Electrode potentials and cellsAQA A-Level Chemistry: Revision notes

A half-equation shows one electrode reaction. By the IUPAC convention half-equations are written as reductions, with the oxidised form and electrons on the left:

oxidised form + ne⁻ ⇌ reduced form, for example Zn²⁺(aq) + 2e⁻ ⇌ Zn(s)

When a metal is placed in a solution of its ions, an equilibrium is set up and a potential difference develops between the metal and the solution. This electrode potential, E, cannot be measured on its own, so it is measured by connecting the half-cell to a reference electrode.

The more negative the electrode potential, the greater the tendency for the reaction to go backwards (the metal to lose electrons, so it is a stronger reducing agent). The more positive the potential, the greater the tendency for the species on the left to gain electrons (a stronger oxidising agent).

The standard hydrogen electrode (SHE) is the reference. It consists of a platinum electrode (coated in platinum black) in 1.00 mol dm⁻³ H⁺(aq), with hydrogen gas at 100 kPa bubbled over it at 298 K. Its potential is defined as 0.00 V.

The standard electrode potential, E°, of a half-cell is the EMF of a cell in which it is connected to the SHE, under standard conditions: 298 K, 100 kPa (for gases) and 1.00 mol dm⁻³ solutions of the ions.

The conditions matter because changing the concentration, temperature or pressure changes the position of the electrode equilibrium, and so the electrode potential, E. The Nernst equation (which calculates these changes) is not required.

Half-cells with no solid metal, such as Fe³⁺/Fe²⁺, use an inert platinum electrode in the solution.

A cell is written as a cell diagram, for example:

Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s)

• a single line | shows a phase boundary (between solid and solution)
• a double line || shows the salt bridge
• the more negative electrode (where oxidation occurs) is on the left, and the more positive (where reduction occurs) on the right
• species in the same solution are separated by commas, and for half-cells without a metal an inert Pt electrode is written at the outside, for example Pt(s) | Fe²⁺(aq), Fe³⁺(aq)

The EMF of the cell is E°(right-hand electrode) − E°(left-hand electrode), that is, E°(positive) − E°(negative). For the cell above this is +0.34 − (−0.76) = +1.10 V.

To measure the EMF of a cell such as Zn | Zn²⁺ || Cu²⁺ | Cu:

1. Clean the metal strips with emery paper and place each in a solution of its own ions of concentration 1.00 mol dm⁻³.
2. Join the two beakers with a salt bridge, for example filter paper soaked in saturated potassium nitrate solution.
3. Connect the electrodes with wires through a high-resistance voltmeter.
4. Keep the temperature at 298 K, for example using a water bath, and record the steady reading.

The high resistance means almost no current flows, so the half-cells do not change and the maximum EMF is measured. Potassium nitrate is used because its ions do not react with the half-cell solutions. To measure E° for a single half-cell, connect it to the SHE. Fresh solutions and clean electrodes help give reliable readings.

Listing half-equations in order of E° gives the electrochemical series. A species on the left of a half-equation with a more positive E° is a stronger oxidising agent, and the reduced form of a half-equation with a more negative E° is a stronger reducing agent.

To predict a reaction:

1. Write the two half-equations and their E° values.
2. The half-equation with the more positive E° goes forward (reduction); the other is reversed (oxidation).
3. E°cell = E°(reduction) − E°(oxidation). If E°cell is positive, the reaction is feasible.

Example: Fe³⁺ (+0.77 V) and I₂ (+0.54 V): Fe³⁺ is reduced and I⁻ is oxidised, so E°cell = +0.77 − 0.54 = +0.23 V and 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂ is feasible.

Do not multiply E° values by the coefficients when balancing electrons; E° does not depend on the amount of substance.