high activation energy).

Applications

Storage cells/batteries (non-rechargeable, rechargeable like lithium-ion) and fuel cells (e.g. hydrogen–oxygen fuel cell producing electricity + water, more efficient with fewer pollutants).

Worked example

Given E⦵(Cu²⁺/Cu) = +0.34 V and E⦵(Zn²⁺/Zn) = −0.76 V, find E⦵cell and state the feasible reaction.

E⦵cell = (+0.34) − (−0.76) = +1.10 V (positive → feasible). Zn (more negative) is oxidised; Cu²⁺ is reduced. ✓

Common mistakes

Reversing OIL RIG (oxidation is loss of electrons).
Getting E⦵cell the wrong way round — it's the more positive minus the more negative.
Assuming a positive E⦵cell means the reaction is fast — it only means feasible.

Exam tips

Assign oxidation numbers to identify what's oxidised/reduced.
Combine half-equations so electrons cancel.
Calculate E⦵cell and use its sign for feasibility, noting the rate limitation.

Key facts to remember

OIL RIG; oxidising agent gains electrons, reducing agent loses them; track with oxidation numbers.
E⦵cell = E⦵(positive) − E⦵(negative); positive → feasible; more negative E⦵ = better reducing agent.
Predictions assume standard conditions and ignore rate; applications include batteries and fuel cells.