Electrochemical cells and cell diagramsEdexcel A-Level Chemistry: Revision notes

An electrochemical cell converts the energy of a redox reaction into electrical energy by keeping oxidation and reduction in separate half-cells. Each half-cell contains the oxidised and reduced forms of one species in an equilibrium, for example $\mathrm{Zn^{2+}(aq) + 2e^- \rightleftharpoons Zn(s)}$.

Two types of half-cell are common:

• Metal / metal ion: a strip of the metal in a solution of its own ions (e.g. Cu in Cu²⁺(aq)).
• Ion / ion: two ions of the same element in solution (e.g. Fe³⁺(aq) and Fe²⁺(aq)), with an inert platinum electrode to carry electrons.

The half-cells are joined by a salt bridge (filter paper soaked in potassium nitrate solution) which completes the circuit and keeps each solution electrically neutral. Electrons flow through the external wire from the negative electrode (where oxidation occurs) to the positive electrode (where reduction occurs). The voltmeter must have a high resistance so that almost no current flows.

The potential of a single half-cell cannot be measured alone, so it is measured against a reference. The standard hydrogen electrode (SHE) is defined as having $E^\ominus = 0.00$ V. It is made from hydrogen gas at 100 kPa bubbled over a platinum electrode in a solution containing H⁺(aq) at 1.00 mol dm⁻³, at 298 K.

The standard electrode potential, $E^\ominus$, of a half-cell is the emf of a cell in which that half-cell (under standard conditions) is connected to the SHE. Values are quoted for the reduction half-equation, with the more positive $E^\ominus$ meaning a greater tendency to be reduced.

The standard cell potential combines two electrode potentials:

$E^\ominus_{cell} = E^\ominus_{positive} - E^\ominus_{negative}$

The positive electrode is the half-cell with the more positive $E^\ominus$. $E^\ominus$ values are not multiplied by the number of electrons, because electrode potential does not depend on the amount of substance.

Worked example. Zn and Cu: $E^\ominus$(Cu²⁺/Cu) = +0.34 V, $E^\ominus$(Zn²⁺/Zn) = −0.76 V.

$E^\ominus_{cell} = +0.34 - (-0.76) = +1.10$ V. Copper is the positive electrode; zinc is oxidised and Cu²⁺ is reduced: $\mathrm{Zn + Cu^{2+} \rightarrow Zn^{2+} + Cu}$.

A cell diagram is a shorthand for the cell:

• A single vertical line | shows a phase boundary (e.g. solid electrode | solution).
• A double vertical line || shows the salt bridge.
• Species in the same solution are separated by a comma (e.g. Fe²⁺(aq), Fe³⁺(aq)).
• A platinum electrode (Pt(s)) is written at the outer end of an ion/ion half-cell.
• The negative electrode (more negative $E^\ominus$) is written on the left and the positive electrode on the right, so $E^\ominus_{cell} = E^\ominus_{right} - E^\ominus_{left}$.

Examples: $\mathrm{Zn(s) \mid Zn^{2+}(aq) \parallel Cu^{2+}(aq) \mid Cu(s)}$, $E^\ominus$ = +1.10 V; and $\mathrm{Pt(s) \mid Fe^{2+}(aq), Fe^{3+}(aq) \parallel Ag^{+}(aq) \mid Ag(s)}$, $E^\ominus$ = +0.03 V.

Each half-cell reaction is an equilibrium, e.g. $\mathrm{Cu^{2+}(aq) + 2e^- \rightleftharpoons Cu(s)}$. Changing the conditions shifts the position of equilibrium and so changes the electrode potential. For example, a lower [Cu²⁺] shifts the equilibrium to the left and makes the electrode less positive.

To give a standard value, the measurement must be made at:

• ion concentrations of 1.00 mol dm⁻³
• a temperature of 298 K
• a gas pressure of 100 kPa
• using a high-resistance voltmeter, so almost no current flows and the concentrations stay constant (and the reading is the true emf).

A cell used in different conditions gives a cell potential $E_{cell}$, not $E^\ominus_{cell}$.